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\begin{enumerate}
\item[1] Let $l^2(\mathbb{N})=\{(x_n)_{n=1}^\infty:x_n\in\mathbb{R},\displaystyle\sum_{n=1}^\infty x_n^2<\infty\}$. Consider $d(x,y)=\sqrt{\displaystyle\sum_{n=1}^\infty (x_n-y_n)^2}$.

Let $x,y\in l^2(\mathbb{N})$ s.t. $x\neq y$. Then for some $j$, $x_j\neq y_j$. Thus $(x_j-y_j)^2>0$. Since for all $k$ $(x_k-y_k)^2$ is nonnegative, $\sqrt{\displaystyle\sum_{n=1}^\infty (x_n-y_n)^2}>0$. Furthermore, since for all $k$ $(x_k-x_k)^2=0$, $d(x,x)=0$. Thus $d(x,y)=0$ iff $x=y$.

Since for all $k$ $(x_k-y_k)^2=(y_k-x_k)^2$, $d(x,y)=d(y,x)$.

SHOW HERE THAT $d(x,y)+d(y,z)\geq d(x,z)$

$\therefore d$ is well-defined and $(l^2(\mathbb{N}),d)$ is a metric space.

Now let $\{x_n\}$ be a Cauchy sequence in $l^2(\mathbb{N})$, i.e. $\forall\ \varepsilon\ \exists\ N$ s.t. $d(x_m,x_n)<\varepsilon$ if $m,n>N$.

Each $x_n$ can be written as $\{x_{nj}\}$, $x_{nj}\in\mathbb{R}$, where $\displaystyle\sum_{j=1}^{\infty}x_{nj}^2<\infty$. So for any $m,n>N$, $d(x_m,x_n)=\sqrt{\displaystyle\sum_{j=1}^\infty (x_{nj}-x_{mj})^2}<\varepsilon$.

So $\displaystyle\sum_{j=1}^\infty (x_{nj}-x_{mj})^2<\varepsilon^2$

So for any $j$, $(x_{nj}-x_{mj})^2<\varepsilon^2$. So $|x_{mj}-x_{nj}|<\varepsilon$.

By definition, $\{x_{1j},x_{2j},x_{3j},...\}$ is a Cauchy sequence and therefore converges to $y_j\in\mathbb{R}$

Thus for any $k$, for each $j$ there is some $N_j$ s.t. for $n>N_j$, $|y_j-x_{nj}|<\frac{1}{k^j}$. By taking $N=\sup N_j$, for $n>N$ $|y_j-x_{nj}|<\frac{1}{k^j}$ for all $j$, so $(y_j-x_{nj})^2<\frac{1}{k^j}$ for all $j$. Since $\displaystyle\sum_{j=1}^\infty\frac{1}{k^j}=\frac{1}{k-1}$, $\displaystyle\sum_{j=1}^\infty(y_j-x_{nj})^2<\frac{1}{k-1}$. For any $\varepsilon$, there is $k$ s.t. $\sqrt{\frac{1}{k-1}}<\varepsilon$, so there is $N$ s.t. for $n>N$ $d(x_n,y)<\varepsilon$. Therefore $\{x_n\}$ converges to $y$.

Now let $Z=\{0,0,0,...\}$. For each $x_n$, since $\displaystyle\sum_{n=1}^\infty x_n^2<\infty$, $d(x_n,Z)=\sqrt{\displaystyle\sum_{n=1}^\infty (x_n-0)^2}<\infty$

By the triangle inequality, $d(y,Z)\leq d(y,x_n)+d(x_n,Z)$. Then since for $n>N$ $d(x_n,y)<\varepsilon$ and $d(x_n,Z)<\infty$, $d(y,Z)\leq d(x_n,y)+d(x_n,Z)<\infty$

So $\displaystyle\sum_{n=1}^\infty (y_n-0)^2<\infty.$ Then $\sqrt{\displaystyle\sum_{n=1}^\infty y_n^2}<\infty$. By definition, $y\in l^2(\mathbb{N})$

Since any Cauchy sequence in $l^2(\mathbb{N})$ converges to such a $y$, $l^2(\mathbb{N})$ is complete.
\item[2]
\begin{description}
\item[(i)] Let $X$ be path-connected, i.e. $\forall x,y\in X\ \exists\ f:[0,1]\mapsto X$ s.t. $f(0)=x$ and $f(1)=y$.

