Proof. Suppose $(p_i)$ converges to $p$. Let $U$ be a neighborhood of $p$, then for some $r\gt0$, $B_r(p)\subseteq U$. And there exists $m\in N$, such that for all $n\ge m$, $d(p_n,p)\lt r$, and thus $p_n\in B_r(p)\subseteq U$. Hence $(p_i)$ topologically converges to $p$.

Suppose $(p_i)$ topologically converges to $p$. Let $\varepsilon\gt0$, then there exists $m\in N$, such that for all $n\ge m$, $p_n\in B_\varepsilon(p)$, and thus $d(p_n,p)\lt\varepsilon$. Hence $(p_i)$ converges to $p$. $\blacksquare$

Proof. Suppose $(a_n)$ converges to $p$ and $q$, where $p\neq q$. Then $d(p,q)\gt0$. So there exists $m\in N$, such that for all $n\ge m$, $d(a_n,p)\lt d(p,q)/2$ and $d(a_n,q)\lt d(p,q)/2$. Therefore, $d(p,q)\le d(a_n,p)+d(a_n,q)\lt d(p,q)$, a contradiction. Therefore, if $(a_n)$ converges to $p$ and $q$, then $p=q$. $\blacksquare$

Proof. Let $\lim_{n\to\infty}a_n=z$ and $\lim_{n\to\infty}b_n=w$.

If $c=0$, then the first equation is trivial. Now suppose $c\neq0$, then $\abs{c}\gt0$. Let $\varepsilon\gt0$, then there exists $m\in N$ such that for all $n\ge m$, $d(a_n,z)\lt\varepsilon/\abs{c}$, hence $$d(ca_n,cz)=\norm{c(a_n-z)}=\abs{c}\norm{a_n-z}=\abs{c}d(a_n,z)\lt\varepsilon$$ implying $\lim_{n\to\infty}(ca_n)=cz$.

Let $\varepsilon\gt0$, then there exists $m\in N$ such that for all $n\ge m$, $d(a_n,z)\lt\varepsilon/2$ and $d(b_n,w)\lt\varepsilon/2$, hence $$d(a_n\pm b_n,z\pm w)=\norm{(a_n-z)\pm(b_n-w)}\le\norm{a_n-z}+\norm{b_n-w}=d(a_n,z)+d(b_n,w)\lt\varepsilon$$ implying $\lim_{n\to\infty}(a_n\pm b_n)=z\pm w$.

We will only prove the third one for complex sequences, and the result for real sequences follows trivially. Since $(a_n)$ is convergent, for some $r\gt0$, $\abs{a_n}\lt r$ for all $n$. Let $\varepsilon\gt0$, then there exists $m\in N$ such that for all $n\ge m$, $d(a_n,z)\lt\frac{\varepsilon}{2(\abs{w}+1)}$ and $d(b_n,w)\lt\frac{\varepsilon}{2r}$, hence $$d(a_nb_n,zw)=\abs{a_nb_n-zw}=\abs{(b_n-w)a_n+(a_n-z)w}\le\abs{b_n-w}\abs{a_n}+\abs{a_n-z}\abs{w} \lt\frac{\abs{a_n}\varepsilon}{2r}+\frac{\abs{w}\varepsilon}{2(\abs{w}+1)}\lt\varepsilon$$ implying $\lim_{n\to\infty}(a_nb_n)=zw$. $\blacksquare$

Proof. Let $L$ denote $\lim_{i\to\infty,j\to\infty}a_{i,j}$ and let $L_i$ denote $\lim_{j\to\infty}a_{i,j}$ for each $i$. Let $\varepsilon\gt0$. Then there exist $M,N$ such that for all $i\ge M$ and $j\ge N$, $d(a_{i,j},L)\lt\frac{\varepsilon}{2}$. Let $m\ge M$, then there exists $n\ge N$ such that for all $k\ge n$, $d(a_{m,k},L_m)\lt\frac{\varepsilon}{2}$. Thus $$d(L_m,L)\le d(L_m,a_{m,n})+d(a_{m,n},L)\lt\varepsilon$$ We have shown that $$\lim_{i\to\infty}\lim_{j\to\infty}a_{i,j}=\lim_{i\to\infty,j\to\infty}a_{i,j}$$ $\blacksquare$

Proof. Let $(a_n)$ be a convergent sequence in a metric space $X$, and let $\lim_{n\to\infty}a_n=p$. Let $\varepsilon\gt0$, then there exists $m\in N$ such that for all $n\ge m$, $d(a_n,p)\lt\varepsilon/2$. So for all $n,k\ge m$, $d(a_n,a_k)\le d(a_n,p)+d(a_k,p)\lt\varepsilon$. Hence, $(a_n)$ is a Cauchy sequence. $\blacksquare$

Proof. Since closed and bounded means compact in $R^k$, every member of $\mathcal C$ is compact. Let $\mathcal D$ be the collection of subsets $R^k\setminus E$ where $E\in\mathcal C$, then every member of $\mathcal D$ is open. Fix a member $S$ of $\mathcal C$. Suppose for contradiction that no point of $S$ belong to every member of $\mathcal C$. Then every point of $S$ belong to some member of $\mathcal D$, so $\mathcal D$ is an open cover of $S$. Since $S$ is compact, $\mathcal D$ has a finite subcover $\mathcal D'$. Let $\mathcal C'$ be the collection of subsets $R^k\setminus E$ where $E\in\mathcal D'$, then no point of $S$ belong to every member of $\mathcal C'$. Note that $S\cup\mathcal C'$ is a finite sub-collection of $\mathcal C$, but the intersection of $S\cup\mathcal C'$ is empty, a contradiction. Hence, $\cap\mathcal C$ is non-empty. $\blacksquare$

Proof. Since the intersection of every finite sub-collection of $(K_n)$ is a member of $(K_n)$ and hence non-empty, $\cap K_n$ is non-empty. Since $\cap K_n\subseteq K_1$, $\cap K_n$ is also bounded. If $\cap K_n$ contains more then one point, then $\text{diam } \cap K_n\gt0$. But for each $n$, $\cap K_n\subseteq K_n$, so that $\text{diam } K_n\ge\text{diam } \cap K_n\gt0$. This contradicts the assumption that $\lim_{n\to\infty}\text{diam } K_n=0$. Hence, $\cap K_n$ contains exactly one point. $\blacksquare$

Proof. Since $E\subseteq\overline E$, $\text{diam } E\le\text{diam }\overline E$. Let $p,q\in\overline E$, and let $\varepsilon\gt0$. Then there exist $p',q'\in E$ such that $d(p,p')\lt\varepsilon/2$ and $d(q,q')\lt\varepsilon/2$. We then have $d(p,q)\le d(p,p')+d(p',q')+d(q',q)\lt\varepsilon+d(p',q')\le\varepsilon+\text{diam } E$. Since $\varepsilon$ is arbitrary, $d(p,q)\le\text{diam } E$, so $\text{diam } E$ is an upper bound of the set of all $d(p,q)$ where $p,q\in\overline E$. Hence, $\overline E$ is bounded and $\text{diam }\overline E\le\text{diam } E$. And we conclude that $\text{diam }\overline E=\text{diam } E$. $\blacksquare$

Proof. Let $(\vb x_n)$ be a Cauchy sequence in $R^k$. Then there exists $m\in N$ such that for all $a,b\ge m$, $d(\vb x_a,\vb x_b)\lt1$. So $\vb x_n\in B_1(\vb x_m)$ for all $n\ge m$, implying the subsequence starting from index $m$ is bounded. Since $\{\vb x_1,\ldots,\vb x_{m-1}\}$ is also bounded, the sequence $(\vb x_n)$ is bounded. For $k\in N$, let $E_k$ be the set consisting of all $\vb x_n$ such that $n\ge k$. Since $(\vb x_n)$ is a Cauchy sequence, let $\varepsilon\gt0$, there exists $m\in N$, such that for all $a,b\ge m$, $d(\vb x_a,\vb x_b)\lt\varepsilon/2$. So for all $p,q\in E_m$, $d(p,q)\lt\varepsilon/2$. Hence $\text{diam }E_m\le\varepsilon/2\lt\varepsilon$. Then for all $k\ge m$, $\text{diam }\overline E_k=\text{diam }E_k\le\text{diam }E_m\lt\varepsilon$, so $d(\text{diam }\overline E_k,0)\lt\varepsilon$. Therefore, $\lim_{k\to\infty}\text{diam }\overline E_k=0$. Note that each $E_k$ is non-empty and bounded, so each $\overline E_k$ is non-empty, closed and bounded. Also, $E_{k+1}\subseteq E_k$, so $\overline E_{k+1}\subseteq\overline E_k$. Therefore, $\cap\overline E_k$ consists of exactly one point $\vb p$. Let $\varepsilon\gt0$. Then there exists $m\in N$ such that for all $k\ge m$, $\text{diam }\overline E_k\lt\varepsilon$, so $\text{diam }\overline E_m\lt\varepsilon$. Since $\vb p\in\overline E_m$, we have $d(\vb p,\vb q)\le\text{diam }\overline E_m\lt\varepsilon$ for all $\vb q$ in $\overline E_m$, and hence in $E_m$. Therefore, for all $n\ge m$, $d(\vb p,\vb x_n)\lt\varepsilon$. And we conclude that $\lim_{n\to\infty}\vb x_n=\vb p$. $\blacksquare$

Proof. Let $(a_n)$ be increasing. Suppose $(a_n)$ has a limit at $L$. If there is some $M$ such that $a_M\gt L$, then for all $m\gt M$, $a_m\ge a_M\gt L$, and if we let $\varepsilon=d(a_M,L)$, $a_m\notin B_\varepsilon(L)$ for all $m\gt M$, which is a contradiction to $L$ being the limit of $(a_n)$, so we know $L$ is an upper bound for $(a_n)$. Now let $\varepsilon\gt0$, then there has to be some $M$ such that $a_M\in(L-\varepsilon,L]$, because otherwise, $a_n\notin B_\varepsilon(L)$ for all $n$, then $L$ will not be the limit of $(a_n)$. Since $\varepsilon\gt0$ is arbitrarily chosen, for every $s\lt L$, we can always choose $\varepsilon=L-s$, then there exists $a_M\gt L-\varepsilon=s$, so $s$ is not an upper bound of $(a_n)$. Hence, we conclude that $L$ is the supremum of $(a_n)$.

Now suppose $(a_n)$ has a supremum $L$, then let $\varepsilon\gt0$, there has to be some $M$ such that $a_M\in(L-\varepsilon,L]$, because otherwise, $L-\varepsilon$ is an upper bound of $(a_n)$, meaning $L$ is not a supremum. Since $a_M\in(L-\varepsilon,L]$, for all $m\gt M$, we have $a_m\ge a_M\gt L-\varepsilon$, and $a_m\le L$ because $L$ is the supremum. Then by the definition of sequence limit, $L$ is the limit of $(a_n)$.

The case for decreasing sequences can be proven by a symmetric argument. $\blacksquare$

Proof. Let us call a positive-integer-valued index $n$ of a real-valued sequence $(x_n)$ a "peak" when $x_k\lt x_n$ for every $k\gt n$. Suppose first that there exists a peak in the sequence.

Suppose for every peak in the sequence, there exists another peak with greater index. By axiom of choice, there exists a function for every peak to find a peak with greater index. Since there exists a peak in the sequence, by recursion theorem, we can take that peak as the initial element, and recursively define a decreasing subsequence of the sequence $(x_n)$.

Suppose there exists a peak in the sequence, such that there is no peak with greater index. Let $N$ be the index of that peak, then for every $n\gt N$, there exists $k\gt n$ such that $x_k\ge x_n$. By axiom of choice, there exists a function for every $n\gt N$ to find a $k\gt n$ such that $x_k\ge x_n$. And by the recursion theorem, we can take $N+1$ as the initial element, and recursively define an increasing subsequence of the sequence $(x_n)$.

Now suppose there is no peak in the sequence, then we can use the above procedure to define an increasing subsequence of the sequence $(x_n)$, with $N$ being $0$.

Therefore, in every case, the sequence $(x_n)$ has either an increasing subsequence or a decreasing subsequence. $\blacksquare$

Proof. Let $(\vb x_n)$ be a bounded sequence in $R^m$, then its component sequences $({x_i}_n)$, such that $\vb x_n=({x_1}_n,\ldots,{x_m}_n)$, are bounded sequences of $R$. We will construct a convergent subsequence of $(\vb x_n)$ by the following process:

• Denote $(\vb x_n)$ as $S_0$.
• For each coordinate $i$ in order, extract a subsequence $S_i$ of $S_{i-1}$ such that the $i$th components form an increasing or decreasing component subsequence $({x_i}_{n_j})$. Since $({x_i}_{n_j})$ is bounded, it has both a supremum and an infimum, so it is convergent to either of them depending on whether its increasing or decreasing. Note that for every coordinate prior to $i$, the resulting component subsequence also converges.

The resulting subsequence $S_m$ has the property that every component subsequence converges. Therefore, $S_m$ itself converges. $\blacksquare$

Proof. For base case, $\sum_{i=0}^n a_i=a_0=\frac{(0+1)(a_0+a_0)}{2}$. For inductive step, suppose $\sum_{i=0}^n a_i=\frac{(n+1)(a_0+a_n)}{2}$, then $$\sum_{i=0}^{n+1} a_i=\frac{(n+1)(a_0+a_n)}{2}+a_{n+1}=\frac{(n+1)a_0+(n+1)a_n+(a_0+a_n+(n+2)c)}{2}=\frac{(n+2)a_0+(n+2)(a_n+c)}{2}=\frac{(n+2)(a_0+a_{n+1})}{2}$$ $\blacksquare$

Proof. For base case, $\sum_{i=0}^n a_i=a_0=\frac{a_0(1-c^1)}{1-c}$. For inductive step, suppose $\sum_{i=0}^n a_i=\frac{a_0(1-c^{n+1})}{1-c}$, then $$\sum_{i=0}^{n+1} a_i=\frac{a_0(1-c^{n+1})}{1-c}+a_{n+1}=\frac{a_0-a_{n+1}+a_{n+1}-a_{n+2}}{1-c}=\frac{a_0-a_{n+2}}{1-c}$$ $\blacksquare$

Proof. For every $\varepsilon\gt0$, let $r=\frac{1}{c}-1$, then $r\gt0$. Let $N$ be a natural number greater than $\frac{a_0}{\varepsilon r}$, then $N\ge1$, and for all $n\ge N$, $$d(a_n,0)=\abs{a_n}=a_0c^n=\frac{a_0}{(1+r)^n}\le\frac{a_0}{1+nr}\lt\frac{a_0}{nr}\le\frac{a_0}{Nr}\lt\varepsilon$$ Hence $(a_n)$ converges to $0$. $\blacksquare$

Proof. $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\p{\sum_{k=0}^na_k-\sum_{k=0}^{n-1}a_k} =\lim_{n\to\infty}\sum_{k=0}^na_k-\lim_{n\to\infty}\sum_{k=0}^{n-1}a_k=\sum a_n-\sum a_n=0$$ $\blacksquare$

Proof. We will only show for the complex case. The real case trivially follows. If $\sum\abs{b_n}$ exists, the sequence $(\sum_{i=0}^n\abs{b_i})$ converges and is thus a Cauchy sequence. So given $\varepsilon\gt0$, there exists $M\ge N$ such that $m\ge n\ge M$ implies $$\abs{\sum_{i=0}^m\abs{a_i}-\sum_{i=0}^n\abs{a_i}}=\sum_{i=n+1}^m\abs{a_i}\le\sum_{i=n+1}^m\abs{b_i}=\abs{\sum_{i=0}^m\abs{b_i}-\sum_{i=0}^n\abs{b_i}}\lt\varepsilon$$ Hence, the sequence $(\sum_{i=0}^n\abs{a_i})$ is a Cauchy sequence and thus converges. Therefore, $\sum\abs{a_i}$ exists. $\blacksquare$

Proof. Let $c\in S$, then for all $n$, $\sup_{m\ge n}a_m\ge c$, so $\limsup_{n\to\infty}a_n=\inf_{n\in N}\sup_{m\ge n}a_m\ge c$. This implies that $\limsup_{n\to\infty}a_n$ is an upper bound of $S$.

Now let $c=\limsup_{n\to\infty}a_n$. If $c\in R$, then for all $M$ and $n\neq0$, there exists $m\gt M$ such that $c\le\sup_{k\ge m}a_k\lt c+1/n$, so there exists $k\ge m\gt M$ such that $c-1/n\lt a_k\lt c+1/n$. By axiom of choice and recursion theorem, we can define a subsequence $(a_{n_k})$ of $(a_n)$ such that $c-1/k\lt a_{n_k}\lt c+1/k$, which converges to $c$, implying $c\in S$. If $c=\pm\infty$, the increasing/decreasing unbounded subsequence is constructed similarly.

Therefore, $\limsup_{n\to\infty}a_n=\sup S$. By a symmetric argument, $\liminf_{n\to\infty}a_n=\inf S$. $\blacksquare$

Proof. We will only show for the complex case. The real case trivially follows. Let $r=\limsup_{n\to\infty}\abs{a_n}^{1/n}\lt1$, then clearly, $r\in[0,1)$. Let $s=\frac{r+1}{2}$, then $s\in(r,1)$, and there exists $N_0\neq0$ such that $\sup_{n\ge N_0}\abs{a_n}^{1/n}\lt s$, implying for all $n\ge N_0$, $\abs{a_n}^{1/n}\lt s$, and hence $$\abs{a_n}=(\abs{a_n}^{1/n})^n\lt s^n=\abs{s^n}\lt 1$$ Note that $(s^n)$ is a geometric sequence, so $$\sum\abs{s^n}=\sum s^n=\lim_{n\to\infty}\frac{1-s^{n+1}}{1-s}=\frac{1}{1-s}$$ By comparison test, $\sum\abs{a_n}$ exists. $\blacksquare$

Proof. We will only show for the complex case. The real case trivially follows. Let $r=\limsup_{n\to\infty}\abs{\frac{a_{n+1}}{a_n}}\lt1$, then clearly $r\in[0,1)$. Let $s=\frac{r+1}{2}$, then $s\in(r,1)$, and there exists $N_0$ such that $$\sup_{n\ge N_0}\abs{\frac{a_{n+1}}{a_n}}\lt s$$ implying for all $n\ge N_0$, $$\abs{\frac{a_{n+1}}{a_n}}\lt s\lt1$$ Consider the subsequence $(a_{N_0+n})$. If we define $b_n=a_{N_0}s^n$, then by induction, $\abs{a_{N_0+n}}\le\abs{b_n}$ for all $n$. Since $(\abs{b_n})$ is a geometric sequence, $$\sum\abs{b_n}=\lim_{n\to\infty}\frac{\abs{a_{N_0}}(1-s^{n+1})}{1-s}=\frac{\abs{a_{N_0}}}{1-s}$$ By comparison test, $\sum\abs{a_{N_0+n}}$ exists, hence $\sum\abs{a_n}$ exists. $\blacksquare$

Proof. Suppose $\sum\norm{a_n}$ exists, then the sequence $(\sum_{i=0}^n\norm{a_i})$ converges and is thus a Cauchy sequence. So given $\varepsilon\gt0$, there exists $N$ such that $m\ge n\ge N$ implies $$\norm{\sum_{i=0}^ma_i-\sum_{i=0}^na_i}=\norm{\sum_{i=n+1}^ma_i}\le\sum_{i=n+1}^m\norm{a_i}=\abs{\sum_{i=0}^m\norm{a_i}-\sum_{i=0}^n\norm{a_i}}\lt\varepsilon$$ Hence, the sequence $(\sum_{i=0}^na_i)$ is a Cauchy sequence and thus converges. Therefore, $\sum a_n$ exists.

Now suppose $(b_m)$ is a reindex of $(a_n)$, meaning $b=a\circ\sigma$ for some bijection $\sigma:N\to N$. Let $\varepsilon\gt0$. Then there exists $N$ such that $\norm{\sum_{i=0}^Na_i-\sum a_n}\lt\frac{\varepsilon}{2}$ and $\abs{\sum_{i=0}^N\norm{a_i}-\sum\norm{a_n}}\lt\frac{\varepsilon}{2}$. And there exists $M$ such that $\{i:i\le N\}\subseteq\{\sigma(j):j\le M\}$. Let $m\ge M$, then $$ \norm{\sum_{j=0}^mb_j-\sum a_n} \le\norm{\sum_{j=0}^mb_j-\sum_{i=0}^Na_i}+\norm{\sum_{i=0}^Na_i-\sum a_n} \lt\norm{\sum_{i\in\{\sigma(j):j\le m\}}a_i-\sum_{i=0}^Na_i}+\frac{\varepsilon}{2} \le\p{\sum_{i\in\{\sigma(j):j\le m\}}\norm{a_i}-\sum_{i=0}^N\norm{a_i}}+\frac{\varepsilon}{2} \le\p{\sum\norm{a_n}-\sum_{i=0}^N\norm{a_i}}+\frac{\varepsilon}{2} \lt\varepsilon $$ Thus $\sum b_m$ exists and $\sum b_m=\sum a_n$. $\blacksquare$

Proof. Suppose $f(x)\to q$ and $f(x)\to r$ as $x\to p$, with $q\neq r$. Then $d_Y(q,r)\gt0$. So there exists $\delta\gt0$ such that for all $x\in U$ with $0\lt d_X(x,p)\lt\delta$, we have $d_Y(f(x),q)\lt d_Y(q,r)/2$ and $d_Y(f(x),r)\lt d_Y(q,r)/2$. Hence $d_Y(q,r)\le d_Y(f(x),q)+d_Y(f(x),r)\lt d_Y(q,r)$, a contradiction. Therefore, if $f(x)\to q$ and $f(x)\to r$ as $x\to p$, then $q=r$. $\blacksquare$

Proof. Suppose $q=\lim_{x\to p}f(x)$. Then $p$ is a limit point of $U_f$, implying $p$ is also a limit point of $U_g$. Let $\varepsilon\gt0$. Then there exists $\delta\gt0$ such that for all $x\in U_f$ with $0\lt d_X(x,p)\lt\delta$, we have $d_Y(f(x),q)\lt\varepsilon$. Since $N$ is a neighborhood of $p$, it contains some $B_r(p)$ where $r\gt0$. Let $\delta^*=\inf(\delta,r)$, then for all $x\in U_g$ with $0\lt d_X(x,p)\lt\delta^*$, $x\in N\setminus\{p\}$, thus $x\in U_f$ and $g(x)=f(x)$. Since we also have $0\lt d_X(x,p)\lt\delta$, we have $d_Y(g(x),q)=d_Y(f(x),q)\lt\varepsilon$. Therefore, $\lim_{x\to p}g(x)=q=\lim_{x\to p}f(x)$. $\blacksquare$

Proof. Suppose $f$ is continuous. Let $B$ be an open subset of $Y$ and let $A$ be the preimage of $f$ on $B$. Let $p\in A$, then $f(p)\in B$. Since $B$ is an open subset of $Y$, there exists $\varepsilon\gt0$ such that $B_\varepsilon(f(p))\subseteq B$. Since $f$ is continuous, there exists $\delta\gt0$ such that such that for all $x\in U$ with $x\in B_\delta(p)$, $f(x)\in B_\varepsilon(f(p))$, hence $f(x)\in B$, implying $x\in A$. Therefore, $B_\delta(p)\cap U\subseteq A$. Note that $B_\delta(p)\cap U$ is an open ball of $U$ as a metric subspace. Since $p$ is chosen arbitrarily in $A$, $A$ is open with respect to the standard topology of $U$ as a metric subspace, and is therefore open with respect to the topology of $U$ as a topological subspace.

Suppose for every open subset $B$ of $Y$, the preimage $A$ of $f$ on $B$ is an open subset of $U$ as a topological subspace of $X$, then $A$ is open with respect to the standard topology of $U$ as a metric subspace. Let $p\in U$ and $\varepsilon\gt0$. Then $B_\varepsilon(f(p))$ is an open subset of $Y$, so the preimage $A$ of $f$ on $B_\varepsilon(f(p))$ is open with respect to the standard topology of $U$ as a metric subspace. Since $f(p)\in B_\varepsilon(f(p))$, $p\in A$, so there exists $\delta\gt0$ such that $B_\delta(p)\cap U\subseteq A$, implying for all $x\in U$ with $x\in B_\delta(p)$, $x\in A$, thus $f(x)\in B_\varepsilon(f(p))$. Hence $f$ is continuous at $p$. Since $p$ is chosen arbitrarily in $U$, $f$ is continuous. $\blacksquare$

Proof. This directly follows from the definitions of limit and continuity. $\blacksquare$

Proof. Suppose $f$ is continuous at $p$. Let $\varepsilon\gt0$. Then there exists $\delta\gt0$ such that for all $x\in U_f$ with $d_X(x,p)\lt\delta$, we have $d_Y(f(x),q)\lt\varepsilon$. Since $N$ is a neighborhood of $p$, it contains some $B_r(p)$ where $r\gt0$. Let $\delta^*=\inf(\delta,r)$, then for all $x\in U_g$ with $d_X(x,p)\lt\delta^*$, $x\in N$, thus $x\in U_f$ and $g(x)=f(x)$. Since we also have $d_X(x,p)\lt\delta$, we have $d_Y(g(x),q)=d_Y(f(x),q)\lt\varepsilon$. Therefore, $g$ is continuous at $p$. $\blacksquare$

Proof. Let $p\in V$ and $\varepsilon\gt0$. Then there exists $\delta\gt0$ such that for all $x\in U$ with $d_X(x,p)\lt\delta$, $d_Y(f(x),f(p))\lt\varepsilon$. Note that for all $x\in V$ with $d_X(x,p)\lt\delta$, $x\in U$ and $d_X(x,p)\lt\delta$, and thus $d_Y(f(x),f(p))\lt\varepsilon$, hence $d_Y(f|_V(x),f|_V(p))\lt\varepsilon$. Therefore, $f|_V:V\to Y$ is continuous at $p$. Since $p\in V$ is arbitrary, $f|_V$ is continuous. $\blacksquare$

Proof. Let $\lim_{x\to a}f(x)=\vb b$ and $\lim_{x\to a}g(x)=\vb d$. If $c=0$, the first equation is trivially true. Let $c\neq 0$, for every $\varepsilon\gt 0$, there exists $\delta\gt 0$ such that $x\in U\cap(B_\delta(a)\setminus\{a\})$ implies $\norm{f(x)-\vb b}\lt\varepsilon/\abs{c}$, which implies $\norm{cf(x)-c\vb b}=\abs{c}\norm{f(x)-\vb b}\lt\varepsilon$, proving the first equation.

