MATH 131BH: Analysis (Honors)

Spring 2025, I. Kim

We follow UCLA’s 131BH curriculum; the syllabus is linked here. We follow Rudin’s Principles of Mathematical Analysis (3e). Please let me know at kennyguo@ucla.edu if you spot any errors. Thanks!

Homework Solutions: (in progress)

• Homework 1 (compactness)
• Homework 2 (continuity, connectedness)
• Homework 3 (differentiability)
• Homework 4 (integrability)
• Homework 5 (integrability)
• Midterm :)
• Homework 6 (sequences of functions, uniform convergence)
• Homework 7 (Arzela-Ascoli, Stone-Weierstrass)
• Homework 8 (multivariable differentiability)
• Homework 9 (inverse, implicit function theorem)

Homework 1

1. Show that in the metric space \((l^2, d_2)\), the set \(E:=\{x=(x(n))_{n \geq 1} : \sum_{n\geq1} n |x(n)|^2 \leq 1\}\) is compact.

Proof.

Let \((x_k)_{k \geq 1} \subseteq E\). We have for all \(k \geq 1\), \(\sum_{n\geq 1} n|x_k(n)|^2 \leq 1\), so namely for all \(n \geq 1\), we have \(|x_k(n)| \leq 1/\sqrt{n}\).

In particular, for a fixed \(n \geq 1\), we have that \((x_k(n))_{k \geq 1}\) is bounded, and by Bolzano-Weierstrass, there exists a convergent subsequence. Continuing along for all \(n\), by the diagonal argument, we construct a subsequence, call it \((x_{k_j}(n))_{j \geq 1}\), such that for all \(n \geq 1\), \(\lim_{j \rightarrow \infty} x_{k_j}(n)\) exists, call it \(x_\infty(n)\), and let \(x_\infty = (x_\infty(n))_{n\geq 1}\). Thus, we show a) \(x_\infty \in E\) and b) \(x_{k_j} \rightarrow_{j \rightarrow \infty} x_\infty\).

1. Note that: \[ \sum_{n\geq 1} n |x_\infty(n)|^2 = \sum_{n\geq 1} n \lim_{j\rightarrow \infty} |x_{k_j}(n)|^2 = \lim_{N\rightarrow \infty} \sum_{n=1}^N n \lim_{j\rightarrow \infty} |x_{k_j}(n)|^2 = \lim_{N\rightarrow \infty} \lim_{j\rightarrow \infty} \underbrace{\sum_{n=1}^N n |x_{k_j}(n)|^2}_{\leq 1 \text{ since } x_{k_j} \in E \text{ for all } j} \leq 1 \] and thus, \(x_\infty \in E\).

2. Let \(\varepsilon > 0\). For any \(N \geq 1\), note we have the inequality: \[ \sum_{n=N}^\infty |x_{k_j}(n) - x_\infty(n)|^2 \leq 2 \sum_{n=N}^\infty |x_{k_j}(n)|^2 + 2 \sum_{n=N}^\infty |x_{\infty}(n)|^2 \] Furthermore, for all \(x \in E\), we have: \[ N \sum_{n=N}^\infty |x(n)|^2 \leq \sum_{n=N}^\infty n|x(n)|^2 \leq 1 \implies \sum_{n=N}^\infty |x(n)|^2 \leq 1/N \] and so for all \(N\geq 1\), \(\sum_{n=N}^\infty |x_{k_j}(n) - x_\infty(n)|^2 \leq 2(2/N) = 4/N\).

With this, fix an \(N_\varepsilon\) s.t. \(4/N_\varepsilon < \varepsilon^2/2\), so that \(\sum_{n=N_\varepsilon}^\infty |x_{k_j}(n) - x_\infty(n)|^2 \leq \varepsilon^2/2\).

Now, since for \(n = 1, \ldots, N_\varepsilon - 1\), we have that \(x_{k_j}(n) \rightarrow_{j\rightarrow \infty} x_\infty(n)\), we can choose a \(j_\varepsilon\) such that \(\sum_{n=1}^{N_\varepsilon - 1} |x_{k_j}(n) - x_\infty(n)|^2 \leq \varepsilon^2/2\). Thus, let \(j \geq j_\varepsilon\), and we see \[ \sum_{n=1}^\infty |x_{k_j}(n) - x_\infty(n)|^2 = \sum_{n=1}^{N_\varepsilon -1} |x_{k_j}(n) - x_\infty(n)|^2 + \sum_{n=N_\varepsilon}^\infty |x_{k_j}(n) - x_\infty(n)|^2 < \varepsilon^2/2 + \varepsilon^2/2 = \varepsilon^2 \] and so \(d_2(x_{k_j}, x_\infty) < \varepsilon\). Thus, \(x_{k_j} \rightarrow_{j \rightarrow \infty} x_\infty\), and so \(E\) is sequentially compact, and thus, compact. \(\square\)

Remark. This is a proof that doesn’t rely explicitly on Heine-Borel, as we didn’t have this theorem at our disposal.