Assume $X$ is not connected, i.e. $X=A\sqcup B$, where $A$ and $B$ are both open and closed.

Let $x\in A, y\in B$. Then $f^{-1}(A)$ and $f^{-1}(B)$ are non-empty. Since for any $z\in [0,1]$ $f(z)\in A$ or $f(z)\in B$, $[0,1]=f^{-1}(A)\cup f^{-1}(B)$. Since $f(z)\in A\implies f(z)\notin B$, $[0,1]=f^{-1}(A)\sqcup f^{-1}(B)$. By Rudin Theorem 4.8 $f^{-1}(A)$ and $f^{-1}(B)$ are closed, and by the corrollary $f^{-1}(A)$ and $f^{-1}(B)$ are open. Thus $[0,1]$ is not connected. But by Theorem 2.47, $[0,1]$ is connected. So the assumption was false and $X$ is connected.
\item[(ii)] Let $A$ and $B$ be separated subsets of some $\mathbb{R}^k$, suppose $a\in A,b\in B$, and define $p(t)=(1-t)a+tb$ for $t\in\mathbb{R}$. Put $A_0=p^{-1}(A),B_0=p^{-1}(B)$. (Thus $t\in A_0$ iff $p(t)\in A$).
\begin{description}
\item[(a)] Suppose $A_0$ and $B_0$ are not separated. Then for some $t\in A_0$ $t\in\bar{B_0}$ or vice versa; assume WLOG the former. Since by Theorem 4.9 $p$ is continuous, for any $c\in B_0$, for all $\varepsilon$ there is $\delta$ s.t. $|p(t)-p(c)|<\varepsilon$ if $|t-c|<\delta$. Because $t\in\bar{B_0}$, there is some $c\in B_0$ s.t. $|t-c|<\delta$. Thus $p(t)\in\bar{B}$. But $p(t)\in A$, contradicting that $A$ and $B$ are separated.
\item[(b)]  Assume $\forall\ t\in(0,1),\ t\in A_0\cup B_0$; then since $A_0$ and $B_0$ are not separated, for some $t\in A_0$ $t\in\bar{B_0}$ or vice versa. But in part (a) it was demonstrated that this contradicts that $A$ and $B$ are separated, so there must be $t\in(0,1)$ s.t. $t\notin A_0\cup B_0$
\item[(c)] Assume $X$ is a convex subset of $\mathbb{R}^k$, i.e. for any $a,b\in X$ $p(t)=(1-t)a+tb\in X$ when $t\in (0,1)$. If $X=A\cup B$ and $A$ and $B$ are separated, for some $t\in (0,1)$ by part (b) $t\notin(p^{-1}(A)\cup p^{-1}(B)$, contradicting the assumption. Thus any $A$ and $B$ s.t. $X=A\cup B$ are not separated, so $X$ is connected if $X$ is convex.
\end{description}
\item[(iii)] Let $D\subset\mathbb{R}^2$ be countable. Let $x,y\in D^C$. Let $l\subset\mathbb{R}^2$ be a line through $x$. Since there are uncountable such lines intersecting only at x and only countable elements in $D$, there is at least one such $l\subset D^C$. Now let $m\subset\mathbb{R}^2$ be a line through $y$ and some point of $l$. Again, since there are uncountable such lines intersecting only at $y$ and only countable elements in $D$, there must be such a line $m\subset\mathbb{R}^2$ Then let $p=m\cap l$. So the segments $[x,p]$ and $[p,y]$ are contained in $D^C$.