For every $\varepsilon\gt 0$, there exists $\delta\gt 0$ such that $x\in W\cap(B_\delta(a)\setminus\{a\})$ implies both $\norm{f(x)-\vb b}\lt\varepsilon/2$ and $\norm{g(x)-\vb d}\lt\varepsilon/2$, which implies $\norm{(f+g)(x)-(\vb b+\vb d)}\le\norm{f(x)-\vb b}+\norm{g(x)-\vb d}\lt\varepsilon$, proving the addition version of the second equation. The subtraction version is proven by having $f-g=f+((-1)\times g)$ and using the proven results.

To prove the third equation, we consider component functions $f_i,g_i$. Let $\varepsilon\gt 0$, there exists $\delta\gt 0$ such that $x\in W\cap(B_\delta(a)\setminus\{a\})$ implies both $$\abs{f_i(x)-b_i}\le\norm{f(x)-\vb b}\lt\frac{\inf(1,\varepsilon)}{2(\abs{d_i}+1)}\le\frac{1}{2}$$ and $$\abs{g_i(x)-d_i}\le\norm{g(x)-\vb d}\lt\frac{\inf(1,\varepsilon)}{2(\abs{b_i}+1)}\le\frac{1}{2}$$ which implies $$\abs{g_i(x)}\le\abs{g_i(x)-d_i}+\abs{d_i}\lt\frac{1}{2}+\abs{d_i}$$ Now we have $$\abs{f_i(x)g_i(x)-b_id_i} =\abs{f_i(x)g_i(x)-b_ig_i(x)+b_ig_i(x)-b_id_i} \le\abs{g_i(x)}\abs{f_i(x)-b_i}+\abs{b_i}\abs{g_i(x)-d_i} \lt\cfrac{(\frac{1}{2}+\abs{d_i})\inf(1,\varepsilon)}{2(\abs{d_i}+1)}+\cfrac{\abs{b_i}\inf(1,\varepsilon)}{2(\abs{b_i}+1)} \lt\inf(1,\varepsilon) \le\varepsilon$$ So $$\lim_{x\to a}(f_ig_i)(x)=b_id_i$$ And we conclude $$\lim_{x\to a}(f\cdot g)(x)=\lim_{x\to a}(\sum_if_ig_i)(x) =\sum_i\lim_{x\to a}(f_ig_i)(x)=\sum_ib_id_i=\vb b\cdot\vb d=\lim_{x\to a}f(x)\cdot \lim_{x\to a}g(x)$$ $\blacksquare$

Proof. If $a$ is an isolated point in $U$, then $cf$ is continuous at $a$. If $a$ is a limit point of $U$, then $\lim_{x\to a}(cf)(x)=c\lim_{x\to a}f(x)=cf(a)=(cf)(a)$, so $cf$ is continuous at $a$.

If $a$ is an isolated point in $W$, then $f\pm g$ and $f\cdot g$ are continuous at $a$. If $a$ is a limit point of $W$, then $a$ is a limit point of both $U$ and $V$. So $\lim_{x\to a}(f\pm g)(x) =\lim_{x\to a}f(x)\pm \lim_{x\to a}g(x) =f(a)\pm g(a) =(f\pm g)(a)$, and $\lim_{x\to a}(f\cdot g)(x) =\lim_{x\to a}f(x)\cdot \lim_{x\to a}g(x) =f(a)\cdot g(a) =(f\cdot g)(a)$, implying both $f\pm g$ and $f\cdot g$ are continuous at $a$. $\blacksquare$

Proof. Suppose $\lim_{x\to a}f_i(x)=c_i$ for every integer $i$ from $1$ to $n$, then for every $\varepsilon\gt 0$, there exists $\delta\gt 0$, such that $x\in U\cap(B_\delta(a)\setminus\{a\})$ implies $\abs{f_i(x)-c_i}\lt\varepsilon/\sqrt n$ for all $i$, which implies $\norm{f(x)-(c_1,\ldots,c_n)}=\sqrt{\sum_i\abs{f_i(x)-c_i}^2}\lt\sqrt{\sum_i(\varepsilon/\sqrt n)^2}=\varepsilon$. Hence we have $\lim_{x\to a}f(x)=(c_1,\ldots,c_n)$.

Now suppose we have $\lim_{x\to a}f(x)=\vb d=(d_1,\ldots,d_n)$ instead. Then for every $\varepsilon\gt 0$, there exists $\delta\gt 0$, such that $x\in U\cap(B_\delta(a)\setminus\{a\})$ implies $\abs{f_i(x)-d_i}\le\norm{f(x)-\vb d}\lt\varepsilon$ for all $i$. Hence we have $\lim_{x\to a}f_i(x)=d_i$. $\blacksquare$

Proof. If $a$ is an isolated point in $U$, then $f_1,\ldots,f_n$ and $f$ are continuous at $a$.

Suppose $a$ is a limit point of $U$. If $f$ is continuous at $a$, then by limit of component functions, $$\lim_{x\to a}f_i(x)=(\lim_{x\to a}f(x))_i=f(x)_i=f_i(x)$$ for all $i$, meaning all component functions of $f$ are continuous at $a$. If we know all component functions of $f$ are continuous at $a$ instead, then by limit of component functions, $$\lim_{x\to a}f(x)=(\lim_{x\to a}f_1(x),\ldots,\lim_{x\to a}f_n(x))=(f_1(a),\ldots,f_n(a))=f(a)$$ implying $f$ itself is continuous at $a$. $\blacksquare$

Proof. For every $\varepsilon\lt 0$ there exists $\delta\gt 0$ such that $y\in U\cap B_\delta(b)$ implies $f(y)\in B_\varepsilon(f(b))$. But for that $\delta$ there also exists $r\gt\eta\gt 0$ such that $x\in V\cap(B_\eta(a)\setminus\{a\})$ implies $g(x)\in B_\delta(b)$ and $g(x)\in U$, which implies $f(g(x))\in B_\varepsilon(f(b))$, meaning $\lim_{x\to a}(f\circ g)(x)=f(b)=f(\lim_{x\to a}g(x))$. $\blacksquare$

Proof. By limit of composite functions, we have $$\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))=f(g(a))$$ implying $f\circ g$ is continuous at $a$. $\blacksquare$

Proof. Suppose $a\gt0$, fix $\varepsilon\gt0$, let $\delta=\inf(\frac{a^2\varepsilon}{2},\frac{a}{2})$, then for all $\abs{x-a}\lt\delta$, we have $x\gt a-\delta\ge\frac{a}{2}\gt0$, and $\abs{\frac{1}{x}-\frac{1}{a}}=\frac{\abs{x-a}}{xa}\le\frac{2}{a^2}\abs{x-a}\lt\frac{2\delta}{a^2}\le\varepsilon$. Hence $\frac{1}{x}$ is continuous on positive real numbers.

Suppose $a\lt0$. note that, let $f(x)=\frac{1}{x}$ and $g(x)=-x$, we have $\frac{1}{x}=(g\circ f\circ g)(x)$. Clearly, $g$ is continuous on $R$. Since $a\lt0$, we have $g(a)\gt0$, so $f$ is continuous at $g(a)$. By continuity of composite functions, $(g\circ f\circ g)$ is continuous at $a$, and so is $\frac{1}{x}$. We have shown that $\frac{1}{x}$ is continuous on non-zero real numbers. $\blacksquare$

Proof. Since $\lim_{x\to a}f(x)=b\neq0$, we have $\lim_{x\to b}\frac{1}{x}=\frac{1}{b}$. By limit of composite functions, $\lim_{x\to a}(\frac{1}{f})(x)=\frac{1}{b}$. $\blacksquare$

Proof. Suppose $B_r(a)\subseteq N$. For every $\varepsilon\gt 0$, there exists $r\gt\delta\gt 0$ such that $x\in A\cap(B_{\delta}(a)\setminus\{a\})$ implies $g(x)\in(c-\varepsilon,c+\varepsilon)$ and $x\in C\cap(B_{\delta}(a)\setminus\{a\})$ implies $h(x)\in(c-\varepsilon,c+\varepsilon)$. Then we have $x\in B\cap(B_{\delta}(a)\setminus\{a\})$ implies both $x\in A\cap(B_{\delta}(a)\setminus\{a\})$ and $x\in C\cap(B_{\delta}(a)\setminus\{a\})$, which implies $c-\varepsilon\lt g(x)\le f(x)\le h(x)\lt c+\varepsilon$. This proves that $$\lim_{x\to a}f(x)=c$$ $\blacksquare$

Proof. Let $f:U\to Y$ with $U\subseteq X$, where $X$ and $Y$ are metric spaces. Arbitrarily choose $\varepsilon\gt0$. Since $f$ is continuous, for all $p\in U$, there exists $\phi(p)\gt0$ such that for all $q\in U\cap B_{\phi(p)}(p)$, we have $f(q)\in B_{\varepsilon/2}(f(p))$. The collection of open balls $B_{\phi(p)/2}(p)$ for all $p\in U$ is clearly an open cover of $U$. Since $U$ is compact, there is a finite subcover $B_{\phi(p_i)/2}(p_i)$ where $i=1,\ldots,n$. Let $\delta=\frac{1}{2}\inf(\phi(p_1),\ldots,\phi(p_n))$, then $\delta\gt0$. Now let $p,q\in U$ such that $d(p,q)\lt\delta$. Since there exists $i\in\{1,\ldots,n\}$ such that $p\in B_{\phi(p_i)/2}(p_i)$, we have $d(p,p_i)\lt\phi(p_i)/2$, and thus $d(f(p),f(p_i))\lt\varepsilon/2$. Hence $d(q,p_i)\le d(q,p)+d(p,p_i)\lt\delta+\phi(p_i)/2\le\phi(p_i)$. So we have $d(f(q),f(p_i))\lt\varepsilon/2$. Therefore, $d(f(p),f(q))\le d(f(p),f(p_i))+d(f(p_i),f(q))\lt\varepsilon$. We have shown that $f$ is uniformly continuous. $\blacksquare$

Proof. Uniqueness of derivative and partial derivative follows directly from uniqueness of limit.

For total derivative, Suppose $A$ and $B$ are both total derivatives of $f$ at $\vb p$, an interior point of $U$, then there exists $r\gt0$ such that $B_r(\vb p)\subseteq U$. Let $B_r(\vb 0)\setminus\{\vb 0\}$ be the domain for $\vb h$, then $\vb p+\vb h\in B_r(\vb p)\subseteq U$ and the limits for $A$ and $B$ are both still $0$ for this domain for $\vb h$, due to localness of limit. Since $$\frac{\norm{f(\vb p+\vb h)-f(\vb p)-A(\vb h)}}{\norm{\vb h}}+\frac{\norm{f(\vb p+\vb h)-f(\vb p)-B(\vb h)}}{\norm{\vb h}} \ge\frac{\norm{B(\vb h)-A(\vb h)}}{\norm{\vb h}} \ge0$$ and $$\lim_{\vb h\to\vb 0}\p{\frac{\norm{f(\vb p+\vb h)-f(\vb p)-A(\vb h)}}{\norm{\vb h}}+\frac{\norm{f(\vb p+\vb h)-f(\vb p)-B(\vb h)}}{\norm{\vb h}}}=0$$ by squeeze theorem, $$\lim_{\vb h\to\vb 0}\frac{\norm{(B-A)(\vb h)}}{\norm{\vb h}}=\lim_{\vb h\to\vb 0}\frac{\norm{B(\vb h)-A(\vb h)}}{\norm{\vb h}}=0$$ where $B-A$ is a linear map.

Suppose there exists $\vb x\in R^n\setminus\{\vb 0\}$ such that $(B-A)(\vb x)\neq\vb 0$. Let $\varepsilon=\frac{\norm{(B-A)(\vb x)}}{\Vert\vb x\Vert}$, then $\varepsilon\gt0$, and for all $\delta\gt0$, we have $\frac{\delta\vb x}{2\Vert\vb x\Vert}\in B_\delta(\vb 0)\setminus\{\vb 0\}$, such that $$\abs{\frac{\norm{(B-A)(\frac{\delta\vb x}{2\Vert\vb x\Vert})}}{\norm{\frac{\delta\vb x}{2\Vert\vb x\Vert}}}-0} =\frac{\frac{\delta}{2\Vert\vb x\Vert}\norm{(B-A)(\vb x)}}{\frac{\delta}{2}} =\frac{\norm{(B-A)(\vb x)}}{\Vert\vb x\Vert} \ge\varepsilon$$ We have shown that $0$ is not the limit of $\frac{\norm{(B-A)(\vb h)}}{\norm{\vb h}}$ at $\vb h\to\vb 0$, which is a contradiction. Therefore, for all $\vb x\in R^n$, $(B-A)(\vb x)=\vb 0$, implying $A=B$. $\blacksquare$

Proof. We will regard $1$-vectors as scalars.

Suppose $f$ is differentiable at $p$, then $g(h)=f'(p)h$ is a linear map. We have $\lim_{h\to 0}\frac{f(p+h)-f(p)-g(h)}{h}=\lim_{h\to 0}\frac{f(p+h)-f(p)}{h}-\lim_{h\to 0}\frac{g(h)}{h}=f'(p)-f'(p)=0$, Hence $\lim_{h\to 0}\frac{\abs{f(p+h)-f(p)-g(h)}}{\abs{h}}=\lim_{h\to 0}\abs{\frac{f(p+h)-f(p)-g(h)}{h}}=0$, implying $f$ is totally differentiable at $p$. Also, $Df(p)(h)=g(h)=f'(p)h$.

Suppose $f$ is totally differentiable at $p$, then for some linear map $g$, $\lim_{h\to 0}\abs{\frac{f(p+h)-f(p)-g(h)}{h}}=\lim_{h\to 0}\frac{\abs{f(p+h)-f(p)-g(h)}}{\abs{h}}=0$. Hence $\lim_{h\to 0}\frac{f(p+h)-f(p)-g(h)}{h}=0$. Since $g(h)=hg(1)$, $\lim_{h\to 0}\frac{g(h)}{h}=g(1)$, so $\lim_{h\to 0}\frac{f(p+h)-f(p)}{h}=\lim_{h\to 0}\frac{f(p+h)-f(p)-g(h)}{h}+\lim_{h\to 0}\frac{g(h)}{h}=g(1)$, implying $f$ is differentiable. Also, $f'(p)=g(1)=Df(p)(1)$. $\blacksquare$

Proof. Let $f$ be linear, then for all $\vb p\in R^n$, $$\lim_{\vb h\to \vb 0}\frac{\Vert f(\vb p+\vb h)-f(\vb p)-f(\vb h)\Vert}{\Vert\vb h\Vert} =\lim_{\vb h\to \vb 0}\frac{\Vert \vb 0\Vert}{\Vert\vb h\Vert}=0$$ so $f$ is differentiable. $\blacksquare$

Proof. Suppose $f$ is linear and let $A$ be its matrix representation, which is $m\times n$. Let $M$ be the maximum among absolute values of entries of $A$. If $M=0$, then $A$ is a zero matrix, so $f$ is zero everywhere, implying it is continuous at $\vb 0$. Now suppose $M\neq0$. For every $\varepsilon\gt0$, let $\delta=\varepsilon/(nM\sqrt m)$, then for every $\vb x\in R^n$ such that $\Vert\vb x\Vert\lt\delta$, by triangle inequality, we have $$\norm{A\vb x} =\sqrt{\sum_{j=1}^m\p{\sum_{i=1}^n(a_{ji}x_i)}^2} =\sqrt{\sum_{j=1}^m\abs{\sum_{i=1}^n(a_{ji}x_i)}^2} \le\sqrt{\sum_{j=1}^m\p{\sum_{i=1}^n\abs{a_{ji}x_i}}^2} \le\sqrt{\sum_{j=1}^m\p{\sum_{i=1}^nM\Vert\vb x\Vert}^2}$$ $$=\sqrt{\sum_{j=1}^m\p{nM\Vert\vb x\Vert}^2} =\sqrt{m\p{nM\Vert\vb x\Vert}^2} =\sqrt{m}nM\Vert\vb x\Vert \lt \sqrt{m}nM\delta =\varepsilon$$ so $\lim_{\vb x\to \vb 0}f(\vb x)=\vb 0=f(\vb 0)$, implying $f$ is continuous at $\vb 0$. $\blacksquare$

Proof. Let $f$ be differentiable at $\vb p$. Let $B_r(\vb p)\subseteq U$. Since $\lim_{\vb h\to \vb 0}\frac{\Vert f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)\Vert}{\Vert\vb h\Vert}=0$ and $\lim_{\vb h\to \vb 0}\Vert\vb h\Vert=0$, we have $\lim_{\vb h\to \vb 0}\Vert f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)\Vert=0$.

Because $Df(\vb p)$ is a linear map, it is continuous at $\vb 0$, so $\lim_{\vb h\to \vb 0}Df(\vb p)(\vb h)=Df(\vb p)(\vb 0)=\vb 0$, and thus $\lim_{\vb h\to \vb 0}\Vert Df(\vb p)(\vb h)\Vert=0$. For $\vb h\in B_r(\vb 0)\setminus\{\vb 0\}$, let $g(\vb h)=\Vert f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)\Vert+\Vert Df(\vb p)(\vb h)\Vert$, then $\lim_{\vb h\to \vb 0}g(\vb h)=0$ and $\Vert f(\vb p+\vb h)-f(\vb p)\Vert \le g(\vb h)$ by triangle inequality.

For every $\varepsilon\gt 0$, there exists $r\gt\delta\gt 0$ such that $\vb h\in B_\delta(\vb 0)\setminus\{\vb 0\}$ implies $g(\vb h)\in B_\varepsilon(0)$, then with the same $\delta$, we also have $\vb x\in B_\delta(\vb p)\setminus\{\vb p\}$ implies $\vb x-\vb p\in B_\delta(\vb 0)\setminus\{\vb 0\}$ and $\Vert f(\vb x)-f(\vb p)\Vert =\Vert f(\vb p+(\vb x-\vb p))-f(\vb p)\Vert \le g(\vb x-\vb p) \in B_\varepsilon(0)$, or $f(\vb x)\in B_\varepsilon(f(\vb p))$.

So we have $\lim_{\vb x\to \vb p}f(\vb x)=f(\vb p)$, which means $f$ is continuous at $\vb p$. $\blacksquare$

Proof. $$(\Vert\vb v\Vert)^2=\sum_i\abs{v_i}^2\le\sum_i\sum_j\abs{v_i}\abs{v_j}=(\sum_i\abs{v_i})^2$$ therefore $\Vert\vb v\Vert\le\sum_i\abs{v_i}$. $\blacksquare$

Proof. Suppose $Df(\vb p)$ exists. Then we have $$\lim_{\vb h\to \vb 0}\frac{\Vert f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)\Vert}{\Vert\vb h\Vert}=0$$ For all $j\in\{1,\ldots,m\}$, by squeeze theorem, we have $$\lim_{\vb h\to \vb 0}\frac{|f_j(\vb p+\vb h)-f_j(\vb p)-\p{Df(\vb p)(\vb h)}_j|}{\Vert\vb h\Vert} =\lim_{\vb h\to \vb 0}\frac{|\p{f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)}_j|}{\Vert\vb h\Vert} =0$$ Note that $\p{Df(\vb p)(\vb h)}_j$ is a linear map from $R^n$ to $R$. This shows that $Df_j(\vb p)$ exists, and $Df_j(\vb p)(\vb h)=\p{Df(\vb p)(\vb h)}_j$.

Now suppose all $Df_j(\vb p)$ exists. Define an $m\times n$ matrix $L$ by $L_{j,*}=Df_j(\vb p)$, where $L$ can be regarded as a matrix or a linear map, interchangeably. Then we have $$\lim_{\vb h\to \vb 0} \sum_j\frac{\abs{(f(\vb p+\vb h)-f(\vb p)-L(\vb h))_j}}{\norm{\vb h}} =\sum_j\lim_{\vb h\to \vb 0} \frac{\abs{(f(\vb p+\vb h)-f(\vb p)-L(\vb h))_j}}{\norm{\vb h}} =\sum_j\lim_{\vb h\to \vb 0} \frac{\abs{f_j(\vb p+\vb h)-f_j(\vb p)-Df_j(\vb p)(\vb h)}}{\norm{\vb h}} =0$$ Let $B_r(\vb p)\subseteq U$. Then for $\vb h\in B_r(\vb 0)\setminus\{\vb 0\}$, we have $$0 \le\frac{\norm{f(\vb p+\vb h)-f(\vb p)-L(\vb h)}}{\norm{\vb h}} \le\frac{\sum_j\abs{(f(\vb p+\vb h)-f(\vb p)-L(\vb h))_j}}{\norm{\vb h}}$$ Then by squeeze theorem, $$\lim_{\vb h\to \vb 0} \frac{\norm{f(\vb p+\vb h)-f(\vb p)-L(\vb h)}}{\norm{\vb h}}=0$$ Hence $Df(\vb p)=L$. $\blacksquare$

Proof. Suppose $Df(\vb p)$ exists. Then all $Df_j(\vb p)$ exist, so we have $$\lim_{\vb h\to \vb 0}\frac{|f_j(\vb p+\vb h)-f_j(\vb p)-Df_j(\vb p)(\vb h)|}{\Vert\vb h\Vert}=0$$ Then for all $i\in\{1,\ldots,n\}$, we have $$\lim_{h\to 0}\abs{\frac{f_j(\vb p+h\vb e_i)-f_j(\vb p)-hDf_j(\vb p)(\vb e_i)}{h}} =\lim_{h\to 0}\frac{|f_j(\vb p+h\vb e_i)-f_j(\vb p)-Df_j(\vb p)(h\vb e_i)|}{\Vert h\vb e_i\Vert} =0$$ which implies $$\lim_{h\to 0}\frac{f_j(\vb p+h\vb e_i)-f_j(\vb p)-hDf_j(\vb p)(\vb e_i)}{h}=0$$ Since $$\lim_{h\to 0}\frac{hDf_j(\vb p)(\vb e_i)}{h}=Df_j(\vb p)(\vb e_i)$$ we conclude that $$\pdv{f_j}{x_i}(\vb p) =\lim_{h\to 0}\frac{f_j(\vb p+h\vb e_i)-f_j(\vb p)}{h} =\lim_{h\to 0}\frac{f_j(\vb p+h\vb e_i)-f_j(\vb p)-hDf_j(\vb p)(\vb e_i)}{h}+\lim_{h\to 0}\frac{hDf_j(\vb p)(\vb e_i)}{h} =Df_j(\vb p)(\vb e_i) =(Df(\vb p)(\vb e_i))_j =Df(\vb p)_{ji}$$ $\blacksquare$

Proof. For the first equation, $$\lim_{\vb h\to \vb 0}\frac{\norm{(cf)(\vb p+\vb h)-(cf)(\vb p)-cDf(\vb p)(\vb h)}}{\norm{\vb h}} =\lim_{\vb h\to \vb 0}\frac{\abs{c}\norm{f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)}}{\norm{\vb h}} =\abs{c}\lim_{\vb h\to \vb 0}\frac{\norm{f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)}}{\norm{\vb h}} =0$$ Hence $D(cf)(\vb p)=cDf(\vb p)$.

Since $\vb p$ is an interior point of both $U$ and $V$, it is an interior point of $U\cap V$. For the second equation, let $B_r(\vb p)\subseteq U\cap V$, and $\vb h\in B_r(\vb 0)\setminus\{\vb 0\}$, then $$\frac{\norm{f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)}}{\norm{\vb h}}+\frac{\norm{g(\vb p+\vb h)-g(\vb p)-Dg(\vb p)(\vb h)}}{\norm{\vb h}} \ge\frac{\norm{(f\pm g)(\vb p+\vb h)-(f\pm g)(\vb p)-(Df(\vb p)\pm Dg(\vb p))(\vb h)}}{\norm{\vb h}} \ge0$$ Since $$\lim_{\vb h\to \vb 0}\p{\frac{\norm{f(\vb p+\vb h)-f(\vb p)-Df(\vb p)(\vb h)}}{\norm{\vb h}}+\frac{\norm{g(\vb p+\vb h)-g(\vb p)-Dg(\vb p)(\vb h)}}{\norm{\vb h}}}=0$$ by squeeze theorem, $$\lim_{\vb h\to \vb 0}\frac{\norm{(f\pm g)(\vb p+\vb h)-(f\pm g)(\vb p)-(Df(\vb p)\pm Dg(\vb p))(\vb h)}}{\norm{\vb h}}=0$$ Hence $D(f\pm g)(\vb p)=Df(\vb p)\pm Dg(\vb p)$.

