Real Analysis Review

Since $f$ is injective, $f'$ is nonvanishing in $\Omega \coloneqq{\left\lvert {z} \right\rvert} \leq r$. A computation: $$\begin{align*} \mu(f({\mathbb{D}}_r)) &= \int_{{\mathbb{D}}_r} {\left\lvert {f'(z)} \right\rvert}\,dz\\ &= \int_{{\mathbb{D}}_r} f'(z) \overline{f'(z)} \,dz\\ &= \int_{{\mathbb{D}}_r} \qty{\sum_{j\geq 1} jc_j z^{j-1} } \overline{ \qty{\sum_{k\geq 1} kc_k z^{k-1}}} \,dz\\ &= \int_{{\mathbb{D}}_r} \qty{\sum_{j\geq 1} j c_j z^{j-1} } { \qty{\sum_{k\geq 1} k \overline{c_k} \overline{z}^{k-1} }} \,dz\\ &= \int_{{\mathbb{D}}_r} \sum_{j, k\geq 1} jk c_j \overline{c_k} z^{j-1}\overline{z}^{k-1}\,dz\\ &= \int_0^R\int_0^{2\pi} \sum_{j, k\geq 1} jk c_j \overline{c_k} (re^{it})^{j-1} (re^{-it})^{k-1} r\,dr\,dt\\ &= \int_0^R\int_0^{2\pi} \sum_{j, k\geq 1} jk c_j \overline{c_k} r^{j+k-1} e^{i(j-k)t} \,dr\,dt\\ &= \int_0^R \sum_{k\geq 1} k^2 {\left\lvert {c_k} \right\rvert}^2 r^{2k-1} \cdot 2\pi \,dt\\ &= \sum_{k\geq 1} k^2 {\left\lvert {c_k} \right\rvert}^2 {r^{2k} \over 2k}\Big|_{r=0}^R \\ &= \pi \sum_{k\geq 1} k {\left\lvert {c_k} \right\rvert}^2 R^{2k} .\end{align*}$$

See above solution: all goes identically up until the integral over $r$ values, just replace $\int_0^R$ with $\int_r^R$.

With some substitution, one can compute $$\begin{align*} {\left\lvert {x_n - x_{n-1}} \right\rvert} = {\left\lvert {{1\over 2} x_{n-1} + {1\over 2} x_{n-2} - x_{n-1}} \right\rvert} = {1\over 2} {\left\lvert {x_{n-1} - x_{n-2}} \right\rvert} ,\end{align*}$$ which holds for all $n$. This is enough to show that the sequence is contractive, i.e.  $$\begin{align*} {\left\lvert {x_n - x_{n-1}} \right\rvert} = c {\left\lvert {x_{n-1} - x_{n-2}} \right\rvert} && c\in (0, 1) .\end{align*}$$

Apply this recursively yields $$\begin{align*} {\left\lvert {x_n - x_{n-1}} \right\rvert} = \qty{1\over 2}^{n-1} {\left\lvert {b-a} \right\rvert} \overset{n\to\infty}\longrightarrow 0 ,\end{align*}$$ since ${\left\lvert {b-a} \right\rvert}$ is a constant. So in fact $x_n$ is convergent and thus Cauchy convergent.

Note: to compare ${\left\lvert {x_i - x_j} \right\rvert}$ directly, assume $i>j$ and apply the above estimate $i-j+1$ on ${\left\lvert {x_i - x_{i-1}} \right\rvert}, {\left\lvert {x_{i-1} - x_{i-2}} \right\rvert}, \cdots$ to reduce to this case. This yields something like $$\begin{align*} {\left\lvert {x_i - x_j} \right\rvert} = \qty{1\over 2}^{i-j+1}{\left\lvert {x_{j} - x_{j-1}} \right\rvert} = \qty{1\over 2}^{i-j+1} \qty{1\over 2}^{j-1} {\left\lvert {b-a} \right\rvert}\to 0 .\end{align*}$$ One could equivalently use the triangle inequality and a partial geometric sum to write $$\begin{align*} {\left\lvert {x_i - x_j} \right\rvert} \leq \sum_{j\leq k \leq i-1} {\left\lvert {x_{k+1} - x_{k}} \right\rvert} \implies {\left\lvert {x_i - x_j} \right\rvert} \leq c^j\qty{1\over 1-c}{\left\lvert {b-a} \right\rvert} .\end{align*}$$