Setting $f(t)=\left\{ \begin{array}{l}(x_1+2(p_1-x_1)t,x_2+2(p_2-y_2)t), t<\frac{1}{2} \\ (p_1+2(y_1-p_1)t,p_2+2(y_2-p_2)t), t\geq\frac{1}{2} \end{array}\right.$, we see $f:[0,1]\mapsto D^C$ is a continuous path, so $D^C$ is path-connected.
\end{description}
\item[3]
\begin{description}
\item[Rudin Ch 4 no 1] Let $f:\mathbb{R}\mapsto\mathbb{R}$ s.t. $f(x)= \left\{ \begin{array}{l} 1, x \neq 0 \\ 0, x = 0 \end{array} \right.$

For $x>0,\ \exists\ \delta$ s.t. $\forall\ h<\delta,\ \ x-h > 0\ \text{and}\ x+h>0.$

So $|f(x+h)-f(x-h)|=|1-1|=0\ \forall\ h<\delta.$

$\therefore \displaystyle \lim_{h\to0}|f(x+h)-f(x-h)|=0\ \text{for}\ x>0$

Similarly, $\forall\ x<0,\ \exists\ \delta$ s.t. $\forall\ h<\delta,\ x+h<0\ \text{and}\ x-h<0.$

So $|f(x+h)-f(x-h)|=|1-1|=0\ \forall\ h<\delta.$

$\therefore \displaystyle \lim_{h\to0}|f(x+h)-f(x-h)|=0\ \text{for}\ x<0$

For $\varepsilon>0\ \exists\ h$ s.t. $0<h<\varepsilon.\ \text{So}\ |f(0+h)-f(0-h)|=0.$

$\therefore \displaystyle \lim_{h\to0}|f(0+h)-f(0-h)|=0$

Thus for every $x\in\mathbb{R},\ \displaystyle \lim_{h\to0}|f(x+h)-f(x-h)|=0\ \text{for}\ x<0$

But $\forall\ \delta>0,\ |f(0+\delta)-f(0)|=|1-0|=1.$ So $f(x)$ is discontinuous at 0.

$\therefore\ \displaystyle \lim_{h\to0}|f(x+h)-f(x-h)|=0$ does not imply that $f$ is continuous.

\item[Rudin Ch 4 no 3] Let $f:X\mapsto\mathbb{R}$ be continuous, and $Z(f)=\{p\in X:f(p)=0\}$.

Let $Y(f)=X-Z(f),$ and $x\in Y(f)$, i.e., $f(x)\neq 0$.

For $\varepsilon<|\frac{f(x)}{2}|,\ \exists\ \delta$ s.t. $f(B(x,\delta))\subset B(f(x),\varepsilon).$

Since $ 0\notin B(f(x),\varepsilon),\ 0\notin f(B(x,\delta)).$ So $B(x,\delta) \subset Y(f).$

$\therefore Y(f)$ is open. By definition, $Z(f)$ is closed.

\item[Rudin Ch 4 no 6] Let $E\subset\mathbb{R}$ and $f:E\mapsto\mathbb{R}$. Let $g(x)=(x,f(x))\in\mathbb{R}^2$
\begin{description}
\item[($\Rightarrow$)] Suppose $E$ is compact, and $f$ is continuous on $E$.

Let A be a limit point of g(E). Then $\exists\ \{a_n\}\subset g(E)$ s.t. $a_n\to A.$

Then each $a_n = (x_n,y_n), x_n\in E, y_n=f(x_n).$

Since $\{a_n\}$ converges, $\forall \varepsilon\ \exists N$ s.t. $m,n>N \implies |a_m-a_n|<\varepsilon$.

Thus $\sqrt{(x_m-x_n)^2 + (y_m-y_n)^2}<\varepsilon.$

So $|x_m-x_n|<\varepsilon$. Therefore $x_n\to x$.
Since $E$ is compact, $x\in E$.

For all $\varepsilon$, $\exists\ \delta$ s.t. $|f(x)-f(c)| < \varepsilon$ if $|x-c|<\delta$, and $\exists\ N$ s.t. $|x-x_n|<\delta$ if $n>N$.

Thus for $n>N$ $|f(x_n)-f(x)|=|y_n-f(x)|<\varepsilon$. So $y_n\to f(x).$

Therefore $A=(x,f(x))\in g(E)$.

Since $A$ is any limit point, $g(E)$ is closed.

E is compact, and by Theorem 4.14 $f(E)$ is compact. So $E$ and $f(E)$ are bounded. Then $\exists$ $M,N$ s.t. $\forall\ x\in E, |x|<M$ and $|f(x)|<N$.