For the third equation, let $j\in\{1,\ldots,m\}$, then both $f_j$ and $g_j$ are differentiable at $\vb p$. For $\vb h\in B_r(\vb 0)\setminus\{\vb 0\}$, let $$\varepsilon_j(\vb h)=\cfrac{f_j(\vb p+\vb h)-f_j(\vb p)-Df_j(\vb p)(\vb h)}{\norm{\vb h}} \quad\text{and}\quad\eta_j(\vb h)=\cfrac{g_j(\vb p+\vb h)-g_j(\vb p)-Dg_j(\vb p)(\vb h)}{\norm{\vb h}}$$ then $$f_j(\vb p+\vb h)-f_j(\vb p)=Df_j(\vb p)(\vb h)+\norm{\vb h}\varepsilon_j(\vb h) \quad\text{and}\quad g_j(\vb p+\vb h)-g_j(\vb p)=Dg_j(\vb p)(\vb h)+\norm{\vb h}\eta_j(\vb h)$$ Hence $$\frac{f_j(\vb p+\vb h)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p)}{\norm{\vb h}} =\frac{f_j(\vb p+\vb h)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p+\vb h)+f_j(\vb p)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p)}{\norm{\vb h}}$$ $$=\frac{\p{f_j(\vb p+\vb h)-f_j(\vb p)}g_j(\vb p+\vb h)+\p{g_j(\vb p+\vb h)-g_j(\vb p)}f_j(\vb p)}{\norm{\vb h}} =\frac{\p{Df_j(\vb p)(\vb h)+\norm{\vb h}\varepsilon_j(\vb h)}g_j(\vb p+\vb h)+\p{Dg_j(\vb p)(\vb h)+\norm{\vb h}\eta_j(\vb h)}f_j(\vb p)}{\norm{\vb h}}$$ $$=\frac{Df_j(\vb p)(\vb h)\p{g_j(\vb p)+Dg_j(\vb p)(\vb h)+\norm{\vb h}\eta_j(\vb h)}+Dg_j(\vb p)(\vb h)f_j(\vb p)}{\norm{\vb h}}+\varepsilon_j(\vb h)g_j(\vb p+\vb h)+\eta_j(\vb h)f_j(\vb p)$$ $$=\frac{Df_j(\vb p)(\vb h)g_j(\vb p)+Dg_j(\vb p)(\vb h)f_j(\vb p)}{\norm{\vb h}}+\frac{Df_j(\vb p)(\vb h)Dg_j(\vb p)(\vb h)}{\norm{\vb h}}+Df_j(\vb p)(\vb h)\eta_j(\vb h)+\varepsilon_j(\vb h)g_j(\vb p+\vb h)+\eta_j(\vb h)f_j(\vb p)$$ By Cauchy-Schwarz inequality, $$\abs{\frac{Df_j(\vb p)(\vb h)Dg_j(\vb p)(\vb h)}{\norm{\vb h}}} =\frac{\abs{(Df_j(\vb p))^T\cdot\vb h}\abs{Dg_j(\vb p)(\vb h)}}{\norm{\vb h}} \le\frac{\norm{(Df_j(\vb p))^T}\norm{\vb h}\abs{Dg_j(\vb p)(\vb h)}}{\norm{\vb h}} =\norm{(Df_j(\vb p))^T}\abs{Dg_j(\vb p)(\vb h)}$$ Since $Df_j(\vb p),Dg_j(\vb p)$ are both linear, they are continuous at $\vb0$, so $$\lim_{\vb h\to \vb 0}Df_j(\vb p)(\vb h)=0 \quad\text{and}\quad \lim_{\vb h\to \vb 0}Dg_j(\vb p)(\vb h)=0$$ implying $$\lim_{\vb h\to \vb 0}\abs{Dg_j(\vb p)(\vb h)}=0$$ So $$\lim_{\vb h\to \vb 0}\norm{(Df_j(\vb p))^T}\abs{Dg_j(\vb p)(\vb h)} =\norm{(Df_j(\vb p))^T}\lim_{\vb h\to \vb 0}\abs{Dg_j(\vb p)(\vb h)} =0$$ By squeeze theorem, $$\lim_{\vb h\to \vb 0}\abs{\frac{Df_j(\vb p)(\vb h)Dg_j(\vb p)(\vb h)}{\norm{\vb h}}}=0$$ implying $$\lim_{\vb h\to \vb 0}\frac{Df_j(\vb p)(\vb h)Dg_j(\vb p)(\vb h)}{\norm{\vb h}}=0$$ Note that $$\lim_{\vb h\to \vb 0}\abs{\varepsilon_j(\vb h)}=0 \quad\text{and}\quad \lim_{\vb h\to \vb 0}\abs{\eta_j(\vb h)}=0$$ implying $$\lim_{\vb h\to \vb 0}\varepsilon_j(\vb h)=0 \quad\text{and}\quad \lim_{\vb h\to \vb 0}\eta_j(\vb h)=0$$ Since $g_j$ is differentiable at $\vb p$, it is continuous at $\vb p$, thus $$\lim_{\vb h\to \vb 0}g_j(\vb p+\vb h)=g_j(\vb p)$$ Now we have $$\lim_{\vb h\to \vb 0}\p{\frac{Df_j(\vb p)(\vb h)Dg_j(\vb p)(\vb h)}{\norm{\vb h}}+Df_j(\vb p)(\vb h)\eta_j(\vb h)+\varepsilon_j(\vb h)g_j(\vb p+\vb h)+\eta_j(\vb h)f_j(\vb p)}$$ $$=\lim_{\vb h\to \vb 0}\frac{Df_j(\vb p)(\vb h)Dg_j(\vb p)(\vb h)}{\norm{\vb h}}+\lim_{\vb h\to \vb 0}Df_j(\vb p)(\vb h)\eta_j(\vb h)+\lim_{\vb h\to \vb 0}\varepsilon_j(\vb h)g_j(\vb p+\vb h)+\lim_{\vb h\to \vb 0}\eta_j(\vb h)f_j(\vb p)$$ $$=\lim_{\vb h\to \vb 0}Df_j(\vb p)(\vb h)\lim_{\vb h\to \vb 0}\eta_j(\vb h)+\lim_{\vb h\to \vb 0}\varepsilon_j(\vb h)\lim_{\vb h\to \vb 0}g_j(\vb p+\vb h)+f_j(\vb p)\lim_{\vb h\to \vb 0}\eta_j(\vb h) =0$$ Then we have $$\lim_{\vb h\to \vb 0}\frac{\abs{f_j(\vb p+\vb h)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p)-(Df_j(\vb p)(\vb h)g_j(\vb p)+Dg_j(\vb p)(\vb h)f_j(\vb p))}}{\norm{\vb h}}$$ $$=\lim_{\vb h\to \vb 0}\abs{\frac{Df_j(\vb p)(\vb h)Dg_j(\vb p)(\vb h)}{\norm{\vb h}}+Df_j(\vb p)(\vb h)\eta_j(\vb h)+\varepsilon_j(\vb h)g_j(\vb p+\vb h)+\eta_j(\vb h)f_j(\vb p)} =0$$ Since $$\frac{\sum_{j=1}^n\abs{f_j(\vb p+\vb h)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p)-(Df_j(\vb p)(\vb h)g_j(\vb p)+Dg_j(\vb p)(\vb h)f_j(\vb p))}}{\norm{\vb h}}$$ $$\ge\frac{\abs{\sum_{j=1}^n\p{f_j(\vb p+\vb h)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p)-(Df_j(\vb p)(\vb h)g_j(\vb p)+Dg_j(\vb p)(\vb h)f_j(\vb p))}}}{\norm{\vb h}}$$ $$=\frac{\abs{(f\cdot g)(\vb p+\vb h)-(f\cdot g)(\vb p)-(Df(\vb p)(\vb h)\cdot g(\vb p)+Dg(\vb p)(\vb h)\cdot f(\vb p))}}{\norm{\vb h}}$$ and we know that $$\lim_{\vb h\to \vb 0}\frac{\sum_{j=1}^n\abs{f_j(\vb p+\vb h)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p)-(Df_j(\vb p)(\vb h)g_j(\vb p)+Dg_j(\vb p)(\vb h)f_j(\vb p))}}{\norm{\vb h}}$$ $$=\sum_{j=1}^n\lim_{\vb h\to \vb 0}\frac{\abs{f_j(\vb p+\vb h)g_j(\vb p+\vb h)-f_j(\vb p)g_j(\vb p)-(Df_j(\vb p)(\vb h)g_j(\vb p)+Dg_j(\vb p)(\vb h)f_j(\vb p))}}{\norm{\vb h}} =0$$ by squeeze theorem, $$\lim_{\vb h\to \vb 0}\frac{\abs{(f\cdot g)(\vb p+\vb h)-(f\cdot g)(\vb p)-(Df(\vb p)(\vb h)\cdot g(\vb p)+Dg(\vb p)(\vb h)\cdot f(\vb p))}}{\norm{\vb h}}=0$$ Clearly, $Df(\vb p)(\vb h)\cdot g(\vb p)+Dg(\vb p)(\vb h)\cdot f(\vb p)$, as a function of $\vb h$, is linear. Hence $D(f\cdot g)(\vb p)$ exists and is defined by $$D(f\cdot g)(\vb p)(\vb h)=Df(\vb p)(\vb h)\cdot g(\vb p)+Dg(\vb p)(\vb h)\cdot f(\vb p)$$ $\blacksquare$

Proof. $$J(cf)_{ji}(\vb p)=D(cf)(\vb p)_{ji}=(cDf(\vb p))_{ji}=cDf(\vb p)_{ji}=cJf_{ji}(\vb p)$$
$$J(f\pm g)_{ji}(\vb p)=D(f\pm g)(\vb p)_{ji}=(Df(\vb p)\pm Dg(\vb p))_{ji}=Df(\vb p)_{ji}\pm Dg(\vb p)_{ji}=Jf_{ji}(\vb p)\pm Jg_{ji}(\vb p)=(Jf_{ji}\pm Jg_{ji})(\vb p)=(Jf\pm Jg)_{ji}(\vb p)$$
$$ J(f\cdot g)^T_i(\vb p) =J(f\cdot g)_i(\vb p) =D(f\cdot g)(\vb p)_i =D(f\cdot g)(\vb p)(\vb e_i) =Df(\vb p)(\vb e_i)\cdot g(\vb p)+Dg(\vb p)(\vb e_i)\cdot f(\vb p) =\sum_j(Df(\vb p)(\vb e_i))_jg(\vb p)_j+\sum_j(Dg(\vb p)(\vb e_i))_jf(\vb p)_j $$ $$ =\sum_jDf(\vb p)_{ji}g(\vb p)_j+\sum_jDg(\vb p)_{ji}f(\vb p)_j =\sum_j(Jf)^T_{ij}(\vb p)g_j(\vb p)+\sum_j(Jg)^T_{ij}(\vb p)f_j(\vb p) =((Jf)^Tg+(Jg)^Tf)_i(\vb p) $$ $\blacksquare$

Proof. Since $g$ is differentiable at $\vb p$ and $f$ is differentiable at $g(\vb p)$, there exists $s\gt0$ such that $B_s(\vb p)\subseteq V$ and $B_s(g(\vb p))\subseteq U$. Since $g$ is differentiable at $\vb p$, it is also continuous at $\vb p$. So there exists $0\lt r\lt s$ such that $\vb x\in B_r(\vb p)$ implies $g(\vb x)\in B_s(g(\vb p))$. For all $\vb h\in B_r(\vb 0)$ and $\vb k\in B_s(\vb 0)$, define $$u(\vb h)=g(\vb p+\vb h)-g(\vb p)-Dg(\vb p)(\vb h) \quad\text{and}\quad v(\vb k)=f(g(\vb p)+\vb k)-f(g(\vb p))-Df(g(\vb p))(\vb k)$$ Also define $\varepsilon(\vb h)$ and $\eta(\vb k)$ such that when $\vb h\neq\vb 0$, $\varepsilon(\vb h)=\frac{\norm{u(\vb h)}}{\norm{\vb h}}$, when $\vb k\neq\vb 0$, $\eta(\vb k)=\frac{\norm{v(\vb k)}}{\norm{\vb k}}$, and $\varepsilon(\vb 0)=\eta(\vb 0)=0$. Then $$\lim_{\vb h\to\vb 0}\varepsilon(\vb h)=0 \quad\text{and}\quad \lim_{\vb k\to\vb 0}\eta(\vb k)=0$$ and $$\norm{u(\vb h)}=\varepsilon(\vb h)\norm{\vb h} \quad\text{and}\quad \norm{v(\vb k)}=\eta(\vb k)\norm{\vb k}$$ where $\vb h$ or $\vb k$ may or may not be $\vb 0$. Given $\vb h\in B_r(\vb 0)$, let $\vb k=g(\vb p+\vb h)-g(\vb p)$, then $\vb k\in B_s(\vb 0)$, and since $g$ is continuous at $\vb p$, $\lim_{\vb h\to\vb 0}\vb k=\vb 0$. Now we have $$\norm{\vb k}=\norm{u(\vb h)+Dg(\vb p)(\vb h)}\le\norm{u(\vb h)}+\norm{Dg(\vb p)(\vb h)}\le(\varepsilon(\vb h)+\norm{Dg(\vb p)})\norm{\vb h}$$ and $$(f\circ g)(\vb p+\vb h)-(f\circ g)(\vb p)-(Df(g(\vb p))\circ Dg(\vb p))(\vb h) =f(g(\vb p)+\vb k)-f(g(\vb p))-(Df(g(\vb p))\circ Dg(\vb p))(\vb h)$$ $$=v(\vb k)+Df(g(\vb p))(\vb k)-Df(g(\vb p))(Dg(\vb p)(\vb h)) =v(\vb k)+Df(g(\vb p))(g(\vb p+\vb h)-g(\vb p)-Dg(\vb p)(\vb h)) =v(\vb k)+Df(g(\vb p))(u(\vb h))$$ and hence $$\norm{(f\circ g)(\vb p+\vb h)-(f\circ g)(\vb p)-(Df(g(\vb p))\circ Dg(\vb p))(\vb h)} =\norm{v(\vb k)+Df(g(\vb p))(u(\vb h))}$$ $$\le\norm{v(\vb k)}+\norm{Df(g(\vb p))(u(\vb h))} \le\eta(\vb k)\norm{\vb k}+\norm{Df(g(\vb p))}\norm{u(\vb h)} \le\eta(\vb k)(\varepsilon(\vb h)+\norm{Dg(\vb p)})\norm{\vb h}+\norm{Df(g(\vb p))}\varepsilon(\vb h)\norm{\vb h}$$ Thus, when $\vb h\neq\vb 0$, $$\frac{\norm{(f\circ g)(\vb p+\vb h)-(f\circ g)(\vb p)-(Df(g(\vb p))\circ Dg(\vb p))(\vb h)}}{\norm{\vb h}} \le\eta(\vb k)(\varepsilon(\vb h)+\norm{Dg(\vb p)})+\norm{Df(g(\vb p))}\varepsilon(\vb h)$$ Note that $\lim_{\vb h\to\vb 0}\varepsilon(\vb h)=0$ and, since $\lim_{\vb k\to\vb 0}\eta(\vb k)=0=\eta(\vb 0)$, by limit of composite functions, $\lim_{\vb h\to\vb 0}\eta(\vb k)=\eta(\lim_{\vb h\to\vb 0}\vb k)=\eta(\vb 0)=0$. We have $$\lim_{\vb h\to\vb 0}\p{\eta(\vb k)(\varepsilon(\vb h)+\norm{Dg(\vb p)})\norm{\vb h}+\norm{Df(g(\vb p))}\varepsilon(\vb h)\norm{\vb h}}=0$$ Therefore, by squeeze theorem, $$\lim_{\vb h\to\vb 0}\frac{\norm{(f\circ g)(\vb p+\vb h)-(f\circ g)(\vb p)-(Df(g(\vb p))\circ Dg(\vb p))(\vb h)}}{\norm{\vb h}}=0$$ We have shown that $D(f\circ g)(\vb p)$ exists and $D(f\circ g)(\vb p)=Df(g(\vb p))\circ Dg(\vb p)$. $\blacksquare$

Proof. $$J(f\circ g)_{ki}(\vb p)=D(f\circ g)(\vb p)_{ki}=(Df(g(\vb p))\circ Dg(\vb p))_{ki}=(Df(g(\vb p))Dg(\vb p))_{ki}$$ $$=\sum_jDf(g(\vb p))_{kj}Dg(\vb p)_{ji}=\sum_j(Jf_{kj}\circ g)(\vb p)Jg_{ji}(\vb p)=(\sum_j(Jf\circ g)_{kj}Jg_{ji})(\vb p)=((Jf\circ g)Jg)_{ki}(\vb p)$$ $\blacksquare$

Proof. For $n=0$, $\lim_{h\to 0}\frac{(p+h)-p}{h}=1$, thus $\dv{}{x}x^{0+1}=1=(0+1)x^0$. Suppose $\dv{}{x}x^{k+1}=(k+1)x^k$, then $$ \dv{}{x}x^{(k+1)+1} =\dv{}{x}x^{k+1}x+x^{k+1}\dv{}{x}x =(k+1)x^{k+1}+x^{k+1} =((k+1)+1)x^{k+1} $$ By induction, for all $n\in N$, $$\dv{}{x}x^{n+1}=(n+1)x^n$$ $\blacksquare$

Proof. Suppose $f$ has a local maximum at $x$. There exists $\delta\gt0$ such that $(x-\delta,x+\delta)\subseteq(a,b)$ and for all $y\in(x-\delta,x+\delta)$, $f(y)\le f(x)$. Note that for all $h\in(-\delta,0)$, $\frac{f(x+h)-f(x)}{h}\ge0$, and for all $h\in(0,\delta)$, $\frac{f(x+h)-f(x)}{h}\le0$. Hence $f'(x)$ cannot take a value greater than $0$ or less than $0$. Since $f'(x)\in R$, we have $f'(x)=0$. The case where $f$ has a local minimum at $x$ can be proven similarly. $\blacksquare$

Proof. Suppose $f$ is not bounded above, then for every positive natural number $n$, there exists $x_n\in U$ such that $f(x_n)\gt n$. By axiom of choice, this defines a sequence $(x_n)$ of $U$ such that $f(x_n)\gt n$. Then $(x_n)$ is bounded, and by Bolzano-Weierstrass theorem, it has a convergent subsequence $(x_{n_k})$. Suppose $(x_{n_k})$ converges to $c$, since $U$ is closed, $c\in U$. Since $f$ is continuous at $c$, $\lim_{x\to c}f(x)=f(c)$, so $\lim_{n\to\infty}f(x_{n_k})=f(c)$, a contradiction to $x_{n_k}\gt n_k\ge k$. Therefore, $f$ is bounded above. By a symmetric argument, $f$ is also bounded below. $\blacksquare$

Proof. By the above proposition, $f$ is bounded. By completeness of real numbers, $f$ has a supremum $M$. Let $n$ be a natural number. Since $M$ is the supremum of $f$, $M-1/n$ is not an upper bound of $f$. Therefore, for every positive natural number $n$, there exists $x_n\in U$ such that $M\ge f(x_n)\gt M-1/n$. By axiom of choice, this defines a sequence $(x_n)$ of $U$ such that $f(x_n)$ converges to $M$. Since $U$ is bounded, $(x_n)$ is bounded. By Bolzano-Weierstrass theorem, there exists a convergent subsequence $(x_{n_k})$ of $(x_n)$. Suppose $(x_{n_k})$ converges to $c$, then $c\in U$ as $U$ is closed. Since $f$ is continuous at $c$, $\lim_{x\to c}f(x)=f(c)$, so $\lim_{n\to\infty}f(x_{n_k})=f(c)$, which implies $f(c)=M$. Therefore, $f$ attains maximum in $U$. By a symmetric argument, $f$ also attains minimum in $U$. $\blacksquare$

Proof. Since $f(a)$ and $f(b)$ are already taken by $a$ and $b$, we only need to consider values strictly in between. Suppose $f(a)=f(b)$, then the theorem is trivially true.

Now Suppose $f(a)\lt f(b)$, for any $u\in(f(a),f(b))$, let $S$ be the set of all $x\in [a,b]$ such that $f(x)\le u$. Then $S$ is non-empty since $a$ is an element of $S$. Because $S$ is also bounded, by completeness of real numbers, its supremum, denoted $c$, exists, and clearly, $c\in[a,b]$. So $\lim_{x\to c}f(x)=f(c)$. This means that for every $\varepsilon\gt 0$, there exists $\delta\gt 0$ such that $x\in[a,b]\cap(c-\delta,c+\delta)$ implies $f(x)\in (f(c)-\varepsilon,f(c)+\varepsilon)$, that is, $f(c)\in(f(x)-\varepsilon,f(x)+\varepsilon)$. By definition of supremum, there exists $s\in S$ such that $s\in(c-\delta,c]$, and we also have $s\in[a,b]$, so $f(c)\lt f(s)+\varepsilon\le u+\varepsilon$. Now if $c\lt b$, take $r\in[a,b]\cap(c,c+\delta)$, then $r\notin S$, so $f(c)\gt f(r)-\varepsilon\gt u-\varepsilon$; otherwise, we have $c=b$, so $f(c)=f(b)\gt u\gt u-\varepsilon$. So we have $f(c)\in(u-\varepsilon,u+\varepsilon)$. Since $\varepsilon\gt0$ is arbitrarily chosen, we conclude that $f(c)=u$.

For the case $f(a)\gt f(b)$, take $g=-f$, then $g(a)\lt g(b)$, so for any $u\in(f(b),f(a))$, we have $-u\in(g(a),g(b))$, and there exists $c\in[a,b]$ such that $g(c)=-u$, hence $f(c)=-g(c)=-(-u)=u$. $\blacksquare$

Proof. Suppose $f$ has a local maximum at $x$. There exists $\delta\gt0$ such that $(x-\delta,x+\delta)\subseteq(a,b)$ and for all $y\in(x-\delta,x+\delta)$, $f(y)\le f(x)$. Note that for all $h\in(-\delta,0)$, $\frac{f(x+h)-f(x)}{h}\ge0$, and for all $h\in(0,\delta)$, $\frac{f(x+h)-f(x)}{h}\le0$. Hence $f'(x)$ cannot take a value greater than $0$ or less than $0$. Since $f'(x)\in R$, we have $f'(x)=0$. The case where $f$ has a local minimum at $x$ can be proven similarly. $\blacksquare$

Proof. Let $h(t)=(f(b)-f(a))g(t)-(g(b)-g(a))f(t)$ for $t\in[a,b]$. Then $h$ is continuous on $[a,b]$ and differentiable in $(a,b)$, and $h(a)=f(b)g(a)-f(a)g(b)=h(b)$.

Suppose for all $t\in(a,b)$, $h(t)=h(a)$, then let $c=(b-a)/2$, we have $c\in(a,b)$ and $h'(c)=0$. Now suppose there exists $t\in(a,b)$ such that $h(t)\neq h(a)$. Consider the case where $h(t)\gt h(a)$. by extreme value theorem, there exists $c\in[a,b]$ such that $h$ attains maximum at $c$. Since $h(t)\gt h(a)=h(b)$, $c\in(a,b)$. So the restriction $h|_{(a,b)}$ of $h$ has $h|_{(a,b)}'(c)=0$, hence $h'(c)=0$. By a symmetric argument, for the case where $h(t)\lt h(a)$, there exists $c\in(a,b)$ such that $h'(c)=0$.

Therefore, in every case, there exists $c\in(a,b)$ such that $h'(c)=0$. And for that $c$, we have $(f(b)-f(a))g'(c)-(g(b)-g(a))f'(c)=h'(c)=0$, so $(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c)$. $\blacksquare$

Proof. Take $g(x)=x$. Since $g$ is linear, it is differentiable and hence continuous. So the restriction $g|_{[a,b]}$ of $g$ is continuous on $[a,b]$ and differentiable in $(a,b)$. Since $\lim_{h=0}\frac{g(x+h)-g(x)}{h}=\lim_{h=0}\frac{h}{h}=1$ for all $x\in R$, we have $g|_{[a,b]}'(x)=1$ for all $x\in(a,b)$. By Cauchy's mean value theorem, there exists $x\in(a,b)$ such that $(f(b)-f(a))g'(x)=(g(b)-g(a))f'(x)$, implying $(f(b)-f(a))=(b-a)f'(x)$ for that $x$. $\blacksquare$

Proof. Let $\vb a,\vb b\in B_r(\vb p)$. Define $s(t)=(1-t)\vb a+t\vb b$. When $t\in[0,1]$, we have $$d(\vb p,s(t))=\norm{s(t)-\vb p}=\norm{(1-t)(\vb a-\vb p)+t(\vb b-\vb p)}\le\norm{(1-t)(\vb a-\vb p)}+\norm{t(\vb b-\vb p)}$$ $$=\abs{1-t}\norm{\vb a-\vb p}+\abs{t}\norm{\vb b-\vb p}\le\sup(\norm{\vb a-\vb p},\norm{\vb b-\vb p})\lt r$$ so $s(t)\in B_r(\vb p)$. $\blacksquare$

Proof. Since $$\sum_i\abs{v_i}^2\le(\sum_i\abs{v_i})^2$$ we have $$\Vert\vb v\Vert=\sqrt{\sum_i\abs{v_i}^2}\le\sum_i\abs{v_i}$$
By Cauchy-Schwarz inequality, $$\sum_i\abs{v_i}=\sum_i(\abs{v_i}1)\le\sqrt{\sum_i\abs{v_i}^2}\sqrt{n}=\sqrt{n}\Vert\vb v\Vert$$ $\blacksquare$

Proof. Suppose $f$ is continuously differentiable. Then all partial derivatives of $f$ exist on $U$. Given a partial derivative $D_if_j$, for all $\vb x,\vb y\in U$, $$d(D_if_j(\vb x),D_if_j(\vb y))=\norm{(Df(\vb x)\vb e_i)_j-(Df(\vb y)\vb e_i)_j} =\norm{((Df(\vb x)-Df(\vb y))\vb e_i)_j}\le\norm{(Df(\vb x)-Df(\vb y))\vb e_i}\le\norm{Df(\vb x)-Df(\vb y)}=d(Df(\vb x),Df(\vb y))$$ Hence for all $\vb p\in U$, for all $\varepsilon\gt0$, there exists $\delta\gt0$ such that $B_\delta(\vb p)\subseteq U$ and for all $\vb q\in B_\delta(\vb p)$, $Df(\vb q)\in B_\varepsilon(Df(\vb p))$, hence $D_if_j(\vb q)\in B_\varepsilon(D_if_j(\vb p))$, implying $D_if_j$ is continuous in $U$.