Computing its limit: the usual trick of setting $L \coloneqq\lim x_n = \lim x_{n-1} = \lim x_{n-2}$ only yields $L = {L + L \over 2}$ here, and thus no information. Instead assume $x_n = r^n$ is geometric, then $$\begin{align*} 2x_n - x_{n-1} - x_{n-2} = 0 \implies 2r^n - r^{n-1} - r^{n-2} = 0 \implies 2r^2 - r - 1 = 0 \iff (2r+1)(r-1) = 0 \implies r = -1/2, 1 .\end{align*}$$ Write a general solution as $$\begin{align*} x_n = c_1 (-1/2)^n + c_2 (1)^n = c_1 (-1/2)^n + c_2 ,\end{align*}$$ and solve for initial conditions: $$\begin{align*} x_0: \quad a &= c_1 + c_2 \\ x_1: \quad b &= (-1/2)c_1 + c_2 \\ \\ \implies { \begin{bmatrix} {1} & {1} \\ {-1/2} & {1} \end{bmatrix} } \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} &= \begin{bmatrix} a \\ b \end{bmatrix} \\ \implies \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} &= {1\over 1 + (1/2)} { \begin{bmatrix} {1} & {-1} \\ {1/2} & {1} \end{bmatrix} } \begin{bmatrix} a \\ b \end{bmatrix} \\ &= \qty{1\over 3} { \begin{bmatrix} {2} & {-2} \\ {1} & {2} \end{bmatrix} } \begin{bmatrix} a \\ b \end{bmatrix} \\ &= \qty{1\over 3} \begin{bmatrix} 2a-2b \\ a+b \end{bmatrix} .\end{align*}$$

So the general solution is $$\begin{align*} x_n = {2\over 3}(a-b) \qty{-1\over 2}^n + {1\over 3}(a+b)\overset{n\to \infty}\longrightarrow \qty{1\over 3}(a+b) .\end{align*}$$

Fix ${\varepsilon}>0$, we need to find a $\delta = \delta({\varepsilon})$ such that $$\begin{align*} {\left\lvert {x-y} \right\rvert}<\delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}&& \forall x, y\in {\mathbf{R}} .\end{align*}$$ Use that $\lim_x\to \pm \infty f(x) = 0$ to choose $M\gg 0$ such that $$\begin{align*} {\left\lvert {x} \right\rvert} \geq M \implies {\left\lvert {f(x)} \right\rvert} \leq {\varepsilon}/2 ,\end{align*}$$ then $$\begin{align*} {\left\lvert {x} \right\rvert}, {\left\lvert {y} \right\rvert} \geq M \implies {\left\lvert {f(x) - f(y)} \right\rvert} \leq {\left\lvert {f(x)} \right\rvert} + {\left\lvert {f(y)} \right\rvert} \leq {\varepsilon} .\end{align*}$$ So in this region choose (say) $\delta_1 < {\varepsilon}$ to ensure that $B_\delta(x), B_\delta(y) \subseteq [-M, M]^c$. On $[-M, M]$, note that this region is compact and $f$ continuous on a compact set implies uniformly continuous. So use this to choose $\delta_2 = \delta_2({\varepsilon})$ in this region to ensure ${\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}$.

This handles the cases $x, y \in (M, M)^c$, or $x,y\in [M, M]$, so it only remains to handle $x\in [M, M]$ and $y\in (M, M)^c$ (wlog, relabeling $x,y$ if necessary). In this case, use the triangle inequality: $$\begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} &= {\left\lvert {f(x) - f(M) + f(M) -f(y)} \right\rvert} \\ &\leq {\left\lvert {f(x) - f(M)} \right\rvert} + {\left\lvert {f(M) -f(y)} \right\rvert} \\ &\leq {\varepsilon}+ {\left\lvert {f(M)} \right\rvert} + {\left\lvert {f(y)} \right\rvert} \\ &\leq {\varepsilon}+ {\varepsilon}+ {\varepsilon} ,\end{align*}$$ where we’ve used that $M, y\in (M, M)^c$ to apply the first bound and $M, x\in [M, M]$ to apply the second.