Then $|g(x)|=\sqrt{x^2+f(x)^2}<x+f(x)<M+N.$ So $g(E)$ is bounded.

Since $g(E)$ is a closed, bounded subset of $\mathbb{R}^2$, $g(E)$ is compact.

\item[($\Leftarrow$)] Let $E$ and $g(E)$ be compact. Assume $f$ is not continuous on E.

For any x, let $\{a_n\}\subset E$ s.t. $a_n\to x$, and $a_n>x$. By the compactness of $g(E)$ there is some subsequence $\{b_n\}$ s.t. $g(b_n)$ converges to a point in $g(E)$.

Thus $g(b_n)=(b_n,f(b_n))\to$ some $(c,f(c))$. Then for any $\varepsilon$, for some $N$ if $n>N$ $|b_n-c| < \sqrt{(b_n-c)^2+(f(b_n)-f(c))^2} < \varepsilon$, so $b_n\to c$.

Since $b_n\to x,\ g(b_n)\to g(x)$.

Then $|f(b_n)-f(x)|<\sqrt{(b_n-x)^2+(f(b_n)-f(x))^2}<\epsilon$, so $f(b_n)\to f(x)$.

Hence $f(x+) = f(x)$.

Letting $a_n<x$ instead, we see that $f(x-) = f(x)$ as well. Since both the right-hand and left-hand limits exist, there is no discontinuity of the second kind at $x$.
Since $f(x-)=f(x+)=f(x)$, there is no simple discontinuity at $x$. Therefore $f$ is continuous at $x$. Since $x$ is arbitrary, $f$ is continuous at all points in $E$.
\end{description}

\item[Rudin Ch 4 no 7] Let $f(x,y)=\frac{xy^2}{x^2+y^4}$ and $g(x,y)=\frac{xy^2}{x^2+y^6}$

Since $(x-y^2)^2 = x^2-2xy^2+y^4$, $x^2+y^4=(x-y^2)^2+2xy^2$, so $x^2+y^4\geq2xy^2$

Substituting, $\frac{xy^2}{x^2+y^4}\leq\frac{xy^2}{2xy^2}=\frac{1}{2}$. So $f(x,y)$ is bounded by $\frac{1}{2}$

Now consider the series $(x_n,y_n)=(\frac{1}{n^3},\frac{1}{n})$. $g(x_n,y_n)=\frac{(\frac{1}{n^3})(\frac{1}{n})^2}{(\frac{1}{n^3})^2+(\frac{1}{n})^6}=\frac{\frac{1}{n^5}}{\frac{2}{n^6}}=\frac{n}{2}$.
For any $\varepsilon,\delta$ $\exists$ $N$ s.t. for $n>N$, $|(x_n,y_n)|<\delta$ and $g(x_n,y_n)>\varepsilon$. So $g(x,y)$ is unbounded in all neighborhoods of $(0,0)$.

Now instead let $(x_n,y_n)=(\frac{1}{n^2},\frac{1}{n})$. $f(x_n,y_n)=\frac{(\frac{1}{n^2})(\frac{1}{n})^2}{(\frac{1}{n^2})^2+(\frac{1}{n})^4)}=\frac{\frac{1}{n^4}}{\frac{2}{n^4}}=\frac{1}{2}$.
For any $\varepsilon,\delta$ $\exists$ $N$ s.t. for $n>N$, $|(x_n,y_n)|<\delta$ and $g(x_n,y_n)=\frac{1}{2}>g(0,0)+\varepsilon$. So $g(x,y)$ is not continuous at $(0,0)$.

Let $E$ be a line in $\mathbb{R}^2$. i.e., either $E=\{(x,y)\in\mathbb{R}^2:y=ax+b\}$ for some $a\neq0,b$, or $E=\{(x,y)\in\mathbb{R}^2:x=c\}$ for some $c$.