Now suppose all partial derivatives $D_if_j$ exist and are continuous on $U$. Given $\vb p\in U$, for all $\varepsilon\gt0$, let $\eta=\frac{\varepsilon}{2nm}$, define $r\gt0$ such that $B_r(\vb p)\subseteq U$ and for all $\vb q\in B_r(\vb p)$, $D_if_j(\vb q)\in B_{\eta}(D_if_j(\vb p))$. Let $\vb h\in B_r(\vb 0)\setminus\{\vb0\}$. For some scalars $h_i$, $\vb h=\sum_ih_i\vb e_i$. Denote $\sum_{k=1}^ih_k\vb e_k$ by $\vb v_i$ for $i$ from $0$ to $n$, then $$f_j(\vb p+\vb h)-f_j(\vb p)=\sum_i(f_j(\vb p+\vb v_i)-f_j(\vb p+\vb v_{i-1}))$$ Note that for each $\vb v_i$, $d(\vb p+\vb v_i,\vb p)=\norm{(\vb p+\vb v_i)-\vb p}=\norm{\vb v_i}\le\norm{\vb h}\lt r$, so $\vb p+\vb v_i\in B_r(\vb p)$. If we define $s_i(t)=(1-t)(\vb p+\vb v_{i-1})+t(\vb p+\vb v_i)$, then $s_i(t)\in B_r(\vb p)$ for all $t\in[0,1]$. Suppose $h_i\neq0$, then for all $t\in[0,1]$, $$(f_j\circ s_i)'(t) =\lim_{h\to0}\frac{f_j(s_i(t)+hh_i\vb e_i)-f_j(s_i(t))}{h} =h_i\lim_{h\to0}\frac{f_j(s_i(t)+hh_i\vb e_i)-f_j(s_i(t))}{hh_i} =h_iD_if_j(s_i(t))$$ Suppose $h_i=0$, then $f_j\circ s_i$ is a constant function, so we still have $(f_j\circ s_i)'(t)=h_iD_if_j(s_i(t))$ for all $t\in[0,1]$. By mean value theorem, there exists $c_i\in(0,1)$ such that $$h_iD_if_j(s_i(c_i))=(f_j\circ s_i)'(c_i)=(f_j\circ s_i)(1)-(f_j\circ s_i)(0)=f_j(\vb p+\vb v_i)-f_j(\vb p+\vb v_{i-1})$$ Hence $$f_j(\vb p+\vb h)-f_j(\vb p)=\sum_i(h_iD_if_j(s_i(c_i)))$$ And we have $$\norm{f_j(\vb p+\vb h)-f_j(\vb p)-\sum_i(h_iD_if_j(\vb p))} =\norm{\sum_i(h_iD_if_j(s_i(c_i)))-\sum_i(h_iD_if_j(\vb p))} =\norm{\sum_i(h_i(D_if_j(s_i(c_i))-D_if_j(\vb p)))} \le\sum_i\abs{h_i}\norm{D_if_j(s_i(c_i))-D_if_j(\vb p)} \lt\sum_i\abs{h_i}\eta \le\sqrt{n}\norm{\vb h}\eta \lt\norm{\vb h}\varepsilon$$ Therefore, $f_j$ is differentiable at $\vb p$ and $Df_j(\vb p)_i=D_if_j(\vb p)$. Since each component function is differentiable, $f$ is differentiable at $\vb p$ and $Df(\vb p)_{j,*}=Df_j(\vb p)$. For continuity of $Df$, with $\vb p$, $\varepsilon$ and $r$ defined the same way, for all $\vb q\in B_r(\vb p)$, given $\vb x$ such that $\Vert\vb x\Vert\le1$, we have $$\norm{(Df(\vb q)-Df(\vb p))\vb x} \le\sum_j\abs{\sum_i(D_if_j(\vb q)-D_if_j(\vb p))x_i} \le\sum_j\sum_i\abs{(D_if_j(\vb q)-D_if_j(\vb p))}\abs{x_i} \le\sum_j\sum_i\frac{\varepsilon}{2nm}\Vert\vb x\Vert \le\frac{\varepsilon}{2}$$ Hence $d(Df(\vb q),Df(\vb p))=\norm{(Df(\vb q)-Df(\vb p))}\le\frac{\varepsilon}{2}\lt\varepsilon$. $\blacksquare$

Proof. Differentiability of $f\circ g$ follows directly from chain rule.

Note that $J(f\circ g)=(Jf\circ g)Jg$. Thus $$J(f\circ g)_{ki}=\sum_j(Jf\circ g)_{kj}Jg_{ji}=\sum_j(Jf_{kj}\circ g)Jg_{ji}$$ By continuous differentiability of $f$ and $g$, every $Jf_{kj}$ is continuous on $U$ and every $Jg_{ji}$ is continuous on $V$. By differentiability and hence continuity of $g$, and continuity of composite functions, every $Jf_{kj}\circ g$ is continuous on $V$. Hence, by continuity of function operations, every $J(f\circ g)_{ki}$ is continuous on $V$. Therefore, $f\circ g$ is continuously differentiable. $\blacksquare$

Proof. Let $I_1$ and $I_2$ be open intervals obtained by splitting $I$ by $c$. Note that $g'$ is non-zero in $I_1$ and $I_2$. Suppose $a,b\in I_1$ with $g(a)=g(b)=0$. If $a\neq b$, then there exists $x$ strictly between $a$ and $b$ such that $g'(x)=0$, a contradiction. Hence $g$ either has no zero or has a unique zero in $I_1$. Same for $I_2$. Therefore, we can find an open interval $c\in(a,b)\subseteq I$ such that $g$ is non-zero in $(a,b)\setminus\{c\}$.

Suppose $\lim_{x\to c}\frac{f'(x)}{g'(x)}=L\in R$. Then for every $\varepsilon\gt0$, there exists $\delta\gt0$ such that $c+\delta\lt b$ and for all $x\in(c,c+\delta)$, $\frac{f'(x)}{g'(x)}\in(L-\varepsilon/2,L+\varepsilon/2)$. Let $y\in(c,c+\delta)$, then for every $x\in(c,y)$, by Cauchy's mean value theorem, there exists $t\in(x,y)$ such that $(f(y)-f(x))g'(t)=(g(y)-g(x))f'(t)$. Again, since $g'$ is non-zero in $(c,b)$, $g(y)-g(x)\neq0$. Hence $$\frac{f'(t)}{g'(t)}=\frac{f(y)-f(x)}{g(y)-g(x)}$$
Suppose $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0$. Note that we have $$\frac{f'(t)}{g'(t)}=\frac{f(y)-f(x)}{g(y)-g(x)}=\frac{\frac{f(y)}{g(y)}-\frac{f(x)}{g(y)}}{1-\frac{g(x)}{g(y)}}$$ Since $t\in(c,c+\delta)$, we have $$\frac{\frac{f(y)}{g(y)}-\frac{f(x)}{g(y)}}{1-\frac{g(x)}{g(y)}}=\frac{f'(t)}{g'(t)}\in(L-\varepsilon/2,L+\varepsilon/2)$$ Since $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0$, $$\lim_{x\to c}\frac{\frac{f(y)}{g(y)}-\frac{f(x)}{g(y)}}{1-\frac{g(x)}{g(y)}}=\frac{f(y)}{g(y)}$$ and we have $\frac{f(y)}{g(y)}\in[L-\varepsilon/2,L+\varepsilon/2]\subset(L-\varepsilon,L+\varepsilon)$, where $y$ can be arbitrarily chosen in $(c,c+\delta)$. By a symmetric argument, there exists $\delta'\gt0$ such that for all $y\in (c-\delta',c)$, $\frac{f(y)}{g(y)}\in(L-\varepsilon,L+\varepsilon)$. Therefore, let $\delta^*=\inf(\delta,\delta')$, then for all $x\in (c-\delta^*,c+\delta^*)\setminus\{c\}$, $\frac{f(x)}{g(x)}\in(L-\varepsilon,L+\varepsilon)$. We have shown that $\lim_{x\to c}\frac{f(x)}{g(x)}=L$.

Suppose $\lim_{x\to c}g(x)=\pm\infty$. Note that we have $$\frac{f'(t)}{g'(t)}=\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{\frac{f(x)}{g(x)}-\frac{f(y)}{g(x)}}{1-\frac{g(y)}{g(x)}}$$ $$\frac{f'(t)}{g'(t)}\p{1-\frac{g(y)}{g(x)}}=\frac{f(x)}{g(x)}-\frac{f(y)}{g(x)}$$ $$\frac{f'(t)}{g'(t)}=\frac{f(x)}{g(x)}-\p{\frac{f(y)}{g(x)}-\frac{f'(t)}{g'(t)}\frac{g(y)}{g(x)}}$$ And we will denote $\frac{f(y)}{g(x)}-\frac{f'(t)}{g'(t)}\frac{g(y)}{g(x)}$ by $r(x)$, where $y$ is fixed and $t$ depends on $x$, so that $$\frac{f(x)}{g(x)}=\frac{f'(t)}{g'(t)}+r(x)$$ Since $t\in(c,c+\delta)$, $\frac{f'(t)}{g'(t)}\in(L-\varepsilon/2,L+\varepsilon/2)$, so $\frac{f(x)}{g(x)}\in(L-\varepsilon/2+r(x),L+\varepsilon/2+r(x))$. Note that, since $\lim_{x\to c}g(x)=\pm\infty$, we have $\lim_{x\to c}\frac{1}{g(x)}=0$. Combining with the fact that $\frac{f'(t)}{g'(t)}$, as a function of $x$, is bounded in $(c,y)$, we have $$\lim_{x\to c^+}r(x)=0$$ So given the same $\varepsilon$, there exists $\sigma\gt0$ such that $c+\sigma\lt y$ and for all $x\in (c,c+\sigma)$, $r(x)\in(-\varepsilon/2,\varepsilon/2)$. Then for all $x\in (c,c+\sigma)$, $\frac{f(x)}{g(x)}\in(L-\varepsilon,L+\varepsilon)$. By a symmetric argument, there exists $\sigma'\gt0$ such that for all $x\in (c-\sigma',c)$, $\frac{f(x)}{g(x)}\in(L-\varepsilon,L+\varepsilon)$. Therefore, let $\sigma^*=\inf(\sigma,\sigma')$, then for all $x\in (c-\sigma^*,c+\sigma^*)\setminus\{c\}$, $\frac{f(x)}{g(x)}\in(L-\varepsilon,L+\varepsilon)$. We have shown that $\lim_{x\to c}\frac{f(x)}{g(x)}=L$.

The cases where $c=\pm\infty$ or $\lim_{x\to c}\frac{f'(x)}{g'(x)}=\pm\infty$ can be proven similarly. If $c=\pm\infty$, we only need to prove one side of the limit, obviously. $\blacksquare$

Proof. We will only show the smooth case, the $C^k$ case follows similarly.

Note that $(cf)_j=cf_j$. For each depth $k$, suppose $$D_{i_{k-1}}\ldots D_{i_1}(cf)_j=c\alpha$$ where $\alpha:U\to R$ is a smooth. Then $$D_{i_k}\ldots D_{i_1}(cf)_j=D_{i_k}(c\alpha)=J(c\alpha)_{i_k}=cJ\alpha_{i_k}=cD_{i_k}\alpha$$ which takes the form $c\beta$ where $\beta:U\to R$ is a smooth. Hence $cf$ is smooth.

Note that $(f+g)_j=f_j+g_j$. For each depth $k$, suppose $$D_{i_{k-1}}\ldots D_{i_1}(f+g)_j=\alpha+\beta$$ where $\alpha:U\to R$ and $\beta:U\to R$ are smooth. Then $$D_{i_k}\ldots D_{i_1}(f+g)_j=D_{i_k}(\alpha+\beta)=J(\alpha+\beta)_{i_k}=J\alpha_{i_k}+J\beta_{i_k}=D_{i_k}\alpha+D_{i_k}\beta$$ which takes the form $\alpha'+\beta'$ where $\alpha':U\to R$ and $\beta':U\to R$ are smooth. Hence $f+g$ is smooth.

For each depth $k$, suppose $$D_{i_{k-1}}\ldots D_{i_1}(f\cdot g)=\sum_t(\alpha_t\beta_t)$$ where each $\alpha_t:U\to R$ and $\beta_t:U\to R$ are smooth. Then $$ D_{i_{k}}\ldots D_{i_1}(f\cdot g) =\p{J\p{\sum_t(\alpha_t\beta_t)}}_{i_{k}} =\sum_tJ(\alpha_t\beta_t)_{i_{k}} =\sum_t((J\alpha_t)^T\beta_t+(J\beta_t)^T\alpha_t)^T_{i_{k}} =\sum_t(D_{i_{k}}\alpha_t\beta_t)+\sum_t(D_{i_{k}}\beta_t\alpha_t) $$ which takes the form $\sum_s(\alpha_s\beta_s)$ where each $\alpha_s:U\to R$ and $\beta_s:U\to R$ are smooth. Hence $f\cdot g$ is smooth. $\blacksquare$

Proof. We will only show the smooth case, the $C^k$ case follows similarly.

Note that $(f\circ g)_j=f_j\circ g$. For each depth $k$, suppose $$D_{i_{k-1}}\ldots D_{i_1}(f\circ g)_j=\sum_t((\alpha_t\circ\beta_t)\gamma_t)$$ where $\alpha_t:V\to R$, $\beta_t:U\to V$, $\gamma_t:U\to R$ are smooth. Then $$D_{i_k}\ldots D_{i_1}(f\circ g)_j =\p{J\sum_t((\alpha_t\circ\beta_t)\gamma_t)}_{i_k} =\sum_t(J((\alpha_t\circ\beta_t)\gamma_t))_{i_k} =\sum_t((J(\alpha_t\circ\beta_t)^T\gamma_t+J\gamma_t^T(\alpha_t\circ\beta_t))^T)_{i_k} =\sum_t((J(\alpha_t\circ\beta_t)^T\gamma_t)_{i_k}+(J\gamma_t^T(\alpha_t\circ\beta_t))_{i_k}) $$ $$ =\sum_t(J(\alpha_t\circ\beta_t)_{i_k}\gamma_t+(J\gamma_t)_{i_k}(\alpha_t\circ\beta_t)) =\sum_t\p{\p{\sum_j((J\alpha_t)_j\circ\beta_t)(J\beta_t)_{ji_k}}\gamma_t+(J\gamma_t)_{i_k}(\alpha_t\circ\beta_t)} =\sum_t\sum_j(D_j\alpha_t\circ\beta_t)D_{i_k}{\beta_t}_j\gamma_t+\sum_t(\alpha_t\circ\beta_t)D_{i_k}\gamma_t$$ which takes the form $\sum_s((\alpha_s\circ\beta_s)\gamma_s)$ where $\alpha_s:V\to R$, $\beta_s:U\to V$, $\gamma_s:U\to R$ are smooth. Hence $f\circ g$ is smooth. $\blacksquare$

Proof. Let $\vb p\in U$. Then $$D_iD_jf(\vb p)=\lim_{h\to0}\frac{D_jf(\vb p+h\vb e_i)-D_jf(\vb p)}{h} =\lim_{h\to0}\frac{\lim_{k\to0}\frac{f(\vb p+h\vb e_i+k\vb e_j)-f(\vb p+h\vb e_i)}{k}-\lim_{k\to0}\frac{f(\vb p+k\vb e_j)-f(\vb p)}{k}}{h} =\lim_{h\to0}\lim_{k\to0}\frac{(f(\vb p+h\vb e_i+k\vb e_j)-f(\vb p+h\vb e_i))-(f(\vb p+k\vb e_j)-f(\vb p))}{hk}$$ Now define $u(t)=f(\vb p+t\vb e_i+k\vb e_j)-f(\vb p+t\vb e_i)$, then $u'(t)=D_if(\vb p+t\vb e_i+k\vb e_j)-D_if(\vb p+t\vb e_i)$ and $$D_iD_jf(\vb p)=\lim_{h\to0}\lim_{k\to0}\frac{u(h)-u(0)}{hk}$$ By mean value theorem, there exists $s(h)$ strictly between $0$ and $h$ such that $u(h)-u(0)=hu'(s(h))=h(D_if(\vb p+s(h)\vb e_i+k\vb e_j)-D_if(\vb p+s(h)\vb e_i))$. Note that $\lim_{h\to0}s(h)=0$. Thus $$D_iD_jf(\vb p)=\lim_{h\to0}\lim_{k\to0}\frac{D_if(\vb p+s(h)\vb e_i+k\vb e_j)-D_if(\vb p+s(h)\vb e_i)}{k} =\lim_{h\to0}D_jD_if(\vb p+s(h)\vb e_i) =D_jD_if(\vb p)$$ $\blacksquare$

Proof. If $f$ is smooth by the original definition, it is trivial that $f$ is smooth by the generalized definition.

Suppose $f$ is smooth by the generalized definition. Let $j\in\{1,\ldots,m\}$, let $l\in N$, let $\vb i$ be an $l$-tuple of $\{1,\ldots,n\}$. Define $g_0=f_j$. Let $x\in U$, then there exists $r_x\gt0$ such that $f(u)=f_x(u)$ for all $u\in B_{r_x}(x)\subseteq U\cap U_x$. And there exists an $(l+1)$-tuple of functions $(g_{x,0},\ldots,g_{x,l})$ from $U_x$ to $R$ such that

• $g_{x,0}={f_x}_j$,
• $g_{x,k}=D_{i_k}g_{x,k-1}$ for $k\in\{1,\ldots,l\}$, and
• $g_{x,l}$ is continuous.

And we have $g_{x,0}(u)={f_x}_j(u)=f_j(u)=g_0(u)$ for all $u\in B_{r_x}(x)$. Inductively, for $k$ from $1$ to $l$, suppose for all $x\in U$, for all $u\in B_{r_x}(x)$, $g_{x,k-1}(u)=g_{k-1}(u)$. Let $x\in U$, then $g_{x,k}=D_{i_k}g_{x,k-1}$, by localness of derivative, for all $u\in B_{r_x}(x)$, $D_{i_k}g_{k-1}(u)=g_{x,k}(u)$. Thus $D_{i_k}g_{k-1}(x)=g_{x,k}(x)$. Define $g_k=D_{i_k}g_{k-1}$, then for all $x\in U$, for all $u\in B_{r_x}(x)$, $g_{x,k}(u)=D_{i_k}g_{k-1}(u)=g_k(u)$. By induction, $g_{k}=D_{i_k}g_{k-1}$ for $k\in\{1,\ldots,l\}$. And for all $x\in U$, for all $u\in B_{r_x}(x)$, $g_{x,l}(u)=g_l(u)$. Let $x\in U$, since $g_{x,l}$ is continuous, by localness of continuity, $g_l$ is continuous at $x$. Thus $g_l$ is continuous. We have shown that $f$ is smooth by the original definition. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. If $n=0$ or $l=0$, then $f\circ g$ is smooth by definition. If $m=0$, then $f\circ g$ is constant and thus smooth.

Suppose $l,m,n\neq0$. Let $x\in U$, then there exists an open subset $U_x$ of $R^n$ containing $x$ and a smooth map $g_x:U_x\to R^m$ that agrees with $g$ on $U_x\cap U$. Note that $g(x)\in V$, thus there exists an open subset $V_x$ of $R^m$ containing $g(x)$ and a smooth map $f_x:V_x\to R^l$ that agrees with $f$ on $V_x\cap V$. Then there exists $r_x\gt0$ such that $B_{r_x}(g(x))\subseteq V_x$. Since smoothness implies continuity, there exists $s_x\gt0$ such that $B_{s_x}(x)\subseteq U_x$ and for all $u\in B_{s_x}(x)$, $g_x(u)\in B_{r_x}(g(x))$. By smoothness of composite functions, $f_x|_{B_{r_x}(g(x))}\circ g_x|_{B_{s_x}(x)}$ is smooth. And for all $u\in B_{s_x}(x)\cap U$, we have $(f_x|_{B_{r_x}(g(x))}\circ g_x|_{B_{s_x}(x)})(u)=f_x(g_x(u))=f(g(u))=(f\circ g)(u)$. We have shown that $f\circ g$ is smooth. $\blacksquare$

Proof. This follows directly from generalized smoothness of composite functions. $\blacksquare$

Proof. If $n=0$ or $m=0$, then we trivially have $n=m$. Now suppose $n,m$ are non-zero.

Let $f:U\to V$ be a diffeomorphism. Let $p\in U$. Then $I_n=D(f^{-1}\circ f)(p)=D(f^{-1})(f(p))Df(p)$ and $I_m=D(f\circ f^{-1})(f(p))=Df(p)D(f^{-1})(f(p))$. Thus $n=\tr(D(f^{-1})(f(p))Df(p))=\tr(Df(p)D(f^{-1})(f(p)))=m$. And we have $Df(p)^{-1}=D(f^{-1})(f(p))$. Since $U$ and $V$ are diffeomorphic and $U$ is non-empty, such $f$ and $p$ exist, thus $n=m$. $\blacksquare$

Proof. Let $x^*\in X$. Use recursion theorem to define a function $\phi:N\to X$ such that $\phi(0)=x^*$ and $\phi(n+1)=f(\phi(n))$. Then $\phi$ defines a sequence $(x_n)$ of $X$. Since $f$ is a contraction on $X$, there exists $c\lt1$ such that for all $x,y\in X$, we have $d(f(x),f(y))\le cd(x,y)$. Note that $c\ge0$. By induction, $d(x_n,x_{n+1})\le c^nd(x_0,x_1)$. Suppose $n\lt m$, then $$d(x_n,x_m)\le\sum_{i=n}^{m-1}d(x_i,x_{i+1})\le\sum_{i=n}^{m-1}(c^id(x_0,x_1))=d(x_0,x_1)\sum_{i=n}^{m-1}c^i =d(x_0,x_1)\frac{c^n(1-c^{m-n})}{1-c}\le d(x_0,x_1)\frac{c^n}{1-c}$$ For every $\varepsilon\gt0$, there exists $N_0$ such that $d(x_0,x_1)\frac{c^{N_0}}{1-c}\lt\varepsilon$, so for all $m,n\ge N_0$, $d(x_n,x_m)\le d(x_0,x_1)\frac{c^{N_0}}{1-c}\lt\varepsilon$. Hence $(x_n)$ is a Cauchy sequence, and so $\lim x_n=x$ for some $x\in X$. Now suppose $f(x)\neq x$, then $d(x,f(x))\gt0$, so there exists some $n$ such that $d(x_n,x)\lt\frac{d(x,f(x))}{4}$ and $d(x_n,x_{n+1})\lt\frac{d(x,f(x))}{4}$. But then we have $d(x,x_{n+1})\le d(x_n,x)+d(x_n,x_{n+1})\lt\frac{d(x,f(x))}{2}$, so $$d(f(x_n),f(x))=d(x_{n+1},f(x))\ge d(x,f(x))-d(x,x_{n+1})\gt\frac{d(x,f(x))}{2}\gt\frac{d(x,f(x))}{4}\gt d(x_n,x)$$ a contradiction. Therefore, $f(x)=x$. To show uniqueness, suppose $f(x)=x$ and $f(y)=y$, then $d(x,y)=d(f(x),f(y))\le cd(x,y)$, so $d(x,y)=0$, implying $x=y$. $\blacksquare$

Proof. Let $\vb a,\vb b\in U$. Define $s(t)=(1-t)\vb a+t\vb b$. If $U$ is an open ball $B_r(\vb p)$, then when $t\in[0,1]$, we have $s(t)\in U$. This is trivially true if $U$ is $R^n$. Define $g(t)=f(s(t))$, then when $t\in[0,1]$, $Dg(t)=Df(s(t))Ds(t)=Df(s(t))(\vb b-\vb a)$ since $s(t)$ is the sum of a linear map and a constant map. And we have $\norm{Dg(t)}\le\norm{Df(s(t))}\norm{\vb b-\vb a}\le M\norm{\vb b-\vb a}$.

Let $\vb z=g(1)-g(0)$ and define $c(t)=\vb z\cdot g(t)$ on $t\in[0,1]$. Clearly, $c(t)$ is continuous on $[0,1]$ and differentiable in $(0,1)$. By mean value theorem, there exists $x\in(0,1)$ such that $$c(1)-c(0)=c'(x)=\vb z\cdot Dg(x)$$ But then we also have $$c(1)-c(0)=\vb z\cdot g(1)-\vb z\cdot g(0)=\vb z\cdot\vb z=\Vert\vb z\Vert^2$$ so by Cauchy-Schwarz inequality, $$\Vert\vb z\Vert^2=\vb z\cdot Dg(x)\le\Vert\vb z\Vert\norm{Dg(x)}$$ and we have $\norm{f(\vb b)-f(\vb a)}=\norm{g(1)-g(0)}\le\norm{Dg(x)}\le M\norm{\vb b-\vb a}$. $\blacksquare$

Proof. Let $p\in R\setminus\{0\}$. $$ \lim_{h\to 0}\frac{\frac{1}{p+h}-\frac{1}{p}}{h} =\lim_{h\to 0}\frac{-1}{p(p+h)}\frac{h}{h} =\lim_{h\to 0}\frac{-1}{p(p+h)}\lim_{h\to 0}\frac{h}{h} =\frac{-1}{p^2} $$ Thus $$\dv{}{x}\frac{1}{x}=\frac{-1}{x^2}$$ Given a function of the form $c\p{\frac{1}{x}}^k$ where $c\in R$ and $k\in N\setminus\{0\}$, we have $$ \dv{}{x}c\p{\frac{1}{x}}^k =ck\p{\frac{1}{x}}^{k-1}\frac{-1}{x^2} =-ck\p{\frac{1}{x}}^{k+1} $$ which is again of the form $c\p{\frac{1}{x}}^k$ with $k\neq0$. By induction, every function of the form $c\p{\frac{1}{x}}^k$ with $k\neq0$ is smooth. Since $\frac{1}{x}$ is itself of the form $c\p{\frac{1}{x}}^k$ with $k\neq0$, it is smooth. $\blacksquare$

Proof. This follows from the inverse matrix formula $A^{-1}=\frac{1}{\det(A)}\text{adj}(A)$, the properties of smooth (or $C^k$) functions, and that $\frac{1}{x}$ is smooth. $\blacksquare$

Proof. Whether $f$ is smooth or $C^k$ with $k\gt0$, it is continuously differentiable.