1. Show that if \((K, d)\) is sequentially compact, then \((K,d)\) has a countable base.
2. Using (a), show that every open cover of \(K\) has a countable subcover.

Proof.

1. One can show \(K\) is totally-bounded, that is for all \(r>0\), there exists a finite number of balls such that they form a cover of \(K\), i.e., there exists \(n \in \mathbb{N}\), \(x_1, \ldots, x_n \in K\) such that \(K \subseteq \bigcup_{i=1}^n B_r(x_i)\). To construct our countable base, for all \(k \in \mathbb{N}\), consider the finite cover of \(K\), \(\{B_{1/k}(x_i)\}_{i=1}^{n_k}\), and we claim then \(\bigcup_{k=1}^\infty \{B_{1/k}(x_i)\}_{i=1}^{n_k}\) is a countable (as countable union is countable) base for \(K\).

Let \(x_0 \in K\) and \(x_0 \in G\) open. Thus, there exists some \(N \in \mathbb{N}\) such that \(B_{1/N}(x_0) \subseteq G\). But by total boundedness, we also know we chose \(x_1, \ldots, x_{n_{2N}}\) such that \(x_0 \in \bigcup_{i=1}^{n_{2N}} B_{1/(2N)}(x_i)\). If \(x_0 \in \{x_1, \ldots, x_{n_{2N}}\}\), we are done, since \(B_{1/(2N)}(x_0)\) is a set in our base. If else, then there is a \(x_i\) such that \(x_0 \in B_{1/(2N)}(x_i)\). Let \(y \in B_{1/(2N)}(x_i)\). By triangle inequality, we have: \[ d(y, x_0) \leq d(y, x_i) + d(x_i, x_0) < 1/(2N) + 1/(2N) = 1/N \] and so \(y \in B_{1/(N)}(x_0)\), and we have that \(x_0 \in B_{1/(2N)}(x_i) \subseteq B_{1/(N)}(x_0) \subseteq G\), so our set \(\bigcup_{k=1}^\infty \{B_{1/k}(x_i)\}_{i=1}^{n_k}\) is in fact a (countable) base.

2. Let \(\{G_\alpha\}\) be an open cover for \(K\). Let \(x \in K\), so \(x \in G_\alpha\) open for some \(\alpha\). Using the countable base of balls in part (a), we have that there exists some \(n_x \in \mathbb{N}, x_i\) such that \(x \in B_{1/n_x}(x_i) \subseteq G_\alpha\). Since there are only countably many balls in this base, we can select at most countably many \(G_\alpha\) from \(\{G_\alpha\}\), so that for all \(x \in K\), \(x\) is still contained in one of them, and so we extract our desired countable subcover. \(\square\)

3. Show that \(K\) is sequentially compact if and only if every infinite subset of \(K\) has a limit/accumulation point in \(K\).

Proof.

\((\implies)\) Assume \(K\) is sequentially compact, and let \(A \subseteq K\) be infinite (if can’t find, vacuously true). Thus, we can extract a sequence \(\{a_n\}_{n\geq 1} \subseteq A\) of non-repeating terms. By sequential compactness, there exists a convergent subsequence, call it \(\{a_{k_n}\}_{n\geq 1} \rightarrow_{n\rightarrow \infty} a \in K\). Finally, note that \(a \in A'\). Indeed, let \(r > 0\), and by limit definition, there is an \(n_r \in \mathbb{N}\) such that for all \(n \geq n_r\), \(d(a_n, a) < r\), and so \(a_n \in B_r(a) \cap A \backslash \{a\}\) (since the \(a_n\) are distinct), and so namely it is nonempty, and \(a\) is an accumulation point of \(A\) in \(K\).

\((\impliedby)\) Assume for all infinite \(A \subseteq K\), \(A' \cap K \neq \emptyset\). Let \(\{a_n\}_{n\geq 1} \subseteq K\). If \(\{a_n\}_{n\geq 1}\) infinitely repeats a term, then this will be a (constant) convergent subsequence. Thus, assume doesn’t infinitely repeat terms, and so the set \(\{a_n : n \geq 1\}\) is infinite. By assumption, there exists an \(a \in A' \cap K\).