Let $E=\{(x,y)\in\mathbb{R}^2:x=c\}$. If $c\neq0$, since $(0,0)\notin E$, for all $(x,y)\in E$ $f(x,y)=\frac{cy^2}{c^2+y^4}.$ Since $c^2+y^4>0$, $f(x,y)$ is continuous by Theorem 4.9. Similarly, $g(x,y)=\frac{cy^2}{c^2+y^6}$ which is also continuous by the same theorem. If $c=0$, $f(x,y)=\frac{0}{y^4}=0$ and $g(x,y)=\frac{0}{y^6}=0$ for $(x,y)\neq(0,0)$, and $f(x,y)=g(x,y)=0$ for $(x,y)=(0,0)$. Therefore $f(x,y)=g(x,y)=0$ for all $(x,y)$, so $f$ and $g$ are continuous at all points in E.

Now let $E=\{(x,y)\in\mathbb{R}^2:y=ax+b\},a\neq0$.
If $b\neq0$, $(0,0)\notin E$. So for all $(x,y)\in E$, $f(x,y)=\frac{x(ax+b)^2}{x^2+(ax+b)^4}$ and $g(x,y)=\frac{x(ax+b)^2}{x^2+(ax+b)^6}$ are continuous by Theorem 4.9. Otherwise $y=ax$ for all $(x,y)\in E$. So $f(x,y)=\frac{a^2x^3}{x^2+a^4x^4}=\frac{a^2x}{1+a^4x^2}<a^2x$ and $g(x,y)=\frac{a^2x^3}{x^2+a^6x^6}=\frac{a^2x}{1+a^6x^4}<a^2x$, so $f$ and $g$ are continuous by Theorem 4.9 for $x\neq0$. For any $\varepsilon$, $\exists$ $\delta=\frac{\varepsilon}{a^2}$ s.t. if $|(x,y)|<\delta$, $|x|<|(x,y)|<\frac{\varepsilon}{a^2}$, and $|f(x,y)|=|f(x,ax)|<|a^2\frac{\varepsilon}{a^2}|=\varepsilon$ and $|g(x,y)|=|g(x,ax)|<|a^2\frac{\varepsilon}{a^2}|=\varepsilon$. Therefore $f$ and $g$ are continuous at $(0,0)$, so $f$ and $g$ are continuous at all points in E.

\item[Rudin Ch 4 no 11] Suppose $f:X\mapsto Y$ is uniformly continuous, and $\{x_n\}\subset X$ is a Cauchy sequence.

By the uniform continuity of $f$, for any $\varepsilon$ there is $\delta$ s.t. if $d(p,q)<\delta$, $d(f(p),f(q))<\varepsilon$.

Since ${x_n}$ is Cauchy, there is $N$ s.t. if $m,n>N$, $d(x_m,x_n)<\delta$. So $d(f(x_m),f(x_n))<\varepsilon$. Thus $\{f(x_n)\}$ is a Cauchy sequence in $Y$.

Now let $E\subset X$ be a dense subset, and $f:E\mapsto\mathbb{R}$ be uniformly continuous. By definition, any point $x\in X=\lim \{x_n\},x_n\in E$. Define $g:X\mapsto\mathbb{R}$ s.t. $g(x)=\lim \{f(x_n)\}$ Since $f$ is continuous on E, for any $\{e_n\}\to e\in E$, since $e$ is not a discontinuity $g(e)=\lim \{f(e_n)\}=f(e)$, so $g$ is an extension of $f$ to $X$.

Let $x=\{x_n\}$ be any point in $X$, and $y$ be a point in $E$. Since $f$ is continuous, for any $\varepsilon$ there is $\delta$ s.t. if $|x_n-y|<\delta$, $|f(x_n)-f(y)|<\varepsilon$. If $|x-y|<\delta$, there is $N$ s.t. for $n>N$, $|x_n-y|=\delta$, so $|f(x_n)-f(y)|<\varepsilon$. Then by definition $|\lim\{f(x_n)\}-f(y)|<\varepsilon$. Since $|g(x)-g(y)|=|\lim\{f(x_n)\}-\lim\{f(y)\}|=|\lim\{f(x_n)\}-f(y)|<\varepsilon$, $|g(x)-g(y)|<\varepsilon$ if $|x-y|<\delta$. Therefore $g$ is continuous.

\item[Rudin Ch 4 no 15] Let $f:\mathbb{R}\mapsto\mathbb{R}$ be continuous and open, i.e. $f(V)$ is open if $V$ is open.