Let $A=Df(\vb p)$ and $\varepsilon=\frac{1}{2\norm{A^{-1}}}$. Since $f$ is continuously differentiable at $\vb p$, there exists an open ball $V\subseteq U$ centered at $\vb p$ such that for all $\vb x\in V$, $\norm{Df(\vb x)-A}\lt\varepsilon$. For every $\vb y\in f(V)$, define a function $$\varphi_{\vb y}(\vb x)=\vb x+A^{-1}(\vb y-f(\vb x))$$ on $V$, then $$\norm{D(\varphi_{\vb y})(\vb x)}=\norm{I-A^{-1}Df(\vb x)}=\norm{A^{-1}(A-Df(\vb x))}\le\norm{A^{-1}}\norm{A-Df(\vb x)}\lt\norm{A^{-1}}\varepsilon=\frac{1}{2}$$ For all $\vb x_1,\vb x_2\in V$, if $f(\vb x_1)=f(\vb x_2)$, let $\vb y=f(\vb x_1)=f(\vb x_2)$, then $$\norm{\vb x_1-\vb x_2}=\norm{\varphi_{\vb y}(\vb x_1)-\varphi_{\vb y}(\vb x_2)}\le\frac{1}{2}\norm{\vb x_1-\vb x_2}$$ implying $\vb x_1=\vb x_2$. We have shown that $f|_V:V\to R^n$ is injective, thus $f|_V:V\to f(V)$ is bijective.

Let $\vb y^*\in f(V)$. Then there exists $\vb x^*\in V$ such that $f(\vb x^*)=\vb y^*$. Let $B$ be an open ball centered at $\vb x^*$ with radius $r$ such that $\overline B\subseteq V$. Let $\vb y\in B_{r\varepsilon}(\vb y^*)$. then $$\norm{\varphi_{\vb y}(\vb x^*)-\vb x^*}=\norm{A^{-1}(\vb y-\vb y^*)}\lt\norm{A^{-1}}r\varepsilon=\frac{r}{2}$$ When $\vb x\in\overline B$, $$\norm{\varphi_{\vb y}(\vb x)-\vb x^*}\le\norm{\varphi_{\vb y}(\vb x)-\varphi_{\vb y}(\vb x^*)}+\norm{\varphi_{\vb y}(\vb x^*)-\vb x^*}\lt\frac{1}{2}\norm{\vb x-\vb x^*}+\frac{r}{2}\le r$$ thus $\varphi_{\vb y}(\vb x)\in B$. Hence the restriction of $\varphi_{\vb y}$ on $\overline B$ is a contraction on $\overline B$. Since every Cauchy sequence converges in $R^n$, every Cauchy sequence of $\overline B$ also converges in $R^n$, and hence in $\overline B$. Thus, there exists a unique $\vb x\in\overline B$ such that $\varphi_{\vb y}|_{\overline B}(\vb x)=\vb x$. For this $\vb x$, $A^{-1}(\vb y-f(\vb x))=\vb0$, thus $\vb y-f(\vb x)=\vb 0$ since $A^{-1}$ is bijective. Therefore, $\vb y=f(\vb x)\in f(V)$. We have shown that $f(V)$ is open.

Let $\vb y\in f(V)$ and denote $g(\vb y)$ by $\vb x$. Let $r\gt0$ such that $B_r(\vb y)\subseteq f(V)$ and let $\vb k\in B_r(\vb 0)$, then $\vb y+\vb k\in f(V)$. Let $\vb h=g(\vb y+\vb k)-\vb x$, then $\vb x+\vb h\in V$. Note that for all $\vb k\in B_r(\vb 0)\setminus\{\vb 0\}$, since $g$ is bijective, $g(\vb y)\neq g(\vb y+\vb k)$, thus $\vb h\neq\vb0$. Now we have $$\varphi_{\vb y}(\vb x+\vb h)-\varphi_{\vb y}(\vb x)=\vb h+A^{-1}(f(\vb x)-f(\vb x+\vb h))=\vb h-A^{-1}\vb k$$ Hence $$\norm{\vb h-A^{-1}\vb k}=\norm{\varphi_{\vb y}(\vb x+\vb h)-\varphi_{\vb y}(\vb x)}\le\frac{1}{2}\norm{\vb h}$$ Since $$\norm{\vb h}\le\norm{\vb h-A^{-1}\vb k}+\norm{A^{-1}\vb k}\le\frac{1}{2}\norm{\vb h}+\norm{A^{-1}\vb k}$$ we have $$\norm{\vb h}\le2\norm{A^{-1}\vb k}\le2\norm{A^{-1}}\norm{\vb k}=\frac{\norm{\vb k}}{\varepsilon}$$ For all $\sigma\gt0$, let $\delta=\inf(\sigma\varepsilon,r)$, then for all $\vb k\in B_\delta(\vb0)$, $\norm{\vb h}\le\frac{\norm{\vb k}}{\varepsilon}\lt\sigma$. Hence $\lim_{\vb k\to\vb 0}\vb h=\vb0$.

Note that, since $\vb x\in V$, $d(Df(\vb x),A)=\norm{Df(\vb x)-A}\lt\frac{1}{2\norm{A^{-1}}}\lt\frac{1}{\norm{A^{-1}}}$, and by a lemma proven in the "linear algebra" chapter, $Df(\vb x)$ is invertible, and we denote $Df(\vb x)^{-1}$ by $T$. Since $$g(\vb y+\vb k)-g(\vb y)-T\vb k=\vb h-T\vb k=T(T^{-1}\vb h-\vb k)=-T(f(\vb x+\vb h)-f(\vb x)-Df(\vb x)\vb h)$$ we have $$\frac{\norm{g(\vb y+\vb k)-g(\vb y)-T\vb k}}{\norm{\vb k}}\le\frac{\norm{T}}{\varepsilon}\frac{\norm{f(\vb x+\vb h)-f(\vb x)-Df(\vb x)\vb h}}{\norm{\vb h}}$$ For all $\sigma\gt0$, there exists $\delta\gt0$ with $B_\delta(\vb x)\subseteq V$ such that for all $\vb h^*\in B_\delta(\vb 0)\setminus\{\vb 0\}$, $\frac{\norm{f(\vb x+\vb h^*)-f(\vb x)-Df(\vb x)\vb h^*}}{\norm{\vb h^*}}\in B_\sigma(0)$, and for that $\delta$, there exists $\eta\gt0$ with $\eta\le r$ such that for all $\vb k\in B_\eta(\vb 0)\setminus\{\vb 0\}$, $\vb h\in B_\delta(\vb 0)\setminus\{\vb 0\}$, and hence $\frac{\norm{f(\vb x+\vb h)-f(\vb x)-Df(\vb x)\vb h}}{\norm{\vb h}}\in B_\sigma(0)$. Therefore, $$\lim_{\vb k\to\vb 0}\frac{\norm{f(\vb x+\vb h)-f(\vb x)-Df(\vb x)\vb h}}{\norm{\vb h}}=0$$ Since $\frac{\norm{T}}{\varepsilon}$ does not depend on $\vb k$, we have $$\lim_{\vb k\to\vb 0}\frac{\norm{T}}{\varepsilon}\frac{\norm{f(\vb x+\vb h)-f(\vb x)-Df(\vb x)\vb h}}{\norm{\vb h}}=0$$ By squeeze theorem, $$\lim_{\vb k\to\vb 0}\frac{\norm{g(\vb y+\vb k)-g(\vb y)-T\vb k}}{\norm{\vb k}}=0$$ Therefore, $$Dg(\vb y)=T=Df(g(\vb y))^{-1}$$
Since $g$ is differentiable, it is $C^0$. Suppose $g$ is $C^k$ and $f$ is $C^{k+1}$ where $k\in N$. Then every entry $D_if_j\circ g$ of $Jf\circ g$ is $C^k$. Thus every entry $D_ig_j$ of $Jg$ is $C^k$, implying $g$ is $C^{k+1}$. By induction, if $f$ is $C^k$ for some non-zero $k$, then $g$ is $C^k$; if $f$ is smooth, then $g$ is smooth. $\blacksquare$

Proof. $A(\vb h,\vb k)=\vb0$ if and only if $A_x\vb h+A_y\vb k=\vb0$ if and only if $\vb k=-(A_y)^{-1}A_x\vb h$. $\blacksquare$

Proof. Define $F:U\to R^{n+m}$ by $F(\vb x,\vb y)=(\vb x,f(\vb x,\vb y))$. Then $F$ is smooth (or $C^k$). For some $r\gt0$, we have $B_r(\vb a,\vb b)\subseteq U$. Suppose $(\vb h,\vb k)\in B_r(\vb0,\vb0)$, then $(\vb a+\vb h,\vb b+\vb k)\in U$. Note that $f(\vb a,\vb b)=\vb0$. If we let $r(\vb h,\vb k)=f((\vb a,\vb b)+(\vb h,\vb k))-A(\vb h,\vb k)$, then $$F((\vb a,\vb b)+(\vb h,\vb k))-F(\vb a,\vb b)=(\vb h,f(\vb a+\vb h,\vb b+\vb k))=(\vb h,A(\vb h,\vb k))+(\vb0,r(\vb h,\vb k))$$ hence $$\lim_{(\vb h,\vb k)\to\vb0}\frac{\norm{F((\vb a,\vb b)+(\vb h,\vb k))-F(\vb a,\vb b)-(\vb h,A(\vb h,\vb k))}}{\norm{(\vb h,\vb k)}} =\lim_{(\vb h,\vb k)\to\vb0}\frac{\norm{(\vb0,r(\vb h,\vb k))}}{\norm{(\vb h,\vb k)}} =\lim_{(\vb h,\vb k)\to\vb0}\frac{\norm{r(\vb h,\vb k)}}{\norm{(\vb h,\vb k)}} =0$$ by definition of total derivative. Since $L(\vb h,\vb k)=(\vb h,A(\vb h,\vb k))$ is clearly linear, we have $DF(\vb a,\vb b)=L$. Note that if $DF(\vb a,\vb b)(\vb h,\vb k)=\vb0$, then $(\vb h,A(\vb h,\vb k))=(\vb0,\vb0)$, so $\vb h=\vb0$ and $A(\vb 0,\vb k)=\vb0$, implying $\vb k=-(A_y)^{-1}A_x\vb 0=\vb0$. This shows that $DF(\vb a,\vb b)$ is injective, and hence is invertible.

By inverse function theorem, there exists an open subset $V\subseteq U$ such that $(\vb a,\vb b)\in V$, $F(V)$ is open, and the restriction $F|_V:V\to F(V)$ is bijective. Note that $(\vb a,\vb0)=F(\vb a,\vb b)\in F(V)$. If we define $W$ to be all $\vb x\in R^n$ such that $(\vb x,\vb0)\in F(V)$, then clearly, $W$ is also open, and $\vb a\in W$. For every $\vb x\in W$, $(\vb x,\vb0)\in F(V)$, then there exists $(\vb x',\vb y)\in V$ such that $F(\vb x',\vb y)=(\vb x,\vb0)$, which then implies that $\vb x'=\vb x$ and $f(\vb x,\vb y)=\vb0$. Note that $\vb y\in R^m$ and $(\vb x,\vb y)\in V$. To show uniqueness, suppose $\vb y'\in R^m$ such that $(\vb x,\vb y')\in V$ and $f(\vb x,\vb y')=\vb0$, then $F(\vb x,\vb y')=(\vb x,f(\vb x,\vb y'))=(\vb x,f(\vb x,\vb y))=F(\vb x,\vb y)$. Since $(\vb x,\vb y'),(\vb x,\vb y)\in V$, $F|_V(\vb x,\vb y')=F|_V(\vb x,\vb y)$, and by injectivity of $F|_V$, $(\vb x,\vb y')=(\vb x,\vb y)$, hence $\vb y'=\vb y$. Now define $f^*:W\to R^m$ such that $f^*(\vb x)$ is the unique $\vb y\in R^m$ with $(\vb x,\vb y)\in V$ and $f(\vb x,\vb y)=\vb0$. Then for all $\vb x\in W$, $f(\vb x,f^*(\vb x))=\vb0$. Since $f(\vb a,\vb b)=\vb0$, we have $f^*(\vb a)=\vb b$.

Also by inverse function theorem, denote ${F|_V}^{-1}:F(V)\to V$ by $G$, then $G$ is smooth (or $C^k$), and for all $\vb x\in W$, since $F|_V(\vb x,f^*(\vb x))=(\vb x,\vb 0)$, $G(\vb x,\vb 0)=(\vb x,f^*(\vb x))$. Define $H:W\to V$ by $H(\vb x)=G(\vb x,\vb 0)$, then $H$ is smooth (or $C^k$). Note that for all $\vb x\in W$, $H(\vb x)=G(\vb x,\vb 0)=(\vb x,f^*(\vb x))$. Thus $f^*$ is smooth (or $C^k$). Let $\vb x\in W$, then $DH(\vb x)$ is $(n+m)\times n$ with $DH(\vb x)_{i,j}=I_{ij}$ for all $i\in\{1,\ldots,n\}$ and $DH(\vb x)_{n+i,j}=Df^*(\vb x)_{ij}$ for all $i\in\{1,\ldots,m\}$. Hence for all $\vb h\in R^n$, $DH(\vb x)\vb h=(\vb h,Df^*(\vb x)\vb h)$. Since $(f\circ H)(\vb x)=f(\vb x,f^*(\vb x))=\vb 0$ for all $\vb x\in W$, we have $Df(H(\vb x))DH(\vb x)=D(f\circ H)(\vb x)=0$. With $Df(H(\vb a))=Df(\vb a,\vb b)=A$, we have $ADH(\vb a)=0$. It follows that, for all $\vb h\in R^n$, $$A_x\vb h+A_yDf^*(\vb a)\vb h=A(\vb h,Df^*(\vb a)\vb h)=ADH(\vb a)\vb h=\vb0$$ implying $Df^*(\vb a)\vb h=-(A_y)^{-1}A_x\vb h$. Therefore, $$Df^*(\vb a)=-(A_y)^{-1}A_x$$ $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. By the existence of partition given any closed interval and the boundedness of $f$, $\{U(P,f,\alpha):P\in\mathscr P\}$ and $\{L(P,f,\alpha):P\in\mathscr P\}$ are both non-empty sets of real numbers. Since $f$ is bounded, these sets are both bounded, hence $\inf_{P\in\mathscr P}U(P,f,\alpha)$ and $\sup_{P\in\mathscr P}L(P,f,\alpha)$ exist.

Let $P_1,P_2\in\mathscr P$, let $P^*$ be the common refinement of $P_1$ and $P_2$, then $$L(P_1,f,\alpha)\le L(P^*,f,\alpha)\le U(P^*,f,\alpha)\le U(P_2,f,\alpha)$$ Suppose for contradiction that $\inf_{P\in\mathscr P}U(P,f,\alpha)\lt\sup_{P\in\mathscr P}L(P,f,\alpha)$. Then there exists $c$ such that $\inf_{P\in\mathscr P}U(P,f,\alpha)\lt c\lt\sup_{P\in\mathscr P}L(P,f,\alpha)$, which implies that there exists $P_1\in\mathscr P$ such that $L(P_1,f,\alpha)\gt c$ and there exists $P_2\in\mathscr P$ such that $U(P_2,f,\alpha)\lt c$. But then, we have $U(P_2,f,\alpha)\lt c\lt L(P_1,f,\alpha)$, a contradiction. Hence $\inf_{P\in\mathscr P}U(P,f,\alpha)\ge\sup_{P\in\mathscr P}L(P,f,\alpha)$. $\blacksquare$

Proof. Suppose for every $\varepsilon\gt0$ there exists a partition $P$ such that $U(P,f,\alpha)-L(P,f,\alpha)\lt\varepsilon$. Since for every $P\in\mathscr P$, $$U(P,f,\alpha)\ge\inf_{P\in\mathscr P}U(P,f,\alpha)\ge\sup_{P\in\mathscr P}L(P,f,\alpha)\ge L(P,f,\alpha)$$ we have $$\inf_{P\in\mathscr P}U(P,f,\alpha)-\sup_{P\in\mathscr P}L(P,f,\alpha)\lt\varepsilon$$ Since $\varepsilon\gt0$ is arbitrary, we conclude that $$\inf_{P\in\mathscr P}U(P,f,\alpha)=\sup_{P\in\mathscr P}L(P,f,\alpha)$$
Suppose $\inf_{P\in\mathscr P}U(P,f,\alpha)=I=\sup_{P\in\mathscr P}L(P,f,\alpha)$. Given $\varepsilon\gt0$, by definition of supremum and infimum, there exist $P_1,P_2\in\mathscr P$ such that $$U(P_1,f,\alpha)\lt I+\frac{\varepsilon}{2}$$ $$L(P_2,f,\alpha)\gt I-\frac{\varepsilon}{2}$$ Let $P$ be the common refinement of $P_1$ and $P_2$, then $$U(P,f,\alpha)\le U(P_1,f,\alpha)\lt I+\frac{\varepsilon}{2}\lt L(P_2,f,\alpha)+\varepsilon\le L(P,f,\alpha)+\varepsilon$$ $\blacksquare$

Proof. Suppose $f$ is continuous and $\alpha$ is increasing on $[a,b]$. Since $[a,b]$ is closed and bounded, and hence compact, $f$ is uniformly continuous and bounded on $[a,b]$. For every $\varepsilon\gt0$, if $\alpha(a)\neq\alpha(b)$, let $\delta=\frac{\varepsilon}{2(\alpha(b)-\alpha(a))}$, otherwise let $\delta=1$, then we have $\delta\gt0$ and $$(\alpha(b)-\alpha(a))\delta\lt\varepsilon$$ Note that there exists $\eta\gt0$ such that for all $x,y\in[a,b]$ with $d(x,y)\lt\eta$, we have $d(f(x),f(y))\lt\frac{\delta}{2}$. Let $P$ be a partition with $n$ segments such that $\Delta x_i\lt\eta$, if we fix some $x\in P_i$, then for all $y\in P_i$, we have $d(x,y)\lt\eta$, so $d(f(x),f(y))\lt\frac{\delta}{2}$, or $f(x)-\frac{\delta}{2}\lt f(y)\lt f(x)+\frac{\delta}{2}$, thus $\sup_{y\in P_i}f(y)\le f(x)+\frac{\delta}{2}$ and $\inf_{y\in P_i}f(y)\ge f(x)-\frac{\delta}{2}$, implying $\sup_{y\in P_i}f(y)\le\inf_{y\in P_i}f(y)+\delta$. Hence, $$U(P,f,\alpha)-L(P,f,\alpha) =\sum_i(\sup_{x\in P_i}f(x)-\inf_{x\in P_i}f(x))\Delta\alpha_i \le\sum_i\delta\Delta\alpha_i=\delta(\alpha(b)-\alpha(a))\lt\varepsilon$$ Therefore, $f$ is integrable on $[a,b]$ with respect to $\alpha$. $\blacksquare$

Proof. For every $\varepsilon\gt0$, there exist $P_1,P_2$ such that $U(P_1,f_1,\alpha)-L(P_1,f_1,\alpha)\lt\frac{\varepsilon}{2}$ and $U(P_2,f_2,\alpha)-L(P_2,f_2,\alpha)\lt\frac{\varepsilon}{2}$. Let $P$ be the common refinement of $P_1$ and $P_2$, then $$U(P,f_1+f_2,\alpha)-L(P,f_1+f_2,\alpha) \le(U(P,f_1,\alpha)+U(P,f_2,\alpha))-(L(P,f_1,\alpha)+L(P,f_2,\alpha)) \le(U(P_1,f_1,\alpha)-L(P_1,f_1,\alpha))+(U(P_2,f_2,\alpha)-L(P_2,f_2,\alpha)) \lt\varepsilon$$ so $f_1+f_2$ is integrable on $[a,b]$ with respect to $\alpha$. Then we have $$\int_a^bf_1d\alpha+\int_a^bf_2d\alpha-\varepsilon\lt L(P_1,f_1,\alpha)+L(P_2,f_2,\alpha)\le L(P,f_1+f_2,\alpha) \le\int_a^b(f_1+f_2)d\alpha\le U(P,f_1+f_2,\alpha)\le U(P_1,f_1,\alpha)+U(P_2,f_2,\alpha)\lt \int_a^bf_1d\alpha+\int_a^bf_2d\alpha+\varepsilon$$ Hence $$\int_a^b(f_1+f_2)d\alpha=\int_a^bf_1d\alpha+\int_a^bf_2d\alpha$$
Suppose $c\neq0$. For every $\varepsilon\gt0$, there exist $P$ such that $U(P,f,\alpha)-L(P,f,\alpha)\lt\frac{\varepsilon}{\abs{c}}$, so $U(P,cf,\alpha)-L(P,cf,\alpha)=\abs{c}(U(P,f,\alpha)-L(P,f,\alpha))\lt\varepsilon$, implying $cf$ is integrable on $[a,b]$ with respect to $\alpha$. If $c\gt0$, then we have $$c\int_a^bfd\alpha-\varepsilon\lt cL(P,f,\alpha)=L(P,cf,\alpha)\le\int_a^b(cf)d\alpha\le U(P,cf,\alpha)=cU(P,f,\alpha)\lt c\int_a^bfd\alpha+\varepsilon$$ If $c\lt0$, then we have $$c\int_a^bfd\alpha-\varepsilon\lt cU(P,f,\alpha)=L(P,cf,\alpha)\le\int_a^b(cf)d\alpha\le U(P,cf,\alpha)=cL(P,f,\alpha)\lt c\int_a^bfd\alpha+\varepsilon$$ Suppose $c=0$, both $\int_a^b(cf)d\alpha$ and $c\int_a^bfd\alpha$ are clearly $0$. Hence $$\int_a^b(cf)d\alpha=c\int_a^bfd\alpha$$
For every $\varepsilon\gt0$, there exists $P_1,P_2$ such that $U(P_1,f,\alpha_1)-L(P_1,f,\alpha_1)\lt\frac{\varepsilon}{2}$ and $U(P_2,f,\alpha_2)-L(P_2,f,\alpha_2)\lt\frac{\varepsilon}{2}$. Let $P$ be the common refinement of $P_1$ and $P_2$, then $$U(P,f,\alpha_1+\alpha_2)-L(P,f,\alpha_1+\alpha_2) =(U(P,f,\alpha_1)+U(P,f,\alpha_2))-(L(P,f,\alpha_1)+L(P,f,\alpha_2)) \le(U(P_1,f,\alpha_1)-L(P_1,f,\alpha_1))+(U(P_2,f,\alpha_2)-L(P_2,f,\alpha_2)) \lt\varepsilon$$ so $f$ is integrable on $[a,b]$ with respect to $\alpha_1+\alpha_2$. Then we have $$\int_a^bfd\alpha_1+\int_a^bfd\alpha_2-\varepsilon\lt L(P_1,f,\alpha_1)+L(P_2,f,\alpha_2)\le L(P,f,\alpha_1+\alpha_2)\le \int_a^bfd(\alpha_1+\alpha_2)\le U(P,f,\alpha_1+\alpha_2)\le U(P_1,f,\alpha_1)+U(P_2,f,\alpha_2)\lt\int_a^bfd\alpha_1+\int_a^bfd\alpha_2+\varepsilon$$ Hence $$\int_a^bfd(\alpha_1+\alpha_2)=\int_a^bfd\alpha_1+\int_a^bfd\alpha_2$$
For every $\varepsilon\gt0$, there exist $P$ such that $U(P,f,\alpha)-L(P,f,\alpha)\lt\frac{\varepsilon}{c}$, so $U(P,f,c\alpha)-L(P,f,c\alpha)=c(U(P,f,\alpha)-L(P,f,\alpha))\lt\varepsilon$, implying $f$ is integrable on $[a,b]$ with respect to $c\alpha$. Then we have $$c\int_a^bfd\alpha-\varepsilon\lt cL(P,f,\alpha)=L(P,f,c\alpha)\le\int_a^bfd(c\alpha)\le U(P,f,c\alpha)=cU(P,f,\alpha)\lt c\int_a^bfd\alpha+\varepsilon$$ Hence $$\int_a^bfd(c\alpha)=c\int_a^bfd\alpha$$
For every $\varepsilon\gt0$, there exist $P$ such that $U(P,f,\alpha)-L(P,f,\alpha)\lt\varepsilon$. Let $P^*$ be the refinement of $P$ obtained by adding $c$ in $P$. Let $P_1,P_2$ be partitions of $[a,c],[c,b]$ obtained from $P^*$. Then for $i\in\{1,2\}$, $$U(P_i,f,\alpha)-L(P_i,f,\alpha)\le U(P^*,f,\alpha)-L(P^*,f,\alpha)\le U(P,f,\alpha)-L(P,f,\alpha)\lt\varepsilon$$ so $f$ is integrable on $[a,c]$ and $[c,b]$ with respect to $\alpha$. Then we have $$\int_a^bfd\alpha-\varepsilon\lt L(P,f,\alpha)\le L(P_1,f,\alpha)+L(P_2,f,\alpha)\le \int_a^cfd\alpha+\int_c^bfd\alpha\le U(P_1,f,\alpha)+U(P_2,f,\alpha)\le U(P,f,\alpha)\lt\int_a^bfd\alpha+\varepsilon$$ Hence $$\int_a^cfd\alpha+\int_c^bfd\alpha=\int_a^bfd\alpha$$
For every $\varepsilon\gt0$, there exist $P$ such that $U(P,f_2,\alpha)-L(P,f_2,\alpha)\lt\varepsilon$, so $$\int_a^bf_1d\alpha\le U(P,f_1,\alpha)\le U(P,f_2,\alpha)\lt\int_a^bf_2d\alpha+\varepsilon$$ Hence, $$\int_a^bf_1d\alpha\le\int_a^bf_2d\alpha$$ $\blacksquare$

Proof. Note that $\alpha$ is strictly increasing and $f,g$ are integrable on any sub-interval of $[a,b]$ with respect to $\alpha$ due to their continuity. Let $\varepsilon=g(c)-f(c)$. By continuity of $f$ and $g$, we can find an interval $[a^*,b^*]$, where $a^*\lt b^*$, such that $c\in[a^*,b^*]\subseteq[a,b]$ and for all $x\in[a^*,b^*]$, $f(x)\lt f(c)+\varepsilon/3$ and $g(x)\gt g(c)-\varepsilon/3$. So $$\int_{a^*}^{b^*}fd\alpha\le(f(c)+\varepsilon/3)(\alpha(b^*)-\alpha(a^*))\lt(g(c)-\varepsilon/3)(\alpha(b^*)-\alpha(a^*))\le\int_{a^*}^{b^*}gd\alpha$$ Therefore, $$\int_a^bfd\alpha=\int_a^{a^*}fd\alpha+\int_{a^*}^{b^*}fd\alpha+\int_{b^*}^bfd\alpha \lt\int_a^{a^*}gd\alpha+\int_{a^*}^{b^*}gd\alpha+\int_{b^*}^bgd\alpha=\int_a^bgd\alpha$$ $\blacksquare$

Proof. Let $F+c$ be a function in $F+C$. Then given $x\in U$, $(F+c)'(x)=F'(x)+c'(x)=f(x)+0=f(x)$. Thus $F+c$ is an antiderivative of $f$.