A direct computation: fix ${\varepsilon}>0$ and suppose ${\left\lvert {z-w} \right\rvert} < R$. Then $$\begin{align*} {\left\lvert {z^2-w^2} \right\rvert} &= {\left\lvert {z-w} \right\rvert}{\left\lvert {z+w} \right\rvert} \\ &\leq \delta \qty{{\left\lvert {z} \right\rvert} + {\left\lvert {w} \right\rvert}} \\ &\leq \delta \cdot 2R ,\end{align*}$$ so choose $\delta < { {\varepsilon}\over 2R}$ to get uniform continuity on ${\mathbb{D}}_{R/2}(0)$.

To see $f$ can’t be uniformly continuous on ${\mathbf{C}}$, take ${\varepsilon}\coloneqq c$ any constant and suppose the appropriate $\delta$ exists. We’ll look for a bad pair of $z, w$, so take $w = z + {1\over 2}\delta$. This would imply $$\begin{align*} {\left\lvert {z^2 - w^2} \right\rvert} &= {\left\lvert {z^2 - (z+\delta)^2} \right\rvert} \\ &= {\left\lvert {-2z\delta - \delta^2} \right\rvert} \\ &= {\left\lvert {2z\delta + \delta^2} \right\rvert} \\ &= \delta {\left\lvert {2z + \delta} \right\rvert} \\ &\overset{{\left\lvert {z} \right\rvert}\to\infty}\longrightarrow\infty ,\end{align*}$$ using the $\delta = \delta({\varepsilon})$ can’t depend on $z$ or $w$, and is thus constant in this expression. This contradicts that ${\left\lvert {z^2-w^2} \right\rvert} < {\varepsilon}= c < \infty$.

The standard example: $$\begin{align*} f(x) \coloneqq \begin{cases} x^2\sin\qty{1\over x} & x\neq 0 \\ 0 & x=0. \end{cases} .\end{align*}$$

Away from zero, this is clearly differentiable since we can just compute the derivative by the chain rule. It turns out that $$\begin{align*} f'(x) = \begin{cases} 2x\sin\qty{1\over x} + x^2 \cos\qty{1\over x}\qty{-1\over x^2} = 2x\sin\qty{1\over x} - \cos\qty{1\over x} & x\neq 0 \\ 0 & x=0. \end{cases} .\end{align*}$$ Here we check differentiability and compute the derivative at $x=0$ directly: $$\begin{align*} {f(x) - f(0) \over x-0} = {x^2\sin\qty{1\over x} - 0 \over x-0} = x\sin\qty{1\over x} \overset{x\to 0}\longrightarrow 0 ,\end{align*}$$ using that $-x \leq {\left\lvert {x\sin \qty{1\over x}} \right\rvert}\leq x$.

But now notice that the $\cos\qty{1\over x}$ term in $f'$ isn’t enveloped by an $x^c$ term, so $\lim_{x\to 0} f'(x)$ does not exist for oscillatory reasons:

In particular, $\lim_{x\to 0}f'(x) \neq f'(0) = 0$.

Uniformly convergent means that ${\left\lVert {g_i - g_j} \right\rVert}_{\infty} \to 0$, so $\sup_{x\in X}{\left\lvert {g_i(x)-g_j(x)} \right\rvert} \overset{i, j\to\infty}\longrightarrow 0$. We want to show that given ${\varepsilon}$ we can find $N_0$ such that $i, j > N_0$ yields $$\begin{align*} \sup_{x\in X}{\left\lvert { f\circ g_i(x) - f\circ g_j(x) } \right\rvert} < {\varepsilon} .\end{align*}$$

Fix ${\varepsilon}> 0$, then choose $\delta_1 = \delta_1({\varepsilon})$ by uniform continuity of $f$ to guarantee $$\begin{align*} {\left\lvert {y_1 - y_2} \right\rvert} \leq \delta_1 \implies {\left\lvert {f(y_1) - f(y_2) } \right\rvert} < {\varepsilon}\, \forall y_1, y_2\in X .\end{align*}$$ Now by uniform convergence of $\left\{{g_n}\right\}$, choose $N_0 = N_0(\delta_1)$ such that $$\begin{align*} i, j \geq N_0 \implies {\left\lvert { g_i(x) - g_j(x) } \right\rvert} < \delta_1 \, \forall x\in X .\end{align*}$$