From this, we construct our convergent subsequence. We have \(B_1(a) \cap A \backslash \{a\} \neq \emptyset\), so extract \(a_{k_1}\). Then for \(n \geq 2\), extract \(a_{k_n}\) from \(B_r(a) \cap A \backslash \{a\} \neq \emptyset\), where \(r = \min\{1/n, d(a, a_{k_{n-1}})\}\). It follows that \(\{a_{k_n}\}_{n\geq 1}\) is a convergent subsequence, and thus, \(K\) is sequentially compact. \(\square\).

4. Let \(K, E \subseteq X\), where \(K\) is compact and \(E\) is closed in \((X,d)\).

1. If \(K \cap E = \emptyset\), then show that there is a constant \(c > 0\) such that \[ d(x, E) := \inf\{d(x,y) : y \in E\} \geq c \text{ for all } x \in K.\]
2. Is (a) true if \(K\) is only closed? Give a counterexample.

Proof.

1. Suppose for contradiction that for all \(c > 0\), there exists \(x \in K\) such that \(\inf\{d(x,y) : y \in E\} < c\). For all \(n \geq 1\), let \(c = 1/n\). Thus, there is a \(x_n \in K\) such that \(d(x_n, E) < 1/n\). In particular, \(1/n\) is not a lower bound, so there exists a \(y_n \in E\) such that \(d(x_n, y_n) < 1/n\). With this, we construct arbitrarily close \(\{x_n\}_{n\geq 1} \subseteq K, \{y_n\}_{n\geq 1} \subseteq E\).

Since \(K\) is sequentially compact, there exists a convergent subsequence \(\{x_{k_n}\}_{n\geq 1} \rightarrow x_0 \in K\). Since \(\{y_n\}_{n\geq 1}\) is arbitrarily close, we have that \(\{y_{k_n}\}_{n\geq 1} \rightarrow x_0 \in \bar{E}\) as well. Since \(E\) is closed, \(x_0 \in E\). This contradicts \(K \cap E\) being empty. \(\square\)

2. No. Compactness is necessary, and our counterexample relies on finding arbitrarily close sequences in \(K\) and \(E\), but a convergent subsequence is not necessarily guaranteed (say, approaches infinity) by compactness.

Take \((X,d) = (\mathbb{R}, |\cdot|)\) and consider \(K = \{n+1/n : n\geq 2\}\) and \(E = \{n : n \geq 2\}\). These sets are indeed closed, but neither is compact (consider the open cover of \(\varepsilon = 1/2\) balls around each point, which cannot be reduced to a finite subcover). Finally, note \(K \cap E = \emptyset\).

Considering \(\{x_n\}_{n \geq 2} \subseteq K, x_n = n + 1/n\) and \(\{y_n\}_{n \geq 2} \subseteq E, x_n = n\) (note these are arbitrarily close with no convergent subsequence). Thus, we see for all \(c > 0\), there exists an \(n\) such that \(|x_n - y_n| < c\), which gives the desired negation.

5. Let \(A\) be a subset of a complete metric space. Assume that for all \(\varepsilon > 0\), there exists a compact subset \(A_\varepsilon\) so that for any \(x \in A, d(x, A_\varepsilon) < \varepsilon)\). Show that \(\bar{A}\) is compact.

Proof.

We show sequential compactness. Let \((x_n) \subseteq \bar{A}\).

We first extract a sequence from \(A\) that is arbitrary close to \((x_n)\), so any subsequence will converge to the same limit. Note that for all \(n \geq 1\), \(x_n \in \bar{A}\), which means there exists \((a_{k_n}) \subseteq A\) such that \(a_{k_n} \rightarrow x_n\). In particular, there is an \(a_n \in A\) such that \(d(x_n, a_n) < 1/n\), and so we get \((a_n) \subseteq A\), and it suffices to find a convergent subsequence of this.

Consider \(A_1\), so for all \(a \in A\), \(d(a, A_1) < 1\). Thus, for all \(n \geq 1\), \(\inf\{d(a_n, y) : y \in A_1\} < 1\), so \(1\) is not a lower bound, so there exists \(y \in A_1\) such that \(d(a_n, y) < 1\). Consider the open cover of \(A_1\) given by \(\{B_1(y)\}_{y \in A_1}\), which admits a finite subcover, say \(\{B_1(y_i)\}_{i=1}^n\). Increase the radius and now consider \(\{B_{2}(y_i)\}_{i=1}^n\). We can now see for all \(n \geq 1\), \(a_n \in \bigcap_{i=1}^n B_{2}(y_i)\). This follows from triangle inequality, as there is a \(y \in A_1\) such that \(d(a_n, y) < 1\), and this \(y\) is within some \(B_1(y_i)\) for some \(i\), so \(a_n \in B_2(y_i)\).