Now suppose $G$ is an antiderivative of $f$. Then given $x\in U$, $(G-F)'(x)=G'(x)-F'(x)=f(x)-f(x)=0$. Suppose there exists $a,b\in U$ such that $(G-F)(a)\neq(G-F)(b)$. Since $G-F$ is differentiable and thus continuous on $U$, by mean value theorem, for some $c$ strictly between $a,b$, we have $(G-F)'(c)\neq0$, a contradiction. Thus $G-F$ is a constant function, and $G=F+(G-F)$, implying $G\in F+C$. $\blacksquare$

Proof. For every $\varepsilon\gt0$, there exists $P=(x_0,\ldots,x_n)$ such that $U(P,f,x)-L(P,f,x)\lt\varepsilon$. Since $F$ is differentiable, by mean value theorem, for each $i$ there exists $t_i\in P_i$ such that $$F(x_i)-F(x_{i-1})=F'(t_i)\Delta x_i=f(t_i)\Delta x_i$$ Then $$F(b)-F(a)=\sum_i(F(x_i)-F(x_{i-1}))=\sum_if(t_i)\Delta x_i$$ Since $$U(P,f,x)\ge\sum_if(t_i)\Delta x_i\ge L(P,f,x)$$ and $$U(P,f,x)\ge\int_a^bf(x)dx\ge L(P,f,x)$$ We conclude that $$\abs{(F(b)-F(a))-\int_a^bf(x)dx}=\abs{\sum_if(t_i)\Delta x_i-\int_a^bf(x)dx}\lt\varepsilon$$ implying $$F(b)-F(a)=\int_a^bf(x)dx$$ $\blacksquare$

Proof. Since $\abs{f(t)}\le M$, we have $$-M\le f(t)\le M$$ Then $$-M(\alpha(b)-\alpha(a))=\int_a^b(-M)d\alpha\le\int_a^bfd\alpha\le\int_a^bMd\alpha=M(\alpha(b)-\alpha(a))$$ hence $$\abs{\int_a^bfd\alpha}\le M(\alpha(b)-\alpha(a))$$ $\blacksquare$

Proof. $F$ is well defined since $f$ is continuous and hence integrable on any closed interval that is a subset of $(a,b)$. Fix any $p\in(a,b)$. Let $\varepsilon\gt0$, there exists $\delta\gt0$ such that $(p-\delta,p+\delta)\subseteq(a,b)$ and for all $t\in(p-\delta,p+\delta)$, $f(t)\in(f(p)-\varepsilon/2,f(p)+\varepsilon/2)$. Let $h\in(-\delta,\delta)\setminus\{0\}$, then $$\abs{\frac{F(p+h)-F(p)}{h}-f(p)} =\abs{\frac{\int_p^{p+h}f(t)dt}{h}-f(p)} =\abs{\frac{\int_p^{p+h}f(t)dt}{h}-\frac{\int_p^{p+h}f(p)dt}{h}} =\abs{\frac{\int_p^{p+h}(f(t)-f(p))dt}{h}}$$ Since $\abs{f(t)-f(p)}\lt\varepsilon/2$ for all $t$ between $p$ and $p+h$, we have $$\abs{\int_p^{p+h}(f(t)-f(p))dt}\le\abs{h}\varepsilon/2$$ Hence $$\abs{\frac{F(p+h)-F(p)}{h}-f(p)}=\abs{\frac{\int_p^{p+h}(f(t)-f(p))dt}{h}}\le\varepsilon/2\lt\varepsilon$$ We conclude that $$F'(p)=\lim_{h\to0}\frac{F(p+h)-F(p)}{h}=f(p)$$ We have shown that $F$ is an antiderivative of $f$. $\blacksquare$

Proof. Since $F$ and $G$ are continuously differentiable, $f$ and $g$ are continuous, so $fG$, $Fg$ and $fG+Fg$ are continuous and hence integrable on $[a,b]$. Note that $FG$ is an antiderivative of $fG+Fg$. Therefore, $$\int_a^bfG(x)dx+\int_a^bFg(x)dx=\int_a^b(fG+Fg)(x)dx=FG(b)-FG(a)$$ $\blacksquare$

Proof. $f$ is integrable on $[a,b]$ with respect to $\alpha$ since it is continuous. since $f$ is continuous and $\alpha$ is continuously differentiable, both $\alpha'$ and $f\alpha'$ are continuous and hence integrable on $[a,b]$. Since $f$ is continuous, $\abs{f}$ is also continuous and hence bounded on $[a,b]$. Let $M=\sup_{x\in[a,b]}\abs{f(x)}$, then $M\ge0$.

For every $\varepsilon\gt0$, let $\delta=\frac{\varepsilon}{M}$ if $M\neq0$ and $\delta=1$ if $M=0$, then there exists $P_1$ such that $U(P_1,\alpha',x)-L(P_1,\alpha',x)\lt\delta$ and we have $M(U(P_1,\alpha',x)-L(P_1,\alpha',x))\lt\varepsilon$. Also, there exist $P_2$ such that $U(P_2,f,\alpha)-L(P_2,f,\alpha)\lt\varepsilon$ and $P_3$ such that $U(P_3,f\alpha',x)-L(P_3,f\alpha',x)\lt\varepsilon$. Let $P$ be the common refinement of $P_1,P_2,P_3$, then $$M(U(P,\alpha',x)-L(P,\alpha',x))\lt\varepsilon$$ $$U(P,f,\alpha)-L(P,f,\alpha)\lt\varepsilon$$ $$U(P,f\alpha',x)-L(P,f\alpha',x)\lt\varepsilon$$
By mean value theorem, for each $i$, there exists $t_i\in P_i$ such that $$\Delta\alpha_i=\alpha'(t_i)\Delta x_i$$ If $s_i\in P_i$ for each $i$, then $$\abs{\sum_if(s_i)\Delta\alpha_i-\sum_if(s_i)\alpha'(s_i)\Delta x_i} =\abs{\sum_if(s_i)\alpha'(t_i)\Delta x_i-\sum_if(s_i)\alpha'(s_i)\Delta x_i} \le\sum_i\abs{f(s_i)}\abs{\alpha'(t_i)-\alpha'(s_i)}\Delta x_i \le M\sum_i\abs{\alpha'(t_i)-\alpha'(s_i)}\Delta x_i \le M\sum_i(\sup_{x\in P_i}\alpha'(x)-\inf_{x\in P_i}\alpha'(x))\Delta x_i =M(U(P,\alpha',x)-L(P,\alpha',x)) \lt\varepsilon$$ Hence $$\sum_if(s_i)\Delta\alpha_i \lt\sum_if(s_i)\alpha'(s_i)\Delta x_i+\varepsilon \le U(P,f\alpha',x)+\varepsilon$$ and $$\sum_if(s_i)\alpha'(s_i)\Delta x_i \lt\sum_if(s_i)\Delta\alpha_i+\varepsilon \le U(P,f,\alpha)+\varepsilon$$ Then we have $$U(P,f,\alpha)\le U(P,f\alpha',x)+\varepsilon$$ and $$U(P,f\alpha',x)\le U(P,f,\alpha)+\varepsilon$$ implying $$\abs{U(P,f,\alpha)-U(P,f\alpha',x)}\le\varepsilon$$ Therefore, $$\abs{\int_a^bfd\alpha-\int_a^bf(x)\alpha'(x)dx} \le\p{U(P,f,\alpha)-\int_a^bfd\alpha}+\abs{U(P,f,\alpha)-U(P,f\alpha',x)}+\p{U(P,f\alpha',x)-\int_a^bf(x)\alpha'(x)dx} \lt 3\varepsilon$$ And we conclude that $$\int_a^bfd\alpha=\int_a^bf(x)\alpha'(x)dx$$ $\blacksquare$

Proof. By intermediate value theorem, $\gamma$ defines a bijective restriction $\gamma'$ from $[A,B]$ to $[a,b]$. Define a function $\mathscr F:\mathscr P([a,b])\to\mathscr P([A,B])$ such that $\mathscr F((x_0,\ldots,x_n))=(\gamma'^{-1}(x_0),\ldots,\gamma'^{-1}(x_n))$, then $\mathscr F$ is clearly a bijection.

Given $(x_0,\ldots,x_n)=P\in\mathscr P([a,b])$, define $(y_0,\ldots,y_n)=Q=\mathscr F(P)$, then for each $i$, $\gamma$ defines a bijective restriction from $[y_{i-1},y_i]$ to $[x_{i-1},x_i]$, hence $$U(Q,g,\beta) =\sum_i\p{\sup_{y\in[y_{i-1},y_i]}g(y)}(\beta(y_i)-\beta(y_{i-1})) =\sum_i\p{\sup_{x\in[x_{i-1},x_i]}f(x)}((\alpha\circ\gamma)(y_i)-(\alpha\circ\gamma)(y_{i-1})) =\sum_i\p{\sup_{x\in[x_{i-1},x_i]}f(x)}(\alpha(x_i)-\alpha(x_{i-1})) =U(P,f,\alpha)$$ and by a symmetric argument we have $L(Q,g,\beta)=L(P,f,\alpha)$.

For all $\varepsilon\gt0$, there exists $P\in\mathscr P([a,b])$ such that $U(P,f,\alpha)-L(P,f,\alpha)\lt\varepsilon$. Let $Q=\mathscr F(P)$, then $U(Q,g,\beta)-L(Q,g,\beta)\lt\varepsilon$, so $g$ is integrable on $[A,B]$ with respect to $\beta$. Since $$U(P,f,\alpha)\ge\int_a^bfd\alpha\ge L(P,f,\alpha)$$ and $$U(P,f,\alpha)=U(Q,g,\beta)\ge\int_A^Bgd\beta\ge L(Q,g,\beta)=L(P,f,\alpha)$$ we have $$\abs{\int_a^bfd\alpha-\int_A^Bgd\beta}\lt\varepsilon$$ Hence $$\int_a^bfd\alpha=\int_A^Bgd\beta$$ $\blacksquare$

Proof. Since $f$ is continuous, $f$ is integrable on $[a,b]$ with respect to $\alpha$. Hence $$\int_a^bf(x)dx=\int_A^B(f\circ\gamma)d\gamma$$ Since $f\circ\gamma$ is continuous and defined on $[A,B]$, we also have $$\int_A^B(f\circ\gamma)d\gamma=\int_A^B(f\circ\gamma)(x)\gamma'(x)dx$$ and thus $$\int_a^bf(x)dx=\int_A^B(f\circ\gamma)(x)\gamma'(x)dx$$ $\blacksquare$

Proof. The proof is a simple extension of the single-dimensional case. $\blacksquare$

Proof. The proof is a simple extension of the single-dimensional case. $\blacksquare$

Proof. Let $\vb a_k,\vb b_k$ be the restrictions of $\vb a,\vb b$ on $\{1,\ldots,k\}$. Let $F_k$ denote $$\int_{a_{k+1}}^{b_{k+1}}\ldots\int_{a_n}^{b_n}f(x_1,\ldots,x_n)dx_n\ldots dx_{k+1}$$ Let $k\in\{1,\ldots,n\}$ and suppose for inductive hypothesis that $$\int_{[\vb a,\vb b]}f(x_1,\ldots,x_n)d(x_1,\ldots,x_n)=\int_{[\vb a_k,\vb b_k]}F_kd(x_1,\ldots,x_k)$$ where $F_k$ defines a continuous real-valued function on $[\vb a_k,\vb b_k]$.

Given $\vb x=(x_1,\ldots,x_{k-1})\in[\vb a_{k-1},\vb b_{k-1}]$, denote the map $x\mapsto(x_1,\ldots,x_{k-1},x)$ by $\phi_{\vb x}$. Then $F_k\circ\phi_{\vb x}$ is continuous and hence integrable on $[a_k,b_k]$. Thus $F_{k-1}$ defines a real-valued function on $[\vb a_{k-1},\vb b_{k-1}]$. If $a_k=b_k$, then $F_{k-1}$ is constant and thus continuous. Now suppose $a_k\lt b_k$ and let $d$ denote $b_k-a_k$. Since $F_k$ is continuous and $[\vb a_k,\vb b_k]$ is compact, $F_k$ is uniformly continuous. Let $\vb x=(x_1,\ldots,x_{k-1})\in[\vb a_{k-1},\vb b_{k-1}]$. Let $\varepsilon\gt0$. Then for some $\delta\gt0$, for all $\vb w_1,\vb w_2\in[\vb a_k,\vb b_k]$ with $d(\vb w_1,\vb w_2)\lt\delta$, $d(F_k(\vb w_1),F_k(\vb w_2))\lt\frac{\varepsilon}{d}$. Let $\vb y=(y_1,\ldots,y_{k-1})\in[\vb a_{k-1},\vb b_{k-1}]$ with $d(\vb x,\vb y)\lt\delta$. Then for all $x\in[a_k,b_k]$, $d(\phi_{\vb x}(x),\phi_{\vb y}(x))=d(\vb x,\vb y)\lt\delta$, implying $d(F_k(\phi_{\vb x}(x)),F_k(\phi_{\vb y}(x)))\lt\frac{\varepsilon}{d}$. Thus $$\int_{a_k}^{b_k}F_k(\phi_{\vb x}(x))dx-\varepsilon =\int_{a_k}^{b_k}(F_k(\phi_{\vb x}(x))-\frac{\varepsilon}{d})dx \le\int_{a_k}^{b_k}F_k(\phi_{\vb y}(x))dx \le\int_{a_k}^{b_k}(F_k(\phi_{\vb x}(x))+\frac{\varepsilon}{d})dx =\int_{a_k}^{b_k}F_k(\phi_{\vb x}(x))dx+\varepsilon$$ Then $d(F_{k-1}(\vb x),F_{k-1}(\vb y))\lt\varepsilon$. We have shown that $F_{k-1}$ is continuous, and thus integrable.

Let $\varepsilon\gt0$. There exists a partition $P$ of $[\vb a_k,\vb b_k]$, constructed from partitions $P_1,\ldots,P_k$, such that $U(P,F_k)-L(P,F_k)\lt\varepsilon$. Then $P'$ constructed from $P_1,\ldots,P_{k-1}$ is a partition of $[\vb a_{k-1},\vb b_{k-1}]$. Note that $$U(P',F_{k-1}) =\sum_{j_1,\ldots,j_{k-1}}\p{\sup_{\vb x\in P'_{j_1,\ldots,j_{k-1}}}\int_{a_k}^{b_k}(F_k\circ\phi_{\vb x})(x)dx}\prod_{i=1}^{k-1}\Delta {P_i}_{j_i} \le\sum_{j_1,\ldots,j_{k-1}}\p{\sup_{\vb x\in P'_{j_1,\ldots,j_{k-1}}}U(P_k,F_k\circ\phi_{\vb x})}\prod_{i=1}^{k-1}\Delta {P_i}_{j_i} $$ $$ =\sum_{j_1,\ldots,j_{k-1}}\p{\sup_{\vb x\in P'_{j_1,\ldots,j_{k-1}}}\sum_l\p{\sup_{x\in{P_k}_{l}}(F_k\circ\phi_{\vb x})(x)}\Delta{P_i}_{l}}\prod_{i=1}^{k-1}\Delta {P_i}_{j_i} \le\sum_{j_1,\ldots,j_{k-1}}\sum_l\p{\sup_{\vb x\in P'_{j_1,\ldots,j_{k-1}}}\p{\sup_{x\in{P_k}_{l}}(F_k\circ\phi_{\vb x})(x)}\Delta{P_i}_{l}}\prod_{i=1}^{k-1}\Delta {P_i}_{j_i} $$ $$ =\sum_{j_1,\ldots,j_{k-1}}\sum_l\p{\sup_{\vb x\in P'_{j_1,\ldots,j_{k-1}}}\sup_{x\in{P_k}_{l}}(F_k\circ\phi_{\vb x})(x)}\Delta{P_i}_{l}\prod_{i=1}^{k-1}\Delta {P_i}_{j_i} \le\sum_{l_1,\ldots,l_k}\p{\sup_{\vb x\in P_{l_1,\ldots,l_k}}F_k(\vb x)}\prod_{j=1}^k\Delta {P_j}_{l_j} =U(P,F_k)$$ Similarly, $L(P',F_{k-1})\ge L(P,F_k)$. Thus $$\int_{[\vb a_{k-1},\vb b_{k-1}]}F_{k-1}d(x_1,\ldots,x_{k-1})-\varepsilon\le U(P',F_{k-1})-\varepsilon\le U(P,F_k)-\varepsilon\lt\int_{[\vb a_k,\vb b_k]}F_kd(x_1,\ldots,x_k) \lt L(P,F_k)+\varepsilon\le L(P',F_{k-1})+\varepsilon\le\int_{[\vb a_{k-1},\vb b_{k-1}]}F_{k-1}d(x_1,\ldots,x_{k-1})+\varepsilon$$ Therefore, $$\int_{[\vb a,\vb b]}f(x_1,\ldots,x_n)d(x_1,\ldots,x_n)=\int_{[\vb a_k,\vb b_k]}F_kd(x_1,\ldots,x_k)=\int_{[\vb a_{k-1},\vb b_{k-1}]}F_{k-1}d(x_1,\ldots,x_{k-1})$$ We have concluded the inductive step.

Note that $[\vb a_n,\vb b_n]$ is just $[\vb a,\vb b]$ and $F_n$ is just $f(x_1,\ldots,x_n)$. By induction, we have $$\int_{[\vb a,\vb b]}f(x_1,\ldots,x_n)d(x_1,\ldots,x_n)=\int_{[\vb a_0,\vb b_0]}F_0d(x_1,\ldots,x_0)=\int_{a_1}^{b_1}\ldots\int_{a_n}^{b_n}f(x_1,\ldots,x_n)dx_n\ldots dx_1$$ $\blacksquare$

Proof. Let $\vb a_\sigma,\vb b_\sigma$ denote $(a_{\sigma_1},\ldots,a_{\sigma_n}),(b_{\sigma_1},\ldots,b_{\sigma_n})$. Let $\phi$ denote the map $(x_1,\ldots,x_n)\mapsto(x_{\sigma_1},\ldots,x_{\sigma_n})$ from $[\vb a,\vb b]$ to $[\vb a_\sigma,\vb b_\sigma]$. Then $\phi^{-1}$ is the map $(y_1,\ldots,y_n)\mapsto(y_{\sigma^{-1}_1},\ldots,y_{\sigma^{-1}_n})$ from $[\vb a_\sigma,\vb b_\sigma]$ to $[\vb a,\vb b]$. Since $\phi^{-1}$ is continuous, $f\circ\phi^{-1}$ is continuous and thus integrable. Let $\varepsilon\gt0$. There exists a partition $P$ of $[\vb a,\vb b]$, constructed from partitions $P_1,\ldots,P_n$, such that $U(P,f)-L(P,f)\lt\varepsilon$. Let $P'$ denote the partition of $[\vb a_\sigma,\vb b_\sigma]$ constructed from $P_{\sigma_1},\ldots,P_{\sigma_n}$. Then $$ U(P',f\circ\phi^{-1}) =\sum_{j_1,\ldots,j_n}\p{\sup_{\vb x\in P'_{j_1,\ldots,j_n}}f(x_{\sigma^{-1}_1},\ldots,x_{\sigma^{-1}_n})}\prod_i\Delta {P'_i}_{j_i} =\sum_{j_1}\ldots\sum_{j_n}\p{\sup_{\vb x\in P'_{j_1,\ldots,j_n}}f(x_{\sigma^{-1}_1},\ldots,x_{\sigma^{-1}_n})}\Delta {P'_1}_{j_1}\ldots\Delta {P'_n}_{j_n} $$ $$ =\sum_{j_{\sigma^{-1}_1}}\ldots\sum_{j_{\sigma^{-1}_n}}\p{\sup_{\vb x\in P'_{j_1,\ldots,j_n}}f(x_{\sigma^{-1}_1},\ldots,x_{\sigma^{-1}_n})}\Delta {P'_1}_{j_1}\ldots\Delta {P'_n}_{j_n} =\sum_{l_1}\ldots\sum_{l_n}\p{\sup_{\vb x\in P'_{l_{\sigma_1},\ldots,l_{\sigma_n}}}f(x_{\sigma^{-1}_1},\ldots,x_{\sigma^{-1}_n})}\Delta {P'_1}_{l_{\sigma_1}}\ldots\Delta {P'_n}_{l_{\sigma_n}} $$ $$ =\sum_{l_1}\ldots\sum_{l_n}\p{\sup_{\vb x\in P_{l_1,\ldots,l_n}}f(x_1,\ldots,x_n)}\Delta {P_1}_{l_1}\ldots\Delta {P_n}_{l_n} =\sum_{l_1,\ldots,l_n}\p{\sup_{\vb x\in P_{l_1,\ldots,l_n}}f(x_1,\ldots,x_n)}\prod_j\Delta {P_j}_{l_j} =U(P,f) $$ Similarly, $L(P',f\circ\phi^{-1})=L(P,f)$. Thus $$ \int_{a_1}^{b_1}\ldots\int_{a_n}^{b_n}f(x_1,\ldots,x_n)dx_n\ldots dx_1 =\int_{[\vb a,\vb b]}f(x_1,\ldots,x_n)d(x_1,\ldots,x_n) =\int_{[\vb a_\sigma,\vb b_\sigma]}(f\circ\phi^{-1})(y_1,\ldots,y_n)d(y_1,\ldots,y_n) $$ $$ =\int_{a_{\sigma_1}}^{b_{\sigma_1}}\ldots\int_{a_{\sigma_n}}^{b_{\sigma_n}}(f\circ\phi^{-1})(y_1,\ldots,y_n)dy_n\ldots dy_1 =\int_{a_{\sigma_1}}^{b_{\sigma_1}}\ldots\int_{a_{\sigma_n}}^{b_{\sigma_n}}f(x_1,\ldots,x_n)dx_{\sigma_n}\ldots dx_{\sigma_1} $$ $\blacksquare$

Proof. Suppose $(f_n)$ uniformly converges on $E$, to some $f:E\to Y$, then for all $\varepsilon\gt0$, there exists $m\in N$, such that for all $n\ge m$ and $x\in E$, $d(f_n(x),f(x))\lt\varepsilon$. Note that $x$ is not dependent on $\varepsilon$ or $m$, so fix $x\in E$, then for all $\varepsilon\gt0$, there exists $m\in N$, such that for all $n\ge m$, $d(f_n(x),f(x))\lt\varepsilon$. Hence $(f_n(x))$ converges to $f(x)$. Since $x\in E$ is arbitrary, $(f_n)$ converges to $f$ on $E$. $\blacksquare$

Proof. Suppose $(f_n)$ uniformly converges on $E$, to some $f:E\to Y$, then for all $\varepsilon\gt0$, there exists $l\in N$, such that for all $m,n\ge l$ and $x\in E$, $d(f_m(x),f(x))\lt\varepsilon/2$ and $d(f_n(x),f(x))\lt\varepsilon/2$, so $d(f_m(x),f_n(x))\lt\varepsilon$.

Suppose for all $\varepsilon\gt0$, there exists $l\in N$, such that for all $m,n\ge l$ and $x\in E$, $d(f_m(x),f_n(x))\lt\varepsilon$. Note that $x$ is not dependent on $\varepsilon$ or $l$, so fix $x\in E$, then for all $\varepsilon\gt0$, there exists $l\in N$, such that for all $m,n\ge l$, $d(f_m(x),f_n(x))\lt\varepsilon$. Hence $(f_n(x))$ is Cauchy and, by completeness of $Y$, converges. Thus we can define a function $f:E\to Y$ such that $f(x)=\lim_{n\to\infty}f_n(x)$. Keeping the definitions for $\varepsilon$ and $l$, fix $n\ge l$, then for all $x\in E$, there exists $m\ge l$ such that $d(f_m(x),f(x))\lt\varepsilon$, so $d(f_n(x),f(x))\le d(f_m(x),f_n(x))+d(f_m(x),f(x))\lt2\varepsilon$. This implies that $(f_n)$ uniformly converges to $f$ on $E$. $\blacksquare$

Proof. Let $\varepsilon\gt0$. Since $(f_n)$ uniformly converges to $f$ on $E$, there exists $N$ such that for all $m,n\ge N$ and $x\in E$, $d(f_m(x),f_n(x))\lt\varepsilon/3$. Since $\lim_{x\to p}f_m(x)$ and $\lim_{x\to p}f_n(x)$ exist, there exists $\delta\gt0$ such that for all $x\in E\cap B_\delta(p)\setminus\{p\}$, $d(f_m(x),\lim_{x\to p}f_m(x))\lt\varepsilon/3$ and $d(f_n(x),\lim_{x\to p}f_n(x))\lt\varepsilon/3$. Since $p$ is a limit point of $E$, $E\cap B_\delta(p)\setminus\{p\}$ is non-empty. Hence, $$d(\lim_{x\to p}f_m(x),\lim_{x\to p}f_n(x))\lt\varepsilon$$ implying $(\lim_{x\to p}f_n(x))$ is a Cauchy sequence, and by completeness of $Y$, $(\lim_{x\to p}f_n(x))$ converges.

