Real Analysis Review

Tie’s Extra Questions: Fall 2015 (Computing area) #complex/exercise/completed

Let $$f(z) = \sum_{n=0}^\infty c_n z^n$$ be analytic and one-to-one in $$|z| < 1$$. For $$0<r<1$$, let $$D_r$$ be the disk $$|z|<r$$. Show that the area of $$f(D_r)$$ is finite and is given by \begin{align*} S = \pi \sum_{n=1}^\infty n |c_n|^2 r^{2n} .\end{align*}

Note that in general the area of $$f(D_1)$$ is infinite.

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*}

Tie’s Extra Questions: Fall 2015 (Variant) #complex/exercise/completed

Let $$f(z) = \sum_{n= -\infty}^\infty c_n z^n$$ be analytic and one-to-one in $$r_0< |z| < R_0$$. For $$r_0<r<R<R_0$$, let $$D(r,R)$$ be the annulus $$r<|z|<R$$. Show that the area of $$f(D(r,R))$$ is finite and is given by \begin{align*}S = \pi \sum_{n=- \infty}^\infty n |c_n|^2 (R^{2n} - r^{2n}).\end{align*}

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

Spring 2019.1 #complex/qual/work

Define

\begin{align*} E(z)=e^{x}(\cos y+i \sin y) . \end{align*}

• Show that $$E(z)$$ is the unique function analytic on $$\mathbb{C}$$ that satisfies

\begin{align*} E^{\prime}(z)=E(z), \quad E(0)=1 . \end{align*}

• Conclude from the first part that \begin{align*} E(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{n !} .\end{align*}

Recurrences #complex/qual/work

Let $$x_0 = a, x_1 = b$$, and set \begin{align*} x_n \coloneqq{x_{n-1} + x_{n-2} \over 2} \quad n\geq 2 .\end{align*}

Show that $$\left\{{x_n}\right\}$$ is a Cauchy sequence and find its limit in terms of $$a$$ and $$b$$.

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*}

Uniform continuity #complex/qual/work

Suppose $$f:{\mathbf{R}}\to{\mathbf{R}}$$ is continuous and $$\lim_{x\to \pm \infty} f(x) = 0$$. Prove that $$f$$ is uniformly continuous.

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.

Negating uniform continuity #complex/qual/completed

Tie, Fall 2009

Show that $$f(z) = z^2$$ is uniformly continuous in any open disk $$|z| < R$$, where $$R>0$$ is fixed, but it is not uniformly continuous on $$\mathbb C$$.

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 $${\left\lvert {z^2-w^2} \right\rvert} < {\varepsilon}= c < \infty$$.

Non-continuously differentiable #complex/qual/completed

Give an example of a function $$f:{\mathbf{R}}\to {\mathbf{R}}$$ that is everywhere differentiable but $$f'$$ is not continuous at 0.

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 + uniformly continuous #complex/qual/completed

Suppose $$\left\{{g_n}\right\}$$ is a uniformly convergent sequence of functions from $${\mathbf{R}}$$ to $${\mathbf{R}}$$ and $$f:{\mathbf{R}}\to {\mathbf{R}}$$ is uniformly continuous. Prove that the sequence $$\left\{{f\circ g_n}\right\}$$ is uniformly convergent.

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 $${\varepsilon}$$.

Uniform differentiability #complex/qual/completed

Let $$f$$ be differentiable on $$[a, b]$$. Say that $$f$$ is uniformly differentiable iff

\begin{align*} \forall \varepsilon > 0,\, \exists \delta > 0 \text{ such that } \quad {\left\lvert {x-y} \right\rvert} < \delta \implies {\left\lvert { {f(x) - f(y) \over x-y} - f'(y)} \right\rvert} < {\varepsilon} .\end{align*}

Prove that $$f$$ is uniformly differentiable on $$[a, b] \iff f'$$ is continuous on $$[a, b]$$.

$$\implies$$: 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*}

Inf distance #complex/qual/completed

Suppose $$A, B \subseteq {\mathbf{R}}^n$$ are disjoint and compact. Prove that there exist $$a\in A, b\in B$$ such that \begin{align*} {\left\lVert {a - b} \right\rVert} = \inf\left\{{{\left\lVert {x-y} \right\rVert} {~\mathrel{\Big\vert}~}x\in A,\, y\in B}\right\} .\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 $$A \cap B = \emptyset$$.

Connectedness #complex/qual/completed

Suppose $$A, B\subseteq {\mathbf{R}}^n$$ are connected and not disjoint. Prove that $$A\cup B$$ is also connected.

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.

Pointwise and uniform convergence #complex/qual/completed

Suppose $$\left\{{f_n}\right\}_{n\in {\mathbb{N}}}$$ is a sequence of continuous functions $$f_n: [0, 1]\to {\mathbf{R}}$$ such that \begin{align*} f_n(x) \geq f_{n+1}(x) \geq 0 \quad \forall n\in {\mathbb{N}},\, \forall x\in [0, 1] .\end{align*} Prove that if $$\left\{{f_n}\right\}$$ converges pointwise to $$0$$ on $$[0, 1]$$ then it converges to $$0$$ uniformly on $$[0, 1]$$.

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 $$\sup_{x\in X} {\left\lvert {f_n(x)} \right\rvert} = {\left\lVert {f_n} \right\rVert}_{\infty } < {\varepsilon}$$.

#complex/qual/work

Show that if $$E\subset [0, 1]$$ is uncountable, then there is some $$t\in {\mathbf{R}}$$ such that $$E\cap(-\infty ,t)$$ and $$E\cap(t, \infty)$$ are also uncountable.

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

???

#complex/qual/work

Suppose $$f, g: [0, 1] \to {\mathbf{R}}$$ where $$f$$ is Riemann integrable and for $$x, y\in [0, 1]$$, \begin{align*} {\left\lvert {g(x) - g(y)} \right\rvert} \leq {\left\lvert {f(x) - f(y)} \right\rvert} .\end{align*}

Prove that $$g$$ is Riemann integrable.

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*}

Exercises

Show that $$f(x) = x^n$$ is uniformly continuous on any interval $$[-M, M]$$.

\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*}

Show $$f(x) = x^{-n}$$ for $$n\in {\mathbf{Z}}_{\geq 0}$$ is uniformly continuous on $$[0, \infty)$$.

\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 $$(a+b)^m \geq a^m + b^m$$

Show that $$f'$$ bounded implies $$f$$ is uniformly continuous.

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*}

Show that the Dirichlet function $$f(x) = \chi_{I \cap{\mathbf{Q}}}$$ is not Riemann integrable and is everywhere discontinuous.

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

#complex/exercise/completed #complex/qual/work #complex/qual/completed #complex/exercise/work #todo