In particular, since there are finitely many \(2\)-radius balls, by the pigeonhole principle, there exists a subsequence \((a_{k_n})\) such that \((a_{k_n}) \subseteq B_2(y_i)\) for some \(i\). By triangle inequality, we have for all \(m, n \geq 1\), \(d(a_m, a_n) < 4\).

We continue this process iteratively for all \(N \geq 1\), considering \(A_{1/N}\), extracting a subsequence from the previous subsequence, and by the diagonal argument, we find a subsequence, rename it \((a_{k_n})\), such that for all \(m, n \geq N\), \(d(a_{k_m}, a_{k_n}) < 4/N\). In particular, this sequence is Cauchy, so by completeness, it converges. Since the sequence is also in \(A\), we get its limit is in \(\bar{A}\). Coming back to our original sequence, we can take \((x_{k_n})\), which must converge to the same limit, and thus, we get that \(\bar{A}\) is (sequentially) compact. \(\square\)

Remark. One can again use Heine-Borel. Here is an alternative proof:

Since \(\bar{A}\) is closed and a subset of a complete metric space, we have that \(\bar{A}\) is itself complete. To show total-boundedness, let \(\varepsilon >0\). By assumption, there exists \(A_{\varepsilon/3} \subset A\) such that for all \(x \in A\), \(d(x, A_{\varepsilon/3}) < \varepsilon/3\). Consider the open cover of \(A_{\varepsilon/3}\), \(\{B_{\varepsilon/3}(a)\}_{a \in A_{\varepsilon/3}}\). By compactness, we reduce to finite subcover \(\{B_{\varepsilon/3}(a_i)\}_i^n\).

Now we claim \(\{B_{\varepsilon}(a_i)\}_i^n\) is a finite cover of \(\varepsilon\)-balls for \(\bar{A}\). Let \(x \in \bar{A}\), so there exists \((a_n)_{n \geq 1} \subseteq A\) such that \(a_n \rightarrow x\). In particular, there is some \(a' \in A\) such that \(d(a', x) < \varepsilon/3\). We also have that \(\inf \{d(a', a) : a \in A_{\varepsilon/3}\} < \varepsilon/3\), so there is some \(a \in A_{\varepsilon/3}\) such that \(d(a, a') < \varepsilon/3\). Finally, \(a \in \bigcup_{i=1}^n B_{\varepsilon/3}(a_i)\), so there is some \(i\) such that \(d(a, a_i) < \varepsilon/3\). By triangle inequality, we see: \[ d(x, a_i) \leq d(x, a') + d(a', a) + d(a, a_i) < \varepsilon \] and so \(x \in B_{\varepsilon}(a_i)\) and \(\bar{A} \subseteq \bigcup_{i=1}^n B_{\varepsilon}(a_i)\). Since \(\varepsilon\) was arbitrary, \(\bar{A}\) is thus totally bounded, and we conclude using Heine-Borel. \(\square\)

6. (Walter Rudin, pg. 44, #13) Construct a compact set \(A\) of \(\mathbb{R}\) such that \(A'\) is countable.

Proof.

One such set is \(A = \{0\} \cup \{1/n + 1/k : n\geq 1, k \geq n\}\). One can find the confirmation of this construction here.

7. (Walter Rudin, pg. 44, #16) Consider the metric space \((\mathbb{Q}, |\cdot|)\). Let \(E = \{x \in \mathbb{Q} : 2 < x^2 < 3 \}\). Show \(E\) is closed and bounded, but not compact. Determine if \(E\) is open in \(\mathbb{Q}\).

Proof.

(Bounded) Yes. Take \(M=3\).

(Closed) Yes. Let \(x \in \bar{E}\), so for all \(n \geq 1\), \(B^{\mathbb{Q}}_{1/n}(x) \cap E \neq \emptyset\). Thus, there is a \(y \in B^{\mathbb{Q}}_{1/n}(x) \cap E\). If \(y=x\), then \(x \in E\). Assume \(x>y\) (a similar argument applies for \(<\)). Then we have \(x^2 > y^2 > 2\) and \(x-1/n < y \implies (x-1/n)^2 < y^2 < 3 \rightarrow_{n \rightarrow \infty} x^2 \leq 3\). But since \(x \in \mathbb{Q}\), we must have \(x^2 < 3\), and so \(x \in E\). We see \(\overline{E} = E\), and \(E\) is closed.