Again, let $\varepsilon\gt0$. By uniform convergence of $(f_n)$ and convergence of $(\lim_{x\to p}f_n(x))$, there exists $N$ such that

• for all $x\in E$, $d(f_N(x),f(x))\lt\varepsilon/3$,
• and $d(\lim_{x\to p}f_N(x),\lim_{n\to\infty}\lim_{x\to p}f_n(x))\lt\varepsilon/3$.

Since $\lim_{x\to p}f_N(x)$ exists, there exists $\delta\gt0$ such that for all $x\in E\cap B_\delta(p)\setminus\{p\}$, $d(f_N(x),\lim_{x\to p}f_N(x))\lt\varepsilon/3$. Again, $E\cap B_\delta(p)\setminus\{p\}$ is non-empty. Thus, $$d(f(x),\lim_{n\to\infty}\lim_{x\to p}f_n(x))\lt\varepsilon$$ implying $$\lim_{x\to p}f(x)=\lim_{n\to\infty}\lim_{x\to p}f_n(x)$$ $\blacksquare$

Proof. Let $\varepsilon\gt0$. By convergence, and hence the Cauchy property, of $(f_n(c))$ and uniform convergence of $(f_n')$ on $[a,b]$, there exists $N$ such that for all $m,n\ge N$,

• $\abs{f_m(c)-f_n(c)}\lt\varepsilon/2$ and
• for all $x\in[a,b]$, $\abs{f_m'(x)-f_n'(x)}\lt\frac{\varepsilon}{2(b-a)}$.

By mean value theorem, for all $x,t\in[a,b]$, $$\abs{(f_m(x)-f_n(x))-(f_m(t)-f_n(t))}\le\frac{\abs{x-t}\varepsilon}{2(b-a)}\le\varepsilon/2$$ Hence, for all $x\in[a,b]$, $$\abs{f_m(x)-f_n(x)}\le\abs{(f_m(x)-f_n(x))-(f_m(c)-f_n(c))}+\abs{f_m(c)-f_n(c)}\lt\varepsilon$$ We have shown that $(f_n)$ converges uniformly, and hence converges, on $[a,b]$. Therefore, we can define $f:[a,b]\to R$ such that $$f(x)=\lim_{n\to\infty}(f_n(x))$$ Note that $(f_n)$ converges uniformly to $f$ on $[a,b]$. Now fix $p\in(a,b)$ and define $$\phi_n(h)=\frac{f_n(p+h)-f_n(p)}{h}$$ and $$\phi(h)=\frac{f(p+h)-f(p)}{h}$$ where $h$ ranges in $[a-p,b-p]\setminus\{0\}$ denoted $D$, so $p+h\in[a,b]$. Then we have $$\lim_{h\to0}\phi_n(h)=f'_n(p)$$ Keeping the definitions for $\varepsilon$ and $N$, For all $m,n\ge N$ and $h\in D$, using an inequality above, $$\abs{\phi_m(h)-\phi_n(h)}=\abs{\frac{(f_m(p+h)-f_m(p))-(f_n(p+h)-f_n(p))}{h}}\le\frac{\varepsilon}{2(b-a)}$$ This shows that $(\phi_n)$ converges uniformly on $D$. Now also fix $h\in D$, since $(f_n)$ converges to $f$ on $[a,b]$, $(f_n(p+h))$ converges to $f(p+h)$ and $(f_n(p))$ converges to $f(p)$. so $(\phi_n(h))=\p{\frac{f_n(p+h)-f_n(p)}{h}}$ converges to $\phi(h)=\frac{f(p+h)-f(p)}{h}$. Since $h\in D$ is arbitrary, $(\phi_n)$ converges, and hence uniformly converges, to $\phi$ on $D$. Using the proposition above on $(\phi_n)$ at $0$, we have $$f'(p)=\lim_{h\to0}\phi(h)=\lim_{n\to\infty}\lim_{h\to0}\phi_n(h)=\lim_{n\to\infty}f'_n(p)$$ Since $p\in(a,b)$ is arbitrary, for all $x\in(a,b)$, $$f'(x)=\lim_{n\to\infty}f'_n(x)$$ $\blacksquare$

Proof. If $z=0$, then $\abs{c_nz^n}^{1/n}=0$ for $n\in N^+$, so $\limsup_{n\to\infty}\abs{c_nz^n}^{1/n}=0$. Suppose $z\neq0$, then $$\limsup_{n\to\infty}\abs{c_nz^n}^{1/n} =\limsup_{n\to\infty}\abs{c_n}^{1/n}\abs{z^n}^{1/n} =\limsup_{n\to\infty}\abs{c_n}^{1/n}\abs{z} =\abs{z}\limsup_{n\to\infty}\abs{c_n}^{1/n} =s\abs{z}$$ Therefore, if $\abs{z}\lt r$, then $\limsup_{n\to\infty}\abs{c_nz^n}^{1/n}\lt1$, then by root test, $\sum\abs{c_nz^n}$ exists, hence $\sum c_nz^n$ exists, thus $z\in U$. If $\abs{z}\gt r$, then $\limsup_{n\to\infty}\abs{c_nz^n}^{1/n}\gt1$. Denote $\limsup_{n\to\infty}\abs{c_nz^n}^{1/n}$ by $c$, then there exists a subsequence $\abs{c_{n_k}z^{n_k}}^{1/{n_k}}\to c\gt1$ (or increasing and unbounded above if $c=\infty$). So there are infinitely many $k\neq0$ such that $\abs{c_{n_k}z^{n_k}}^{1/{n_k}}\gt1$, thus $\abs{c_{n_k}z^{n_k}}=\p{\abs{c_{n_k}z^{n_k}}^{1/{n_k}}}^{n_k}\gt1^{n_k}=1$. Hence $\abs{c_nz^n}$ does not converge to $0$, implying $c_nz^n$ does not converge to $0$, and $\sum c_nz^n$ does not exist, thus $z\notin U$. $\blacksquare$

Proof. For $n\in N^+$, define $c_n=\frac{1}{2^{n-1}}$, and define $c_0=1$, then $\abs{\frac{1}{n!}}=\frac{1}{n!}\le c_n=\abs{c_n}$ for all $n\in N$. Since $\sum_{n=0}^\infty\abs{c_n}=3$, by comparison test, $\sum_{n=0}^\infty\abs{\frac{1}{n!}}$ exists, hence $\sum_{n=0}^\infty\frac{1}{n!}$ exists. $\blacksquare$

Proof. The sequence $\sum_{k=0}^n\frac{1}{k!}$ is clearly increasing, so $e\ge\sum_{k=0}^1\frac{1}{k!}=2\gt1$. $\blacksquare$

Proof. Note that $$\p{1+\frac{1}{n}}^n =\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k} =\sum_{k=0}^n\frac{1}{k!}\p{\prod_{i=n-k+1}^ni}\frac{1}{n^k} =\sum_{k=0}^n\frac{1}{k!}\p{\prod_{i=0}^{k-1}(n-i)}\frac{1}{n^k} =\sum_{k=0}^n\frac{1}{k!}\prod_{i=0}^{k-1}\p{1-\frac{i}{n}} \le\sum_{k=0}^n\frac{1}{k!} \le e$$ for all $n\in N^+$. Let $\varepsilon\gt0$, then there exists $m\in N^+$ such that $d(\sum_{k=0}^m\frac{1}{k!},e)\lt\varepsilon/2$. For all $n\ge m$, we have $$\p{1+\frac{1}{n}}^n =\sum_{k=0}^n\frac{1}{k!}\prod_{i=0}^{k-1}\p{1-\frac{i}{n}} \ge\sum_{k=0}^m\frac{1}{k!}\prod_{i=0}^{k-1}\p{1-\frac{i}{n}} \ge\sum_{k=0}^m\frac{1}{k!}\p{1-\frac{m}{n}}^k$$ Clearly, there exists $M\ge m$ such that for all $n\ge M$, $$d\p{\sum_{k=0}^m\frac{1}{k!}\p{1-\frac{m}{n}}^k,\sum_{k=0}^m\frac{1}{k!}}\lt\varepsilon/2$$ Hence $$\p{1+\frac{1}{n}}^n \ge\sum_{k=0}^m\frac{1}{k!}\p{1-\frac{m}{n}}^k \gt\sum_{k=0}^m\frac{1}{k!}-\varepsilon/2$$ implying $$e\ge\p{1+\frac{1}{n}}^n\gt\sum_{k=0}^m\frac{1}{k!}-\varepsilon/2\gt e-\varepsilon$$ Therefore, $$\lim_{n\to\infty}\p{1+\frac{1}{n}}^n=e$$ $\blacksquare$

Proof. Fix $z\in C$ and let $a_n=\frac{z^n}{n!}$. If $z=0$, then $\sum_{n=0}^\infty a_n$ trivially exists. Now suppose $z\neq0$, then $(a_n)$ is a sequence of non-zero complex numbers. Since $$\abs{\frac{a_{n+1}}{a_n}}=\abs{\frac{\frac{z^{n+1}}{{n+1}!}}{\frac{z^n}{n!}}}=\abs{\frac{z}{n+1}}=\frac{\abs{z}}{n+1}$$ we have $$\limsup_{n\to\infty}\abs{\frac{a_{n+1}}{a_n}}=\lim_{n\to\infty}\frac{\abs{z}}{n+1}=0\lt1$$ By ratio test, $\sum_{n=0}^\infty\abs{a_n}$ exists, thus $\sum_{n=0}^\infty a_n$ exists. $\blacksquare$

Proof. Since $\sum\abs{a_n}$ exists, $\sum a_n$ exists. Let $\alpha=\sum\abs{a_n}$, $A=\sum a_n$, $B=\sum b_n$, $A_n=\sum_{k=0}^na_k$, $B_n=\sum_{k=0}^nb_k$, $C_n=\sum_{m=0}^n\sum_{k=0}^ma_kb_{m-k}$, $\beta_n=B_n-B$, $\gamma_n=\sum_{k=0}^na_k\beta_{n-k}$. Then $$C_n =\sum_{m=0}^n\sum_{k=0}^ma_kb_{m-k} =\sum_{k=0}^n\sum_{m=k}^na_kb_{m-k} =\sum_{k=0}^na_k\sum_{m=k}^nb_{m-k} =\sum_{k=0}^na_kB_{n-k} =\sum_{k=0}^na_k(B+\beta_{n-k}) =\sum_{k=0}^na_kB+\sum_{k=0}^na_k\beta_{n-k} =A_nB+\gamma_n$$ Clearly, $\lim_{n\to\infty}\beta_n=0$ Let $\varepsilon\gt0$, and let $\delta=\frac{\varepsilon}{\alpha+1}$, then $\delta\gt0$ and there exists $N$ such that for all $n\ge N$, $\abs{\beta_n}\lt\delta$, implying $$\abs{\gamma_n} =\abs{\sum_{k=0}^n\beta_ka_{n-k}} \le\abs{\sum_{k=0}^N\beta_ka_{n-k}}+\abs{\sum_{k=N+1}^n\beta_ka_{n-k}} \le\sum_{k=0}^N\abs{\beta_k}\abs{a_{n-k}}+\sum_{k=N+1}^n\abs{\beta_k}\abs{a_{n-k}} \le\sum_{k=0}^N\abs{\beta_k}\abs{a_{n-k}}+\delta\sum_{k=0}^{n-N-1}\abs{a_k} \le\sum_{k=0}^N\abs{\beta_k}\abs{a_{n-k}}+\delta\alpha$$ Since $\sum\abs{a_n}$ exists, $\lim_{n\to\infty}\abs{a_n}=0$, implying there exists $M\ge N$ such that for all $k\ge M$, $$\abs{a_k}\lt\frac{\delta}{\sum_{l=0}^N\abs{\beta_l}+1}$$ Then for all $m\ge M+N$, $$\abs{\gamma_m} \le\sum_{k=0}^N\abs{\beta_k}\abs{a_{m-k}}+\delta\alpha \le\frac{\delta}{\sum_{l=0}^N\abs{\beta_l}+1}\sum_{k=0}^N\abs{\beta_k}+\delta\alpha \lt\delta(1+\alpha)=\varepsilon$$ Thus $\lim_{n\to\infty}\gamma_n=0$. And we conclude that $$\sum\sum_{k=0}^na_kb_{n-k}=\lim_{n\to\infty}C_n=B\lim_{n\to\infty}A_n+\lim_{n\to\infty}\gamma_n=AB=\sum a_n\sum b_n$$ $\blacksquare$

Proof. $$\exp(z)\exp(w) =\sum_{n=0}^\infty\frac{z^n}{n!}\sum_{n=0}^\infty\frac{w^n}{n!} =\sum_{n=0}^\infty\sum_{k=0}^n\frac{z^kw^{n-k}}{k!(n-k)!} =\sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^n\frac{n!}{k!(n-k)!}z^kw^{n-k} =\sum_{n=0}^\infty\frac{(z+w)^n}{n!} =\exp(z+w)$$ $\blacksquare$

Proof. $$\exp(nz)=\exp(\sum_{k=1}^nz)=\prod_{k=1}^n\exp(z)=\exp(z)^n$$ $\blacksquare$

Proof. Let $z\in C$, then $$\exp(z)\exp(-z)=\exp(0)=1$$ Hence $\exp(z)\neq0$. Then we have $$\exp(-z)=\frac{1}{\exp(z)}\exp(z)\exp(-z)=\frac{1}{\exp(z)}$$ $\blacksquare$

Proof. Note that $$\sum_{k=0}^n\frac{{\overline z}^k}{k!} =\sum_{k=0}^n\frac{\overline{z^k}}{k!} =\sum_{k=0}^n\overline{\frac{z^k}{k!}} =\overline{\sum_{k=0}^n\frac{z^k}{k!}}$$ Let $\varepsilon\gt0$, there exists $m\in N$ such that for all $n\ge m$, we have $$\abs{\sum_{k=0}^n\frac{z^k}{k!}-\exp(z)}\lt\varepsilon$$ Hence $$\abs{\sum_{k=0}^n\frac{{\overline z}^k}{k!}-\overline{\exp(z)}} =\abs{\overline{\sum_{k=0}^n\frac{z^k}{k!}}-\overline{\exp(z)}} =\abs{\overline{\sum_{k=0}^n\frac{z^k}{k!}-\exp(z)}} =\abs{\sum_{k=0}^n\frac{z^k}{k!}-\exp(z)} \lt\varepsilon$$ Therefore, $$\exp(\overline z)=\sum_{n=0}^\infty\frac{{\overline z}^n}{n!}=\overline{\exp(z)}$$ $\blacksquare$

Proof. Let $x\in R$, then $\sum_{k=0}^n\frac{x^k}{k!}\in R$ for all $n\in N$. If $\exp(x)=\sum_{k=0}^n\frac{x^k}{k!}$ for some $n\in N$, then $\exp(x)\in R$. Now suppose $\exp(x)\neq\sum_{k=0}^n\frac{x^k}{k!}$ for all $n\in N$, then for all $r\gt0$, there exists $n\in N$ such that $d(\exp(x),\sum_{k=0}^n\frac{x^k}{k!})\lt r$, and we have $\sum_{k=0}^n\frac{x^k}{k!}\in R$ and $\sum_{k=0}^n\frac{x^k}{k!}\neq\exp(x)$, hence $\exp(x)$ is a limit point of $R$. Since $R$, as a subset of $C$, is clearly closed, we conclude that $\exp(x)\in R$.

If $x\gt0$, then $\sum_{k=0}^n\frac{x^k}{k!}$ is clearly increasing, so $\exp(x)\ge1\gt0$. If $x\lt0$, then $\exp(-x)\gt0$, so $\exp(x)=\frac{1}{\exp(-x)}\gt0$. If $x=0$, then we have $\exp(x)=1\gt0$. Hence for all $x\in R$, $\exp(x)\gt0$. $\blacksquare$

Proof. $$\exp(1)=\lim_{n\to\infty}\sum_{k=0}^n\frac{1^k}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}=e$$ If $n\in N$, then $$\exp(n)=\exp(1)^n=e^n$$ If $x\in Q^+$, then there exist $m,n\in N^+$ such that $x=\frac{m}{n}$. Note that $$\exp(x)^n=\exp(\frac{m}{n})^n=\exp(m)=e^m$$ combining with $\exp(x)\in R^+$, we conclude that $$\exp(x)=(e^m)^{1/n}=e^x$$ If $x\in Q^-$, then $$\exp(x)=\frac{1}{\exp(-x)}=\frac{1}{e^{-x}}=e^x$$ If $x=0$, then $$\exp(x)=1=e^x$$ $\blacksquare$

Proof. If $r\gt0$, then $\sum_{k=0}^n\frac{r^k}{k!}$ is clearly increasing, so $e^r=\exp(r)\ge(1+r)\gt1$. Let $x\lt y$, then $\frac{e^y}{e^x}=e^ye^{-x}=e^{y-x}\gt1$. Since $e^x\gt0$, we conclude that $e^x\lt e^y$. $\blacksquare$

Proof. Let $\varepsilon\gt0$, then for all $0\lt h\lt\inf(1,\frac{\varepsilon}{e-1})$, $$0 \lt e^h-1 =\lim_{n\to\infty}\sum_{k=0}^n\frac{h^k}{k!}-1 =\lim_{n\to\infty}\sum_{k=1}^n\frac{h^k}{k!} \le\lim_{n\to\infty}\sum_{k=1}^n\frac{h}{k!} =h\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k!} =h\p{\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}-1} =h(e-1) \lt\varepsilon$$ and for all $0\gt h\gt-\inf(1,\frac{\varepsilon}{e-1})$, $$0 \lt 1-e^h =e^h(e^{-h}-1) \lt e^{-h}-1 \lt\varepsilon$$ Therefore, $$\lim_{h\to0}e^h=1$$ $\blacksquare$

Proof. Let $h\neq0$, then $$\frac{e^h-1}{h} =\frac{\lim_{n\to\infty}\sum_{k=1}^n\frac{h^k}{k!}}{h} =\lim_{n\to\infty}\sum_{k=1}^n\frac{h^{k-1}}{k!}$$ Suppose $h\gt0$ and $n\gt0$, then $$1\le\sum_{k=1}^n\frac{h^{k-1}}{k!}\le\sum_{k=1}^n\frac{h^{k-1}}{(k-1)!}\le\sum_{k=0}^{n}\frac{h^k}{k!}$$ Hence $$1\le\frac{e^h-1}{h}\le e^h$$ implying $$\lim_{h\to0^+}\frac{e^h-1}{h}=1$$ Suppose $h\lt0$, since $$\frac{e^h-1}{h}=\frac{1-e^h}{-h}=e^h\frac{e^{-h}-1}{-h}$$ we have $$e^h\le\frac{e^h-1}{h}\le e^he^{-h}=1$$ implying $$\lim_{h\to0^-}\frac{e^h-1}{h}=1$$ Therefore, $$\lim_{h\to0}\frac{e^h-1}{h}=1$$ $\blacksquare$

Proof. Let $p\in R$, then $$\lim_{h\to0}\frac{e^{p+h}-e^p}{h} =\lim_{h\to0}\frac{e^pe^h-e^p}{h} =e^p\lim_{h\to0}\frac{e^h-1}{h} =e^p$$ Hence $e^x$ is differentiable and $\dv{x}e^x=e^x$. $\blacksquare$

Proof. We have already shown that $e^x$ is a function from $R$ to $R^+$. Since $e^x$ is strictly increasing, it is injective. Now let $y\in R^+$. If $y\ge1$, let $n$ be an integer greater than $y$; if $y\lt1$, let $n$ be an integer greater than $1/y$. Then we have $1/n\lt y\lt n$ where $n\ge2$. Note that $e\ge2$. Hence $e^n\ge2^n\gt n\gt y$ and $e^{-n}=\frac{1}{e^n}\lt 1/n\lt y$. Note that $e^x$ is differentiable and hence continuous. By intermediate value theorem, there exists $r\in[-n,n]$ such that $e^r=y$. Hence $e^x$ is surjective. $\blacksquare$

Proof. There exist $u,v\in R$ such that $e^u=x$ and $e^v=y$. Hence $$\log(xy)=\log(e^ue^v)=\log(e^{u+v})=u+v=\log(x)+\log(y)$$
Since $$0=\log(1)=\log(x)+\log(\frac{1}{x})$$ we have $$\log(\frac{1}{x})=-\log(x)$$ $\blacksquare$

Proof. Fix $p\in R^+$. Let $q$ denote $\log(p)$. By inverse function theorem, there exists an open subset $U\subseteq R$ with $q\in U$ such that, let $V$ denote the image of $e^x$ on $U$, then $V$ is open and $e^x|_U:U\to V$ is a bijection; also, $(e^x|_U)^{-1}$ is smooth with $$\dv{x}(e^x|_U)^{-1}(v)=\frac{1}{\dv{e^x}{x}((e^x|_U)^{-1}(v))}$$ for all $v\in V$. Note that, for all $v\in V$, there exists $u\in U$ such that $e^u=e^x|_U(u)=v$, hence $\log(v)=\log(e^u)=u=(e^x|_U)^{-1}(e^x|_U(u))=(e^x|_U)^{-1}(v)$, implying $(e^x|_U)^{-1}$ is a restriction of $\log$ on $V$. Since $p=e^q\in V$, $\log$ is differentiable at $p$ and $$\dv{x}\log(p)=\frac{1}{\dv{e^x}{x}(\log(p))}=\frac{1}{e^{\log(p)}}=\frac{1}{p}$$ Therefore, $$\dv{x}\log(x)=\frac{1}{x}$$ $\blacksquare$

Proof. Define $F(x):R^+\to R$ by $F(x)=\int_1^x\frac{1}{t}dt$, then $F(x)$ is an antiderivative of $\frac{1}{x}$. Since $\dv{x}\log(x)=\frac{1}{x}$, $\log(x)$ is also an antiderivative of $\frac{1}{x}$. Hence $\log(x)=F(x)+C$ for some constant function $C$. Note that $\log(1)=0=F(1)$, implying the constant function is $0$. Thus, $$\log(x)=F(x)=\int_1^x\frac{1}{t}dt$$ $\blacksquare$

Proof. If $x\in Q^+$, then there exist $m,n\in N^+$ such that $x=\frac{m}{n}$. Hence $$(e^{x\log(r)})^n =(e^{\frac{m}{n}\log(r)})^n =e^{m\log(r)} =(e^{\log(r)})^m =r^m$$ implying $$e^{x\log(r)}=(r^m)^{1/n}=r^x$$ If $x\in Q^-$, then $$e^{x\log(r)}=\frac{1}{e^{-x\log(r)}}=\frac{1}{r^{-x}}=r^x$$ If $x=0$, then $$e^{x\log(r)}=1=r^x$$ $\blacksquare$

Proof. $$rs^x=e^{x\log(rs)}=e^{x(\log(r)+\log(s))}=e^{x\log(r)+x\log(s)}=e^{x\log(r)}+e^{x\log(s)}=r^xs^x$$ $\blacksquare$

Proof. $$r^{x+y}=e^{(x+y)\log(r)}=e^{x\log(r)+y\log(r)}=e^{x\log(r)}e^{y\log(r)}=r^xr^y$$ $\blacksquare$

If $r\in R^+\setminus Z$, since $x^r|_+\to0$ as $x\to0$, we have $x^r\to0$ as $x\to0$, implying $x^r$ is continuous at $0$.

If $r\in Z^-$, let $x\neq0$, by induction, $x^r=(-x)^r$ if $r$ is even and $x^r=-(-x)^r$ if $r$ is odd. Denote the restriction of $x^r$ on $R^-$ by $x^r|_-$.

• When $r$ is even, $x^r|_-$ is strictly increasing and bijective from $R^-$ to $R^+$, implying $x^r|_-\to0$ as $x\to-\infty$ and $x^r|_-\to\infty$ as $x\to0$, and thus $x^r\to\infty$ as $x\to0$.
• When $r$ is odd, $x^r|_-$ is strictly decreasing and bijective from $R^-$ to $R^-$, implying $x^r|_-\to0$ as $x\to-\infty$ and $x^r|_-\to-\infty$ as $x\to0$, and thus $x^r$ has no limit at $0$.

Also, $x^r|_-$ is clearly differentiable with $(x^r|_-)'(t)=-(x^r|_+)'(-t)$ if $r$ is even and $(x^r|_-)'(t)=(x^r|_+)'(-t)$ if $r$ is odd.

If $r\in N^+$, similarly, $x^r=(-x)^r$ if $r$ is even and $x^r=-(-x)^r$ if $r$ is odd.

• When $r$ is even, $x^r|_-$ is strictly decreasing and bijective from $R^-$ to $R^+$, implying $x^r|_-\to\infty$ as $x\to-\infty$ and $x^r|_-\to0$ as $x\to0$, and thus $x^r\to0$ as $x\to0$.
• When $r$ is odd, $x^r|_-$ is strictly increasing and bijective from $R^-$ to $R^-$, implying $x^r|_-\to-\infty$ as $x\to-\infty$ and $x^r|_-\to0$ as $x\to0$, and thus $x^r\to0$ as $x\to0$.

Also, $x^r|_-$ is clearly differentiable with $(x^r|_-)'(t)=-(x^r|_+)'(-t)$ if $r$ is even and $(x^r|_-)'(t)=(x^r|_+)'(-t)$ if $r$ is odd. In addition, $x^r$ is clearly differentiable at $0$ with $(x^r)'(0)=\lim_{h\to0}h^{r-1}$, which is $1$ if $r=1$ and $0$ otherwise.