Assume $f$ is not monotonic.
Then for some $a,b\in\mathbb{R},a<b$, where $V$ is the open interval $(a,b)$, $\sup f(V)>f(b)$ and $\inf f(V)<f(b)$ by Theorem 4.29. By Theorem 4.16, for some $p,q\in[a,b]$ $f(p)=\sup f(V)$ and $f(q)=\inf f(V)$. Since $f(p)>f(b)$ and $f(q)<f(b)$, $p\neq b$ and $q\neq b$.

If $p=q=a$, then for all $x\in V$, $f(x)=f(a)$. Therefore $f(V)=\{f(a)\}$, contradicting that $f(V)$ is open.

If $p>a$, then $p\in V$, so $f(p)\in f(V)$. By the openness of $V$, for some $\varepsilon$ $B(f(p),\varepsilon)\subset f(V)$. Then $f(p)+\frac{\varepsilon}{2}\in f(V)$, contradicting that $f(p)=\sup f(V)$.

Therefore $q>a$. Similarly, $q\in V$, so $f(q)\in f(V).$ By the openness of $V$, some $B(f(p),\varepsilon)\subset f(V)$. Then $f(q)-\frac{\varepsilon}{2}\in f(V)$, contradicting that $f(q)=\inf f(V)$.

Therefore the assumption that $f$ is not monotonic must be false.

\item[Rudin Ch 4 no 20] Let $E\subset X\neq\emptyset$, $x\in X$, and define $\rho_E(x)=\displaystyle \inf_{z\in E}d(x,z)$.

\begin{description}
\item[(a)]
\begin{description}
\item[$(\Rightarrow)$] Let $\rho_E(x)=0$. Then for any $\varepsilon$, $\exists$ $z\in E$ s.t. $d(x,z)<\varepsilon$. Therefore $x$ is a limit point of $E$. By definition, $x\in\bar{E}$.
\item[$(\Leftarrow)$] Let $x\in\bar{E}$. Then either $x\in E$ or $x$ is a limit point of $E$. If $x\in E$, since $d(x,x)=0$, $\rho_E(x)=\displaystyle \inf_{z\in E}d(x,z)=0$. If $x$ is a limit point of $E$, for any $\varepsilon$ $\exists$ $z$ s.t. $d(x,z)<\varepsilon$. So $\rho_E(x)=\displaystyle\inf_{z\in E}d(x,z)=0$.
\end{description}
\item[(b)] Let $x,y\in X,$ and let $z\in E$ s.t. $d(y,z)=\displaystyle\inf_{a\in E}d(y,a)$. By definition, $\rho_E(x)\leq d(x,z)$. By the Triangle Inequality, $d(x,z)\leq d(x,z)+d(y,z)=d(x,z)+\rho_E(y)$. So $\rho_E(x)-\rho_E(y)\leq d(x,y)$, proving that $\rho_E$ is continuous, since if $d(x,y)\leq\varepsilon$ then $\rho_E(x)-\rho_E(y)\leq\varepsilon$
\end{description}

\item[Rudin Ch 4 no 23] Let $f$ be convex on $(a,b)$, i.e. $f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$ for $\lambda\in(0,1)$ and $x,y\in(a,b)$. Then $f$ is convex on any subinterval of (a,b), since any $x,y$ in that subinterval are in $(a,b)$

Let $l_{a,b}(x)$ be the line from $(a,f(a))$ to $(b,f(b))$. Then $l_{a,b}(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$. Rearranging, $l_{a,b}(x)=(f(b)-f(a))\frac{x-a}{b-a}+f(a)=f(b)\frac{x-a}{b-a}+f(a)(1-\frac{x-a}{b-a})$

Then $l_{a,b}(x)\geq f(\frac{x-a}{b-a}b+(1-\frac{x-a}{b-a})a)=f(\frac{x-a}{b-a}b+\frac{b-x}{b-a}a)=f(\frac{bx-ab+ab-ax}{b-a})=f(\frac{bx-ax}{b-a})$

$\therefore f(x)\leq l_{a,b}(x)$ when $\frac{x-a}{b-a}\in(0,1)$, which is true when $x\in(a,b)$.