(Compact) No. \(E\) is not sequentially compact as one can take a sequence of rationals in \(E\) that converges (in the reals) to \(\sqrt{2} \notin E\). Thus, any subsequence cannot converge within \(E\).

(Open) Yes. This follows by density of the rationals in the reals.

Homework 2

1. (Walter Rudin, pg. 100, #18) Every rational \(x\) can be written in the form of \(x = m/n\) where \(n>0\), and \(m,n\) are relatively prime. When \(x=0\), we let \(n=1\). Consider the function \(f\) defined on \(\mathbb{R}\) by: \[ f(x) = \begin{cases} 0 & \text{if } x \text{ irrational} \\ 1/n & \text{if } x = m/n \end{cases} \] Prove that \(f\) is continuous at every irrational point, and that \(f\) has a simple discontinuity at every rational point.

Proof.

Let \(x \in \mathbb{R} \backslash \mathbb{Q}\) so \(f(x) = 0\). Let \(\varepsilon > 0\). By Archimedean, there is some natural \(N\) such that \(1/(N+1) < \varepsilon < 1/N\). If we choose \(\delta\) so that for all \(t \in (x-\delta, x+\delta)\), \(t = m/n\) reduced-rational where \(n \geq N+1\), we will have \(f(t) \leq 1/(N+1)\) < $, showing continuity.

For \(n = 1, \ldots, N\), consider \(nx\), and by Archimedean, there exists some \(m_n \in \mathbb{Z}\) such that \(m_n/n \leq x < (m_n + 1)/n\). We set \[ \delta_n = \begin{cases} \min \{|m_n/n-x|, |(m_n+1)/n-x|\} &: x>m_n/n \\ 1/n &: x = m_n/n \end{cases} \] and choose \(\delta = \min\{\delta_i\}_{i=1, \ldots, N}\). Now, let \(t\) be such that \(0<|t-x|<\delta\). If \(t\) irrational, \(f(t) = 0 < \varepsilon\). If \(t\) rational, then by construction, its reduced fraction form must have a denominator greater than or equal to \(N+1\), as desired, so \(f(t) < \varepsilon\). Thus, \(f\) is continuous at all irrational \(x\).

For rational \(x\), we write as reduced form \(m/n\), and so \(f(x) = 1/n\). Note in the last part, we showed for any \(x \in \mathbb{R}\), \(f(t) \rightarrow_{t \rightarrow x} 0 \neq 1/n\), so namely \(f\) is not continuous for \(x\) rational. \(\square\)

2. (Walter Rudin, pg. 100, #23) Recall that \(f:(a,b)\rightarrow \mathbb{R}\) is said to be convex if \[ f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda)f(y)\] for \(x, y \in (a,b), \lambda \in (0,1)\). Prove:

1. Every convex function \(f\) is continuous.
2. Every increasing convex function of a convex function \(f\) is also convex.
3. If \(f\) is convex in \((a,b)\) and \(a<s<t<u<b\), then \[ \frac{f(t)-f(s)}{t-s} \leq \frac{f(u)-f(s)}{u-s} \leq \frac{f(u)-f(t)}{u-t}. \]

Proof.

1. Let \(x \in (a,b)\). Since \((a,b)\) open, we find \(s, u\) such that \(a<s<x<u<b\). For continuity, we will take the one-sided limits to \(x\) (from \(s\) and \(u\)) and show they must converge to \(f(x)\).

   Let \((t_n) \subset (x,u)\) such that \(t_n \rightarrow x\). Note that for all \(t_n\), we can thus write them as \(t_n = \lambda_n x+ (1-\lambda_n)u\), which gives \(\lambda_n = \frac{t_n - u}{x-u}\). By convexity, we have \(f(t_n) \leq \lambda_n f(x)+ (1-\lambda_n)f(u)\), and so taking limsup (as \(\lambda_n \rightarrow 1\)), we get that \(\text{limsup}_{n \rightarrow \infty} f(t_n)= f(x)\). Futhermore, we can also write \(x\) as an interpolation between \(t_n\) and \(s\), i.e. \(x = \lambda_n s + (1-\lambda_n)t_n\), and thus by convexity, \(f(x) \leq \lambda_n f(s) + (1-\lambda_n)f(t_n)\). Again, taking \(n\rightarrow \infty\), we have \(\lambda_n \rightarrow 0\), and not assuming \(f(t_n)\) converges yet, taking the liminf, we see: \(f(x) \leq \text{liminf}_{n \rightarrow \infty} f(t_n)\). Together, the (right-sided) limit must exist and equals \(f(x)\). A similar argument for the left-sided limit can be applied for taking any sequence from \((s, x)\) that converges to \(x\). Thus, \(f(t) \rightarrow f(x)\), showing continuity. \(\square\)