Now writing $y_1 \coloneqq g_i(x), y_2 \coloneqq g_j(x)$, choose $i, j > N_0$ yields $$\begin{align*} {\left\lvert {y_1 - y_2} \right\rvert} \coloneqq{\left\lvert {g_i(x) - g_j(x) } \right\rvert} < \delta_1 \\ \implies {\left\lvert {f(y_1) - f(y_2)} \right\rvert} \coloneqq{\left\lvert {f(g_i(x)) - f(g_j(x))} \right\rvert} < {\varepsilon} ,\end{align*}$$ and taking the supremum over $x\in X$ preserves the inequality since $\delta_1$ and consequently $N_0$ only depend on ${\varepsilon}$.

$\implies$: Fix ${\varepsilon}>0$ and choose $\delta = \delta({\varepsilon})$ to get a bound corresponding to ${\varepsilon}/2$, then for all $x,y$ with ${\left\lvert {x-y} \right\rvert} < \delta$ on $[a, b]$, we have $$\begin{align*} {\left\lvert {f'(x) - f'(y) } \right\rvert} \leq {\left\lvert {f'(x) - {f(x) - f(y) \over x- y} } \right\rvert} + {\left\lvert { {f(x) - f(y) \over x-y} - f'(y)} \right\rvert} < {\varepsilon} .\end{align*}$$ This uses uniformity to bound by ${\varepsilon}/2$ the terms involving $f'(x)$ and $f'(y)$ respectively. So $f'$ is in fact uniformly continuous on $[a, b]$.

$\impliedby$: Let ${\varepsilon}> 0$ and $x,y\in [a, b]$ be arbitrary. Then by the MVT, we can a $\xi\in [x, y]$ with $f'(\xi)(x-y) = f(x) - f(y)$. Then use continuity of $f'$ to choose $\delta = \delta({\varepsilon}, x, y)$ so that ${\left\lvert {x-y} \right\rvert} < \delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}$, and note that ${\left\lvert {x-\xi} \right\rvert} \leq {\left\lvert {x-y} \right\rvert} < \delta$, so $$\begin{align*} {\left\lvert { {f(x) - f(y) \over x-y } - f'(y) } \right\rvert} = {\left\lvert { f'(\xi) - f'(y)} \right\rvert} < {\varepsilon} .\end{align*}$$

Define a function $$\begin{align*} d: A \times B &\to {\mathbf{R}}\\ (x, y) &\mapsto {\left\lVert {x- y} \right\rVert} .\end{align*}$$ Then $d$ is a continuous function on a compact topological space (where the product is compact by Tychonoff), and the extreme value theorem applies: $d$ attains its min/max for some pair $(a, b)$ in its domain.

Note that disjointness just guarantees that ${\left\lVert {a-b} \right\rVert}>0$, since ${\left\lVert {a-b} \right\rVert} = 0 \implies a=b$ and $A \cap B = \emptyset$.

Use that $X$ is connected iff $\mathop{\mathrm{Hom}}_{{\mathsf{Top}}}(X, S^0) = \left\{{c_{-1}, c_1}\right\}$, i.e. every continuous map from $X\to \left\{{-1, 1}\right\}$ is a constant map $x \xrightarrow{c_{-1}} -1$ or $x \xrightarrow{c_1} 1$. Let $f: A\cup B \to S^0$ be arbitrary, and let $f_1 \coloneqq{ \left.{{f}} \right|_{{A}} }$ and $f_2 \coloneqq{ \left.{{f}} \right|_{{B}} }$. By connectedness of $A$, $f_1$ is a constant map, as is $f_2$. On the intersection, for $x\in A \cap B \neq \emptyset$, we have $f_1(x) = f_2(x)$ since $x\in A$ and $x\in B$. So $f_1$ and $f_2$ are constant functions that must map to the same constant, so $f$ is constant and this $A\cup B$ is connected.