If $r=0$, then $x^r$ is a constant function at $1$. $\blacksquare$

Proof. By chain rule, $$\dv{x}x^r =\dv{x}e^{r\log(x)} =e^{r\log(x)}\frac{r}{x} =x^r\frac{r}{x} =rx^{r-1}$$ $\blacksquare$

Proof. Let $r\in N^+$. If $r$ is even, then $(x^r|_-)'(t)=-(x^r|_+)'(-t)=-(rx^{r-1})(-t)=-(r(-t)^{r-1})=rt^{r-1}$; if $r$ is odd, then $(x^r|_-)'(t)=(x^r|_+)'(-t)=(rx^{r-1})(-t)=(r(-t)^{r-1})=rt^{r-1}$. Combining with $(x^r)'(0)=1=(rx^{r-1})(0)$ if $r=1$ and $(x^r)'(0)=0=(rx^{r-1})(0)$ otherwise, We have $(x^r)'=rx^{r-1}$, where $x$ ranges in $R$.

Let $r\in Z^-$. With the same reasoning, we have $(x^r)'=rx^{r-1}$, where $x$ ranges in $R\setminus\{0\}$.

If $r=0$, then $(x^r)'=(1)'=0$, which is not exactly $rx^{r-1}$, because $rx^{r-1}=\frac{0}{x}$ is not defined at $0$. $\blacksquare$

Proof. Suppose $y\gt0$. Using integration by substitution with $u:R^+\to R$ such that $u(t)=t^y$, we have $$\log(x^y) =\int_1^{x^y}\frac{1}{t}dt =\int_1^x\frac{yt^{y-1}}{t^y}dt =y\int_1^x\frac{1}{t}dt =y\log(x)$$ If $y\lt0$, then $$\log(x^y)=\log(e^{y\log(x)})=\log(\frac{1}{e^{-y\log(x)}})=-\log(e^{-y\log(x)})=-\log(x^{-y})=y\log(x)$$ If $y=0$, then $$\log(x^y)=0=y\log(x)$$ $\blacksquare$

Proof. $$r^{xy}=e^{xy\log(r)}=e^{y\log(r^x)}=(r^x)^y$$ $\blacksquare$

Proof. Fix $r\in R^+\setminus\{1\}$, then $r^x$ is bijective from $R$ to $R^+$ and hence has an inverse from $R^+$ to $R$, denoted $g$. Let $y\in R^+$, denote $g(y)$ by $x$, then $$\log_r(y) =\frac{\log(y)}{\log(r)} =\frac{\log(r^x)}{\log(r)} =\frac{x\log(r)}{\log(r)} =x =g(y)$$ Hence $\log_r=g$. $\blacksquare$

Proof. $$\log_r(xy)=\frac{\log(xy)}{\log(r)}=\frac{\log(x)+\log(y)}{\log(r)}=\log_r(x)+\log_r(y)$$ $$\log_r(x^y)=\frac{\log(x^y)}{\log(r)}=\frac{y\log(x)}{\log(r)}=y\log_r(x)$$ $\blacksquare$

Proof. $$\lim_{x\to\infty}\frac{\log(x)}{x} =\lim_{x\to\infty}\frac{(\log(x))'}{(x)'} =\lim_{x\to\infty}\frac{1}{x} =0$$ $$\lim_{n\to\infty}n^{1/n} =\lim_{n\to\infty}\exp(\frac{\log(n)}{n}) =\lim_{x\to\infty}\exp(\frac{\log(x)}{x}) =\exp(\lim_{x\to\infty}\frac{\log(x)}{x}) =\exp(0) =1$$ $\blacksquare$

Proof. Suppose $a_{n_k}b_{n_k}\to c$, since $(1/a_{n_k})(a_{n_k}b_{n_k})\to(1/a)c$, so $b_{n_k}\to c/a$, implying $c/a\le b$, or $c\le ab$. This shows that $ab$ is an upper bound of the collection of subsequential limits of $a_nb_n$.

Note that there exists a subsequence $b_{n_k}\to b$, since $a_{n_k}\to a$, we have $a_{n_k}b_{n_k}\to ab$. This shows that $ab$ is the supremum of the collection of subsequential limits of $a_nb_n$, which is $\limsup a_nb_n$. $\blacksquare$

Proof. Let $r$ be the radius of convergence. If $r=0$, this is trivial, so suppose $r\neq0$. Fix $p\in(-r,r)$. Let $s\in(0,r)$ such that $p\in(-s,s)$. For all $x\in[-s,s]$, $\abs{c_nx^n}\le\abs{c_ns^n}$. Since $s$ is in the interval of convergence, root test shows that $\sum\abs{c_ns^n}$ exists (refer to the proof of radius of convergence). So $(\sum_{k=0}^n\abs{c_ks^k})$ has the Cauchy property. Let $\varepsilon\gt0$, there exists $N$ such that for all $m,n\gt N$ and $x\in[-s,s]$, if $m\ge n$, then $$\abs{\sum_{k=0}^m c_kx^k-\sum_{k=0}^n c_kx^k} =\abs{\sum_{k=n+1}^m c_kx^k} \le\sum_{k=n+1}^m\abs{c_kx^k} \le\sum_{k=n+1}^m\abs{c_ks^k} =\abs{\sum_{k=0}^m\abs{c_ks^k}-\sum_{k=0}^n\abs{c_ks^k}} \lt\varepsilon$$ Hence, $(\sum_{k=0}^n c_kx^k)$ converges uniformly to $f$ on $[-s,s]$.

Note that, since we assumed $r\neq0$, $$\limsup_{n\to\infty}\abs{nc_n}^{1/n} =\limsup_{n\to\infty}\abs{n}^{1/n}\abs{c_n}^{1/n} =\limsup_{n\to\infty}\abs{c_n}^{1/n} \in\{0\}\cup R^+$$ thus the radius of convergence for $\sum_{n=0}^\infty nc_nx^n$ is also $r$. So for all $x\in(-r,r)\setminus\{0\}$, $$\sum_{n=1}^\infty nc_nx^{n-1}=(1/x)\sum_{n=1}^\infty nc_nx^n=(1/x)\sum_{n=0}^\infty nc_nx^n$$, implying the radius of convergence for $\sum_{n=1}^\infty nc_nx^{n-1}$ is no less than $r$. Hence for the same reason as above, $(\sum_{k=1}^n kc_kx^{k-1})$ converges uniformly on $[-s,s]$. Note that $$\dv{x}\sum_{k=0}^n c_kx^k=\sum_{k=1}^n kc_kx^{k-1}$$ Applying a proposition above, we have that $$f'(p)=\sum_{n=1}^\infty nc_np^{n-1}$$ Since $p\in(-r,r)$ is arbitrary, for all $x$ in $(-r,r)$, $$f'(x)=\sum_{n=1}^\infty nc_nx^{n-1}$$ $\blacksquare$

Proof. Fix $x\in R$ and let $a_n=\frac{(-1)^n}{(2n)!}x^{2n}$. If $x=0$, then $\sum_{n=0}^\infty a_n$ trivially exists. Now suppose $x\neq0$, then $(a_n)$ is a sequence of non-zero real numbers. Since $$\abs{\frac{a_{n+1}}{a_n}} =\abs{\frac{\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}}{\frac{(-1)^n}{(2n)!}x^{2n}}} =\abs{\frac{-x^2}{(2n+2)(2n+1)}} =\frac{x^2}{(2n+2)(2n+1)}$$ we have $$\limsup_{n\to\infty}\abs{\frac{a_{n+1}}{a_n}}=\lim_{n\to\infty}\frac{x^2}{(2n+2)(2n+1)}=0\lt1$$ By ratio test, $\sum_{n=0}^\infty\abs{a_n}$ exists, thus $\sum_{n=0}^\infty a_n$ exists.

Fix $x\in R$ and let $b_n=\frac{(-1)^n}{(2n+1)!}x^{2n+1}$. If $x=0$, then $\sum_{n=0}^\infty b_n$ trivially exists. Now suppose $x\neq0$, then $(b_n)$ is a sequence of non-zero real numbers. Since $$\abs{\frac{b_{n+1}}{b_n}} =\abs{\frac{\frac{(-1)^{n+1}}{(2n+3)!}x^{2n+3}}{\frac{(-1)^n}{(2n+1)!}x^{2n+1}}} =\abs{\frac{-x^2}{(2n+3)(2n+2)}} =\frac{x^2}{(2n+3)(2n+2)}$$ we have $$\limsup_{n\to\infty}\abs{\frac{b_{n+1}}{b_n}}=\lim_{n\to\infty}\frac{x^2}{(2n+3)(2n+2)}=0\lt1$$ By ratio test, $\sum_{n=0}^\infty\abs{b_n}$ exists, thus $\sum_{n=0}^\infty b_n$ exists. $\blacksquare$

Proof. Since $\sin$ and $\cos$ exist on $R$, their intervals of convergence are both $R$. So for all $p\in R$, $$(\cos)'(p) =\sum_{n=1}^\infty\frac{(-1)^n}{(2n-1)!}p^{2n-1} =\sum_{n=0}^\infty\frac{(-1)^{n+1}}{(2n+1)!}p^{2n+1} =-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}p^{2n+1} =-\sin(p)$$ $$(\sin)'(p) =\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}p^{2n} =\cos(p)$$ $\blacksquare$

Proof. $$\cos(x)+i\sin(x) =\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}+i\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1} =\sum_{n=0}^\infty\p{\frac{(-1)^n}{(2n)!}x^{2n}+i\frac{(-1)^n}{(2n+1)!}x^{2n+1}} =\sum_{n=0}^\infty\p{\frac{(ix)^{2n}}{(2n)!}+\frac{(ix)^{2n+1}}{(2n+1)!}} =\sum_{n=0}^\infty\frac{(ix)^n}{n!} =\exp(ix)$$ $$\exp(-ix)=\exp(\overline{ix})=\overline{\exp(ix)}$$ $$\frac{1}{2}(\exp(ix)+\exp(-ix)) =\frac{1}{2}(\exp(ix)+\overline{\exp(ix)}) =\Re(\exp(ix)) =\cos(x)$$ $$\frac{1}{2i}(\exp(ix)-\exp(-ix)) =\frac{1}{2i}(\exp(ix)-\overline{\exp(ix)}) =\Im(\exp(ix)) =\sin(x)$$ $\blacksquare$

Proof. Let $x\in R$, then $$\cos(x)^2+\sin(x)^2=\abs{\exp(ix)}^2=\Re\p{\exp(ix)\overline{\exp(ix)}}=\Re\p{\exp(ix)\exp(\overline{ix})} =\Re\p{\exp(ix)\exp(-ix)}=\Re\p{\exp(0)}=1$$ $\blacksquare$

Proof. Suppose $\abs{\cos(x)}\gt1$, then $\cos(x)^2+\sin(x)^2\ge\cos(x)^2=\abs{\cos(x)}^2\gt1$, a contradiction. Hence $\abs{\cos(x)}\le1$. By a symmetric argument, $\abs{\sin(x)}\le1$. $\blacksquare$

Proof. Suppose for contradiction that for all $p\in R^+$, $\cos(p)\neq0$. Note that $\cos(0)=1$ by the defining power series, and $\cos$ is differentiable and hence continuous. So by intermediate value theorem, for all $p\in R^+\cup\{0\}$, $\cos(p)\gt0$. Thus for all $x,y\in R^+\cup\{0\}$ with $x\lt y$, we have $$\sin(y)-\sin(x)=\int_x^y\cos(u)du\gt0$$ implying $\sin$ is strictly increasing in $R^+\cup\{0\}$. So for all $x\in R^+$, $\sin(x)\gt\sin(0)=0$ by the defining power series. Hence let $a=1$ and $b=2(1+1/\sin(1))$, then $a\lt b$ and $$2\lt\sin(a)(b-a)\lt\int_a^b\sin(u)du=(-\cos(b))-(-\cos(a))\le2$$ which is a contradiction. Therefore, there exists $p\in R^+$ such that $\cos(p)=0$. $\blacksquare$

Proof. Since $\{p\in R^+|\cos(p)=0\}$ is non-empty and bounded below, $c=\inf\{p\in R^+|\cos(p)=0\}$ is its infimum in $R$, which implies $\pi=2c\in R$. Now suppose for contradiction that $c\notin\{p\in R^+|\cos(p)=0\}$, then either $c\notin R^+$ or $\cos(c)\neq0$.

• If $c\notin R^+$, then either $c\lt0$ or $c=0$. If $c\lt0$, then $c$ cannot be the infimum of $\{p\in R^+|\cos(p)=0\}$, a contradiction. If $c=0$, then $\cos(c)=1\neq0$, which we will discuss below.
• If $\cos(c)\neq0$, then by continuity of $\cos$, there exists $\delta\gt0$ such that for all $x\in(c-\delta,c+\delta)$, $\cos(x)\neq0$. But since $c$ is the infimum of $\{p\in R^+|\cos(p)=0\}$, there exists $u\in\{p\in R^+|\cos(p)=0\}$ such that $u\in[c,c+\delta)$. Note that $u\in(c-\delta,c+\delta)$ and $\cos(u)=0$, a contradiction.

Therefore, $c\in\{p\in R^+|\cos(p)=0\}$. Hence $$\cos(\pi/2)=\cos(c)=0$$ Since $c$ is the infimum of $\{p\in R^+|\cos(p)=0\}$, $c\ge0$; since $\cos(c)=0\neq1=\cos(0)$, $c\neq0$, and hence $c\gt0$. Thus $$\pi=2c\in R^+$$

Proof. Since $\cos(\pi/2)=0$, $\sin(\pi/2)^2=1-\cos(\pi/2)^2=1$, implying $\sin(\pi/2)=\pm1$. Hence $$\exp(i\pi)=\exp(i\pi/2)^2=(\cos(\pi/2)+i\sin(\pi/2))^2=-1$$ $\blacksquare$

Proof. $$\exp(z+2i\pi)=\exp(z)\exp(i\pi)^2=\exp(z)$$ $$\cos(x+2\pi) =\frac{1}{2}(\exp(ix+2i\pi)+\exp(-ix-2i\pi)) =\frac{1}{2}(\exp(ix)+\exp(-ix)) =\cos(x)$$ $$\sin(x+2\pi) =\frac{1}{2i}(\exp(ix+2i\pi)-\exp(-ix-2i\pi)) =\frac{1}{2i}(\exp(ix)-\exp(-ix)) =\sin(x)$$ $\blacksquare$

Proof. This is obvious from the defining power series. $\blacksquare$

Proof. $$\cos(x+y)+i\sin(x+y)=\exp(ix+iy)=\exp(ix)\exp(iy)=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y)) =(\cos(x)\cos(y)-\sin(x)\sin(y))+i(\cos(x)\sin(y)+\sin(x)\cos(y))$$ $\blacksquare$

Proof. $$\sin(x+\pi/2)=\cos(x)\sin(\pi/2)+\sin(x)\cos(\pi/2)=\cos(x)\sin(\pi/2)$$ We know $\sin(\pi/2)=\pm1$, $\cos(0)=1$, $\cos(\pi/2)=0$, and $\cos(x)\gt0$ for $x\in(0,\pi/2)$. Since $\sin'=\cos$, $\sin$ is strictly increasing in $[0,\pi/2]$, implying $\sin(\pi/2)\gt\sin(0)=0$. Hence $\sin(\pi/2)=1$ and $$\sin(x+\pi/2)=\cos(x)\sin(\pi/2)+\sin(x)\cos(\pi/2)=\cos(x)$$ $\blacksquare$

Proof. We know $\cos(0)=1$, $\cos(\pi/2)=0$, and $\cos(x)\gt0$ for $x\in(0,\pi/2)$. Since $\cos(-x)=\cos(x)$, we know $\cos(x)\gt0$ for $x\in(-\pi/2,\pi/2)$ and $\cos(-\pi/2)=\cos(\pi/2)=0$. Since $\sin'=\cos$, $\sin$ is strictly increasing in $[-\pi/2,\pi/2]$. Hence $\cos$ is strictly increasing in $[-\pi,0]$, and hence strictly decreasing in $[0,\pi]$, thus $\sin$ is strictly decreasing in $[\pi/2,3\pi/2]$. Since $\sin$ and $\cos$ are periodic with period $2\pi$, they are strictly increasing or decreasing in corresponding intervals in other periods. Note that $\cos(\pi)=\cos(\pi/2)^2-\sin(\pi/2)^2=-1$. Hence $\cos((2k+1)\pi)=-1$, and thus $\sin((2k+3/2)\pi)=-1$, for all $k\in Z$. Also, $\cos(2k\pi)=1$, and thus $\sin((2k+1/2)\pi)=1$, for all $k\in Z$.

We know $\cos(\pi/2)=0$, so $\cos(-\pi/2)=0$. Since $\cos$ is strictly increasing in $[-\pi,0]$, $\cos(x)\neq0$ for all $x\in[-\pi,0]\setminus\{-\pi/2\}$; Since $\cos$ is strictly decreasing in $[0,\pi]$, $\cos(x)\neq0$ for all $x\in[0,\pi]\setminus\{\pi/2\}$. We have shown that $\cos(x)=0$ if and only if $x=(k+1/2)\pi$ for some $k\in Z$, hence $\sin(x)=0$ if and only if $x=k\pi$ for some $k\in Z$. $\blacksquare$

Proof. Let $p$ in domain of $\tan$. $$\tan'(p) =\cos(p)\cos(p)^{-1}+\sin(p)(-1)\cos(p)^{-2}(-\sin(p)) =\frac{\cos(p)\cos(p)+\sin(p)\sin(p)}{\cos(p)^2} =\cos(p)^{-2}$$ $\blacksquare$

Proof. Let $p$ in domain of $\tan$. $$\tan(p+\pi)=\frac{\sin(p+\pi)}{\cos(p+\pi)} =\frac{\cos(p)\sin(\pi)+\sin(p)\cos(\pi)}{\cos(p)\cos(\pi)-\sin(p)\sin(\pi)} =\frac{-\sin(p)}{-\cos(p)}=\tan(p)$$ $\blacksquare$

Proof. Since $\sin$ is increasing and positive and $\cos$ is decreasing and positive in $(0,\pi/2)$, $\tan$ is increasing and positive in $(0,\pi/2)$. Clearly, $\tan(-x)=-\tan(x)$ for all $x$ in domain, so $\tan$ is increasing and negative in $(-\pi/2,0)$. Since $\tan(0)=0$, $\tan$ is increasing in $(-\pi/2,\pi/2)$.

Let $r\in R^+$, then there exists $\delta\in(0,\pi/2)$ such that for all $x\in(\pi/2-\delta,\pi/2)$, $\sin(x)\in(1/2,1)$ and $\cos(x)\in(0,1/(2r))$, hence $\tan(x)\gt\frac{1/2}{1/(2r)}=r$. This implies for all $y\in R$, there exist $x\in(0,\pi/2)$ such that $\tan(-x)\lt y\lt\tan(x)$, and by intermediate value theorem and differentiability of $\tan$, there exists $x^*\in(-\pi/2,\pi/2)$ such that $\tan(x^*)=y$. Hence, the image of $\tan$ on $(-\pi/2,\pi/2)$ is $R$.

Since $\tan$ is periodic with period $\pi$, $\tan$ is increasing from $((k-1/2)\pi,(k+1/2)\pi)$ onto $R$ for all $k\in Z$.

From what we have shown above, it is clear that $\tan(x)=0$ if and only if $x=k\pi$ for some $k\in Z$. $\blacksquare$

Proof. Clearly, $\phi(x,y)\in R^+\times(-\pi,\pi]$, hence $\phi$ can be viewed as a function from $R^2\setminus{(0,0)}$ to $R^+\times(-\pi,\pi]$. Define a function $\psi:R^+\times(-\pi,\pi]\to R^2$ by $$\psi(r,\theta)=(r\cos(\theta),r\sin(\theta))$$ Then $$\norm{\psi(r,\theta)}=\sqrt{(r\cos(\theta))^2+(r\sin(\theta))^2}=\abs{r}\gt0$$ implying $\psi(r,\theta)\neq(0,0)$. Hence $\psi$ can be viewed as a function from $R^+\times(-\pi,\pi]$ to $R^2\setminus{(0,0)}$.

Clearly, $$\phi(\psi(r,\theta))=\phi(r\cos(\theta),r\sin(\theta))=(\norm{(r\cos(\theta),r\sin(\theta))},\atan2(r\cos(\theta),r\sin(\theta)))=(r,\theta)$$ Note that for all $\theta$ in domain of $\tan$, $$\cos(\theta)^2=\frac{1}{\tan(\theta)^2+1}$$ and $$\sin(\theta)^2=\frac{\tan(\theta)^2}{\tan(\theta)^2+1}$$ Also note that $\cos(\theta)\gt0$ for all $\theta\in(-\pi/2,\pi/2)$. Hence given $x\neq0$, we have $$\cos(\arctan(y/x))=\frac{1}{\sqrt{\tan(\arctan(y/x))^2+1}}=\frac{\abs{x}}{\sqrt{y^2+x^2}}$$ and $$\sin(\arctan(y/x))=\pm\frac{\abs{\tan(\arctan(y/x))}}{\sqrt{\tan(\arctan(y/x))^2+1}}=\pm\frac{\abs{y}}{\sqrt{y^2+x^2}}$$ where the sign is the same as $y/x$. Then clearly, $$\psi(\phi(x,y))=\psi(\norm{(x,y)},\atan2(x,y))=(\norm{(x,y)}\cos(\atan2(x,y)),\norm{(x,y)}\sin(\atan2(x,y)))=(x,y)$$ Therefore, $\phi$ is a bijection from $R^2\setminus{(0,0)}$ to $R^+\times(-\pi,\pi]$ with $\psi$ being its inverse from $R^+\times(-\pi,\pi]$ to $R^2\setminus{(0,0)}$. $\blacksquare$

Proof. Note that $(a,b)$ is the Cartesian form of $(r,\theta)$. Hence $$r\exp(i\theta)=r\cos(\theta)+ir\sin(\theta)=a+ib=z$$ $\blacksquare$

Proof. By fundamental theorem of calculus, $$f(x)=f(p)+\int_p^xf'(t)dt$$ Let $m\in\{1,\ldots,k\}$ and suppose $$f(x)=\sum_{j=0}^{m-1}\frac{f^{(j)}(p)}{j!}(x-p)^j+\int_p^x\frac{f^{(m)}(t)}{(m-1)!}(x-t)^{m-1}dt$$ Using integration by part with $F(t)=f^{(m)}(t)$ and $G(t)=-\frac{(x-t)^m}{m!}$, and thus $F'(t)=f^{(m+1)}(t)$ and $G'(t)=\frac{(x-t)^{m-1}}{(m-1)!}$, we have $$f(x) =\sum_{j=0}^{m-1}\frac{f^{(j)}(p)}{j!}(x-p)^j+\left.\p{-\frac{f^{(m)}(t)}{m!}(x-t)^m}\right|_{t=p}^{t=x}-\int_p^x\p{-\frac{f^{(m+1)}(t)}{m!}(x-t)^m}dt =\sum_{j=0}^m\frac{f^{(j)}(p)}{j!}(x-p)^j+\int_p^x\frac{f^{(m+1)}(t)}{m!}(x-t)^mdt$$ By induction, $$f(x)=\sum_{j=0}^k\frac{f^{(j)}(p)}{j!}(x-p)^j+\int_p^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt$$ $\blacksquare$

Proof. By fundamental theorem of calculus on component functions, for each $i\in\{1,\ldots,n\}$, $$\left.f(p+t(x-p))\right|_{t=0}^{t=1}=\int_0^1\sum_{i=1}^nD_if(p+t(x-p))(x_i-p_i)dt$$ Thus $$f(x)=f(p)+\sum_{i=1}^n\int_0^1D_if(p+t(x-p))dt(x_i-p_i)=P_0(x)+R_0(x)$$ Let $m\in\{1,\ldots,k\}$ and suppose $$f(x)=P_{m-1}(x)+R_{m-1}(x)$$ Let $I\in\{1,\ldots,n\}^m$, using integration by part with $F_I(t)=D_{I_m}\ldots D_{I_1}f(p+t(x-p))$ and $G_I(t)=-\frac{(1-t)^m}{m}$, and thus $F_I'(t)=\sum_{i=1}^nD_iD_{I_m}\ldots D_{I_1}f(p+t(x-p))(x_i-p_i)$ and $G_I'(t)=(1-t)^{m-1}$, we have $$\left.\p{D_{I_m}\ldots D_{I_1}f(p+t(x-p))\frac{-(1-t)^m}{m}}\right|_{t=0}^{t=1}=\int_0^1D_{I_m}\ldots D_{I_1}f(p+t(x-p))(1-t)^{m-1}dt+\int_0^1\sum_{i=1}^nD_iD_{I_m}\ldots D_{I_1}f(p+t(x-p))(x_i-p_i)\frac{-(1-t)^m}{m}dt$$ Thus $$ f(x) =P_{m-1}(x)+\frac{1}{(m-1)!}\sum_{I\in\{1,\ldots,n\}^m}\int_0^1D_{I_m}\ldots D_{I_1}f(p+t(x-p))(1-t)^{m-1}dt\prod_{i=1}^m(x_{I_i}-p_{I_i}) $$ $$ =P_{m-1}(x)+\frac{1}{m!}\sum_{I\in\{1,\ldots,n\}^m}\p{D_{I_m}\ldots D_{I_1}f(p)+\int_0^1\sum_{i=1}^nD_iD_{I_m}\ldots D_{I_1}f(p+t(x-p))(x_i-p_i)(1-t)^mdt}\prod_{i=1}^m(x_{I_i}-p_{I_i}) $$ $$ =P_m(x)+\frac{1}{m!}\sum_{I\in\{1,\ldots,n\}^m}\sum_{i=1}^n\int_0^1D_iD_{I_m}\ldots D_{I_1}f(p+t(x-p))(1-t)^mdt\prod_{i=1}^m(x_{I_i}-p_{I_i})(x_i-p_i) =P_m(x)+R_m(x) $$ By induction, $$f(x)=P_k(x)+R_k(x)$$ $\blacksquare$