   2. Let \(g: f((a,b)) \rightarrow \mathbb{R}\) be increasing and convex. Let \(x,y \in (a,b), \lambda \in (0,1)\). Then: \[ f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y) \] and by monotonicity and convexity of \(g\), \[ g(f(\lambda x + (1-\lambda)y)) \leq g(\lambda f(x) + (1-\lambda)f(y)) \leq \lambda g(f(x)) + (1-\lambda)g(f(y)) \] and so \(g\circ f\) is convex. \(\square\)

   3. Set \(\lambda \in (0,1)\) such that \(t = \lambda s + (1-\lambda) u\). We see \(\lambda = \frac{t-u}{s-u} = \frac{u-t}{u-s}\). By convexity, we have: \[ f(t) \leq \frac{u-t}{u-s} f(s) + (1-\frac{u-t}{u-s}) f(u) \quad (\star) \] For the first inequality, we get from \(\star\): \[ f(t) \leq \frac{u-t}{u-s} f(s) + \frac{t-s}{u-s} f(u) \] \[ \implies f(t) - f(s) \leq \frac{-(t-s)}{u-s} f(s) + \frac{t-s}{u-s} f(u) \] \[ \implies \frac{f(t)-f(s)}{t-s} \leq \frac{f(u) - f(s)}{u-s}. \]

   For the second inequality, we get from \(\star\): \[-f(t) \geq \frac{t-u}{u-s} f(s) + (\frac{u-t}{u-s}-1) f(u) \] \[ \implies f(u) -f(t) \geq \frac{t-u}{u-s} f(s) + \frac{u-t}{u-s} f(u) \] \[ \implies \frac{f(u)-f(s)}{u-s} \leq \frac{f(u)-f(t)}{u-t}.\] \(\square\)

3. (Walter Rudin, pg. 100, #24) Assume \(f\) is a continuous real-valued function defined in \((a,b)\) such that \[ f\left(\frac{x+y}{2}\right) \leq \frac{f(x) + f(y)}{2} \] for all \(x, y \in (a,b)\). Prove that \(f\) is convex.

Proof.

Let \(x<y \in (a,b), \lambda \in (0,1)\). Consider \(k_1 = \frac{x+y}{2}\). Call \(z=\lambda x+ (1-\lambda)y\). We have two cases: 1) \(\lambda = 1/2\), i.e. \(k_1 = z\), and the convexity holds by assumption. 2) WLOG assume \(k_1 < z\). Then iterate with \(k_2 = \frac{k_1 + y}{2}\). Again, we have 2 cases: 1) \(\lambda = 1/4\), i.e. \(k_2 = z\). Then \[ f(z) \leq \frac{f(k_1)+f(y)}{2} \leq \frac{\frac{f(x)+f(y)}{2}+f(y)}{2} \leq \frac{1}{4}f(x) + \frac{3}{4}f(y)\] and so convexity holds. 2) \(k_2 \neq z\), and we iterate again.

From here, we can show via induction that if \(\lambda = m/2^n\) for any \(n,m \in \mathbb{N}, \lambda \in (0,1)\), this process will terminate and convexity will hold. We just showed the base cases, so assume for all \(m \in \mathbb{N}\), for some \(n \in \mathbb{N}\), \(\lambda = m/2^n\) satisfies convexity. Consider \(n+1\), and let \(m \in \mathbb{N}\). If \(m\) is even, then \(\lambda = \frac{m/2}{2^n}\) which is satisfied by the inductive hypothesis. If \(m\) is odd, consider \(\lambda_1 = \frac{(m-1)/2}{2^n}, \lambda_2 = \frac{(m+1)/2}{2^n}\), which are satisfied by the inductive hypothesis. Also note \(\lambda = \lambda_1/2 + \lambda_2/2\), so we can decompose \[ z= \lambda x+ (1-\lambda)y = \underbrace{(\lambda_1 x + (1-\lambda_1)y)}_{h(\lambda_1)} + \underbrace{(\lambda_2 x + (1-\lambda_2)y)}_{h(\lambda_2)} \] and by assumption, we have: \[ f(z) \leq \frac{f(h(\lambda_1)) + f(h(\lambda_2))}{2} \] and by the inductive hypothesis, we have: \[ \leq \frac{\lambda_1 f(x) + (1-\lambda_1)f(y) + \lambda_2 f(x) + (1-\lambda_2)f(y)}{2} \] \[ \leq \frac{\lambda_1 + \lambda_2}{2} f(x) + \left(1-\frac{\lambda_1 + \lambda_2}{2}\right)f(y) \] \[ \leq \lambda f(x) + (1-\lambda) f(y) \] which thus closes the induction.