Let $c\in(a,b)$, and let $x\in(c,b)$ s.t. $f(x)<l_{a,c}(x)$ Then $l_{a,x}(x)=f(x)<l_{a,c}(x).$ So for all $y>a$, $l_{a,x}(y)<l_{a,c}(y)$. Hence $l_{a,x}(c)<l_{a,c}(c)=f(c)$. But since $c\in(a,x)$, $f(c)\leq l_{a,x}(c)$. Therefore for all $x\in(c,b)\ f(x)\geq l_{a,c}(x)$

Similarly, let $c\in(a,b)$, and let $x\in(a,c)$ s.t. $f(x)<l_{c,b}(x)$ Then $l_{x,b}(x)=f(x)<l_{c,b}(x).$ So for all $y<b$, $l_{x,b}(y)<l_{c,b}(y)$. Hence $l_{x,b}(c)<l_{b,c}(c)=f(c)$. But since $c\in(x,b)$, $f(c)\leq l_{x,b}(c)$. Therefore for all $x\in(a,c)\ f(x)\geq l_{c,b}(x)$

Since $l_{a,c}$ and $l_{c,b}$ are continuous, for any $\varepsilon\ \exists\ \delta$ s.t. if $|x-c|<\delta$ $|l_{a,c}(x)-f(c)|=|l_{a,c}(x)-l_{a,c}(c)|<\varepsilon$ and $|l_{c,b}(x)-f(c)|=|l_{c,b}(x)-l_{c,b}(c)|<\varepsilon$.

Thus if $x>c$, $f(c)-\varepsilon<l_{a,c}(x)\leq f(x)\leq l_{c,b}(x)<f(c)+\varepsilon$

Otherwise, $f(c)-\varepsilon<l_{c,b}(x)\leq f(x)\leq l_{a,c}(x)<f(c)+\varepsilon$

$\therefore |f(x)-f(c)|<\varepsilon$, so $f$ is continuous at $x$. Since $x$ is arbitrary, $f$ is continuous on $(a,b)$

Now let $f$ be convex and $g$ be convex and monotone increasing.

Then $f(\lambda x + (1-\lambda)y) \leq \lambda f(x)+(1-\lambda)f(y)$ by the convexity of $f$

Since $g$ is increasing, $g(f(\lambda+(1-\lambda)y))\leq g(\lambda f(x)+(1-\lambda)f(y))$

Since $g$ is convex $g(f(\lambda+(1-\lambda)y))\leq\lambda g(f(x))+(1-\lambda)g(f(y))$

$\therefore f\circ g$ is convex

Now let $f$ be convex in $(a,b)$ and $a<s<t<u<b$

Assume $\frac{f(t)-f(s)}{t-s}>\frac{f(u)-f(s)}{u-s}$. Then $l_{s,t}(x)>l_{s,u}(x)$ for $x>s$. But for $t<x<u$, $l_{s,t}(x)\leq f(x)\leq l_{t,u}(x).$ So the assumption was false and $\frac{f(t)-f(s)}{t-s}\leq\frac{f(u)-f(s)}{u-s}$

Now assume $\frac{f(u)-f(t)}{u-t}<\frac{f(u)-f(s)}{u-s}$. Then $l_{t,u}(x)>l_{s,u}(x)$4 for $x<u$. But for $s<x<t$, $l_{t,u}(x)\leq f(x)\leq l_{s,u}(x)$. So the assumption was false and $\frac{f(u)-f(t)}{u-t}\geq\frac{f(u)-f(s)}{u-s}$

$\therefore \frac{f(t)-f(s)}{t-s}\leq\frac{f(u)-f(s)}{u-s}\leq\frac{f(u)-f(t)}{u-t}$

\item[Rudin Ch 4 no 24] Let $f:(a,b)\mapsto\mathbb{R}$ be continuous and let $f(\frac{x+y}{2})\leq\frac{f(x)+f(y)}{2}$ for $x,y\in(a,b)$.

Then let $m$ and $n$ be any two points s.t. $a\leq m<n\leq b$. Then for any $x,y\in(m,n)$ it is true that $x,y\in(a,b)$. Therefore it is true that $f(\frac{x+y}{2})\leq\frac{f(x)+f(y)}{2}$.