Let ${\varepsilon}>0$, we want to show that there exists an $N_0$ such that $n\geq N_0$ implies ${\left\lVert {f_n} \right\rVert}_\infty<{\varepsilon}$. Fix $x$, by pointwise convergence pick $M_x = M_x(x, {\varepsilon})$ so that $n\geq M \implies {\left\lvert {f_n(x)} \right\rvert} < {\varepsilon}$. By continuity, this bound holds in some neighborhood $U_x \ni x$. Produce a cover $\left\{{U_x}\right\}_{x\in [0, 1]}\rightrightarrows[0, 1]$; by compactness produce a finite subcover $\left\{{U_1, \cdots, U_m}\right\} \rightrightarrows[0, 1]$. Each $U_i$ corresponds to some $x_i$ and some $M_{x_i}$, so choose $N_0 > \max_{i\leq m} \left\{{M_{x_i}}\right\}$. Then $n\geq N_0 \implies N\geq M_{x_i}$ for each $i$, so ${\left\lvert {f_n(x)} \right\rvert} < {\varepsilon}$ for each $x\in [0, 1]$ since $x\in U_i$ for some $i$. So $\sup_{x\in X} {\left\lvert {f_n(x)} \right\rvert} = {\left\lVert {f_n} \right\rVert}_{\infty } < {\varepsilon}$.

See 3.2.12 of Understanding analysis 2ed. of Abbott. Show something stronger, that the following set is nonempty and open: $$\begin{align*} S \coloneqq\left\{{t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}E \cap(-\infty, t), E \cap(t, \infty) \text{ are uncountable}}\right\} \subseteq {\mathbf{R}} .\end{align*}$$ Write $$\begin{align*} S_- &\coloneqq\left\{{ t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}E \cap(- \infty, t) \text{ is countable}}\right\} \\ S_+ &\coloneqq\left\{{ s\in {\mathbf{R}}{~\mathrel{\Big\vert}~}E \cap(s, \infty) \text{ is countable}}\right\} .\end{align*}$$

Note that $S_- \neq {\mathbf{R}}$ since then we could write $E = \bigcup_{n\in {\mathbf{Z}}} E \cap(- \infty, n)$ as a countable union of countable sets.

Claim: $S = (\sup S_-,, \inf S_+)$.

???

Write $U(f), L(f)$ for the upper and lower sums of $f$, so for $\Pi$ the collection of all partitions of $[0, 1]$, $$\begin{align*} U(f) \coloneqq\inf_{P\in \Pi} U(f, P) && U(f, P) \coloneqq\sum_{k=1}^n \sup_{x\in I_k}f(x) \cdot \mu(I_k) \\ L(f) \coloneqq\sup_{P\in \Pi} L(f, P) && L(f, P) \coloneqq\sum_{k=1}^n \inf_{x\in I_k} f(x) \cdot \mu(I_k) .\end{align*}$$

Note that integrability of $f$ is equivalent to $$\begin{align*} \forall {\varepsilon}\exists P \text{ such that } U(f, P) - L(f, P) < {\varepsilon}\\ \iff \sum_{k=1}^n \qty{ \sup_{x\in I_k} f(x) - \inf_{x\in I_k} f(x)} \mu(I_k) < {\varepsilon} .\end{align*}$$

$$\begin{align*} {\left\lvert {x^n - y^n} \right\rvert} = {\left\lvert {y-x} \right\rvert}{\left\lvert {\sum_{1\leq k \leq n} x^k y^{n-k}} \right\rvert} \leq n M^{n-1}{\left\lvert {y-x} \right\rvert} \overset{y\to x}\longrightarrow 0 .\end{align*}$$

$$\begin{align*} x^{1\over n} - y^{1\over n} \leq (x-y)^{1\over n} \overset{x\to y}\longrightarrow 0 ,\end{align*}$$ using $(a+b)^m \geq a^m + b^m$

Apply the MVT: $$\begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} = {\left\lvert {f(\xi)} \right\rvert} {\left\lvert {x-y} \right\rvert} \overset{y\to x}\longrightarrow 0 .\end{align*}$$

Check $\sup f = 1$ and $\inf f = 0$ on every sub-interval, so $L(f, P) = 0$ and $U(f, P) = 1$ for every partition $P$ of $[0, 1]$.

Discontinuity: #todo