Finally, we consider the case where the iteration infinitely repeats, or \(\lambda\) is not of the form \(m/2^n\). We define \(E:= \{\lambda ' \in (0,1) : \lambda' \text{ satisfies convexity with } x,y\}\). Using the \(k_n\)’s from before, we get a sequence that approaches \(\lambda x + (1-\lambda)y\) as \(n \rightarrow \infty\), and since all of these \(k_n\) were constructed using lambdas in \(E\), we also get a sequence \((\lambda '_n) \subseteq E\) such that \(\lambda '_n \rightarrow \lambda\), so \(\lambda \in \overline{E}\). It suffices to show \(E\) is closed. Note the function \(g(\lambda) = \lambda f(x) + (1-\lambda)f(y) - f(\lambda x + (1-\lambda)y)\) is continuous, and \(E = g^{-1}([0, \infty))\), i.e. the preimage of a closed set under continuous \(g\). \(\square\)

4. (Walter Rudin, pg. 100, #26) Suppose \(X, Y, Z\) are metric spaces, and \(Y\) is compact. Let \(f\) map \(X\) into \(Y\), let \(g\) is a continuous, one-to-one mapping of \(Y\) into \(Z\), and put \(h(x)=g(f(x))\) for \(x \in X.\)

1. Prove that \(f\) is uniformly continuous if \(h\) is uniformly continuous.
2. Prove also that \(f\) is continuous if \(h\) is continuous.
3. Show (by example) that the compactness of \(Y\) cannot be omitted from the hypotheses, even when \(X\) and \(Z\) are compact.

Proof.

1. Since \(g\) is continuous and \(Y\) is compact, \(g(Y)\) is compact, and since \(g\) is injective, \(g^{-1}:g(Y) \rightarrow Y\) is well-defined, and furthermore, continuous on a compact domain, so it is uniformly continuous. Taking the composition of uniformly continuous functions \(h, g^{-1}\) gives a uniformly continuous function. \(\square\)
   2. By part (a), we still have \(g^{-1}\) is continuous. Composing continuous functions is continuous. \(\square\)
   3. Consider \(X = [-1, 1]\) (compact), \[f(x) = \begin{cases} 1/x^2 &: x \neq 0 \\ 0 &: x=0 \end{cases}\], \(Y = \{0\} \cup [1, \infty)\) (not compact), \[g = \begin{cases} 1/\sqrt{y} &: y \in [1, \infty) \\ 0 &: y=0 \end{cases}\] (injective and continuous), and \(Z=[0,1]\) (compact). Indeed, \(h = g\circ f\) is uniformly continuous as \(h(x) = x\), but clearly \(f\) is not continuous, let alone uniformly. \(\square\)

5. Suppose \(f: E \rightarrow Y\) is uniformly continuous, where \(E \subseteq \mathbb{R}^k\) and \(Y\) is a metric space.

1. If \(E\) is bounded in \(\mathbb{R}^k\), prove that \(f(E)\) is bounded in \(Y\).
2. Is the statement true if \(\mathbb{R}^k\) is replaced by an arbitrary metric space \((X,d)\)?

Proof.

1. First observe that \(E\) bounded \(\implies\) \(\bar{E}\) closed and bounded in \(\mathbb{R}^k\), so \(\bar{E}\) is compact. Thus, it is totally bounded, and it holds that \(E\) itself is totally bounded (in \(\mathbb{R}^k\)) as well. Now we fix \(\varepsilon = 1\). By uniform continuity of \(f\), we get a \(\delta > 0\) such that for all \(p,q \in E\), if \(d_E(p,q) < \delta\), then \(d_Y(f(p), f(q)) < 1\) (1). By total boundedness, we can cover \(E\) with \(\{B_\delta(x_i)\}_{i=1}^n\), where \(n \in \mathbb{N}, x_i \in E\). We claim \(\{B_1(f(x_i)\}_{i=1}^n\) is a finite cover for \(f(E)\), thus showing it is bounded. Indeed, let \(y \in f(E)\), so \(f^{-1}(y) \subseteq E\). Thus, for all \(x \in f^{-1}(y), x \in B_\delta(x_i)\) for some \(i\), so \(d_E(x, x_i) < \delta\). By (1), we get \(d_Y(f(x), f(x_i)) < 1\), and since \(f(x) = y\), we have \(y \in B_1(x_i) \subseteq \{B_1(f(x_i)\}_{i=1}^n\). \(\square\)