Then as in problem 23 let $l_{m,n}(x)$ be the line from $(m,f(m))$ to $(n,f(n))$. Then $l_{m,n}(x)=\frac{f(m)-f(n)}{m-n}(m-n)+f(m)=f(n)\frac{x-m}{n-m}+f(m)(1-\frac{x-m}{n-m})$.

If $x=\frac{m+n}{2}$, then $\frac{x-m}{b-n}=\frac{1}{2}$, so $l_{m,n}(x)\geq f(x)$ by the same manipulation as in problem 23. Thus $f(\frac{m+n}{2})\leq l_{m,n}(\frac{m+n}{2})$ for all $a\leq m<n\leq b$.

Let $x\in[p,q]$, $a\leq p<q\leq b$, and set $p_1=p$ and $q_1=q$.

If $x=p_1$ or $x=q_1$ $f(x)=l_{p_1,q_1}(x)$.

Otherwise if $x<\frac{p_1+q_1}{2}$, then set $p_2=p_1$ and $q_2=\frac{p_1+q_1}{2}$. Since $l_{p_2,q_2}(q_2)=f(q_2)\leq l_{p_1,q_1}(q_2)$, for all $y\in[p_2,q_2]$ $l_{p_2,q_2}(y)\leq l_{p_1,q_1}(y)$. Note that $q_2-p_2=\frac{q_1-p_1}{2}$.

Otherwise set $p_2=\frac{p_1+q_1}{2}$ and $q_2=q_1$. Since $l_{p_2,q_2}(q_2)=f(q_2)\leq{p_1,q_1}$, for all $y\in[p_2,q_2]$ $l_{p_2,q_2}(y)\leq l_{p_1,q_1}(y)$. Again, note that $q_2-p_2=\frac{q_1-p_1}{2}$.

Repeating this construction, create the sequences $p_n$ and $q_n$.

Note that for any $j$, by induction $l_{p_j,q_j}(y)\leq l_{p,q}(y)$ for $y\in[p,q]$

If $x=p_j$ or $x=q_j$ for any $j$, then $f(x)=l_{p_j,q_j}(x)\leq l_{p,q}(x)$.

Otherwise, since for each $j$ $p_j<x<q_j$ and $q_j-p_j=\frac{q-p}{2^{j-1}}$, $q_j-x<\frac{q-p}{2^{j-1}}$.

Since $f$ is continuous, for any $\varepsilon$ there is $\delta$ s.t. $|f(x)-f(q_j)|<\varepsilon$ if $q_j-x<\delta$. For any such $\delta$ there is $N$ s.t. $\frac{q-p}{2^{n-1}}<\delta$ for $n>N$, so for $n>N$ $q_n-x<\delta$, and $|f(q_n)-f(x)|<\varepsilon$.

Now, assume $f(x)>l_{p,q}(x)$ Then setting $\varepsilon=\frac{f(x)-l_{p,q}(x)}{2}$, we see that $f(q_n)>l_{p,q}(x)$ for $n>N$, contradicting that $f(q_n)=l_{p_n,q_n}(q_n)\leq l_{p,q}(q_n)$.

$\therefore$ for any $x\in(p,q)\subset(a,b)$, $f(x)\leq l_{p,q}(x)$.

Now let $x,y\in(a,b)$. WLOG, assume $x<y$. For any $\lambda\in(0,1)$, let $c=x+\lambda(y-x)$.

Then $\frac{c-x}{y-x}=\lambda$.

Since $\lambda\in(0,1)$, $c\in(x+0(y-x),x+1(y-x))=(x,y)\subset(a,b)$.

$\therefore f(c)\leq l_{x,y}(c)=f(y)\frac{c-x}{y-x}+f(x)=\frac{c-x}{y-x}f(y)+(1-\frac{c-x}{y-x})f(x)=\lambda f(y)+(1-\lambda)f(x)$

And $c=\frac{yc-xc}{y-x}=\frac{yc-xy+xy-xc}{y-x}=y\frac{c-x}{y-x}+x\frac{y-c}{y-x}=\lambda y+(1-\lambda)x$

Combining these two equations, $f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$ for any $x,y\in(a,b)$ and $\lambda\in(0,1)$

$\therefore$ $f$ is convex.
\end{description}
\end{enumerate}
\end{document}