   2. No, and we exploit that fact that boundedness may not imply total-boundedness in metric spaces other than \(\mathbb{R}^k\). Consider \(f:(\mathbb{R}, d_0) \rightarrow (\mathbb{R}, |\cdot|)\), where \(d_0\) is the discrete metric, and \(f(x) = x\). Set \(E = \mathbb{R}\). We have that \(E\) is bounded in \((\mathbb{R}, d_0)\) (take \(M = 1\)) and \(f\) is uniformly continuous (indeed, for any \(\varepsilon\), one can simply choose \(\delta = 1/2\)). But clearly, \(f(E) = \mathbb{R}\) is not bounded in \((\mathbb{R}, |\cdot|)\).

6. Show that if \(E\) is open and connected in \(\mathbb{R}^k\), then \(E\) is pathwise connected in \(\mathbb{R}^k\).

Proof.

Let \(x \in E\). Define \(A = \{a \in E : \text{there exists a path from $x$ to $a$}\}\). We show \(A\) is “clopen” (closed and open) and nonempty, and by connectedness, \(A = E\), showing path connectedness.

(Nonempty) \(A\) is nonempty as \(x \in A\).

(Open) Let \(a \in A\), so there is a path from \(x\) to \(a\). Since \(E\) open, there is also an \(r > 0\) such that \(B_r(a) \subseteq E\) (we show it also \(\subseteq A\)). Let \(z \in B_r(a)\). We can define a path from \(a\) to \(z\) by considering \(g:[0,1] \rightarrow E\) given by \(g(t) = \vec{a} + t(\vec{z} - \vec{a})\) (i.e. the linear interpolation, which is continuous). Then, stitching this together with our path from \(x\) to \(a\) gives us a path from \(x\) to \(z\), and thus, \(z \in A\). So \(B_r(a) \subseteq A\), and \(A\) is open.

(Closed) Consider the complement \(A^C = B = \{b \in E : \text{there does not exist a path from $x$ to $b$}\}\). Let \(b \in B \subseteq E\) open, so there is an \(r > 0\) such that \(B_r(b) \subseteq E\). Let \(z \in B_r(b)\). Then by before, we know there is a path from \(b\) to \(z\), but if there were to exist a path from \(x\) to \(z\), then we would also have a path from \(x\) to \(b\), contradicting \(b \in B\). Thus, \(z\) is also in \(B\), and so \(B = A^C\) is open, so \(A\) is also closed.

We conclude \(A = E\), and \(E\) is path-connected. \(\square\)

7. Let \(f: [a,b] \rightarrow \mathbb{R}\) be continuous. Suppose that \(f\) has a local maximum at \(x_1\) and \(x_2\), with \(x_1 < x_2\). Show that there must be a third point between \(x_1\) and \(x_2\) where \(f\) has a local minimum.

Proof.

First restrict \(f\) to the compact interval \([x_1, x_2]\). Thus, \(f\) attains a minimum \(x_3 \in [x_1, x_2]\) such that \(f(x_3) = \min\{ f(x) : x \in [x_1, x_2]\}\). Now, we just remove the endpoints. Since \(x_1\) is a local max, we have that there exists an \(\varepsilon >0\) such that if \(| y - x_1| < \varepsilon\), then \(f(y) \leq f(x_1)\). Furthermore, restrict this \(\varepsilon\) so that \(x_1+\varepsilon < x_2\). We have two cases:

Case 1: for all such \(y > x_1\), \(f(y) = f(x_1)\) (i.e. \(f\) is constant on this small interval). Then \(f\) is constant on \([x_1, x_1 + \varepsilon)\), and namely, take \(x^\star = x_1 + \varepsilon/2 (< x_2)\), and thus, for all \(y\) such that \(|y - x^\star| < \varepsilon/2\), \(f(y) \geq f(x^\star)\), and thus, it is a local min in \((x_1, x_2)\).

Case 2: there exists \(y > x_1\) such that \(f(y) < f(x_1)\). Namely, \(x_1\) cannot be a min on \([x_1, x_3]\), so \(x_1 \neq x_3\).

By a symmetric argument, assuming case 2 for both \(x_1\) and \(x_2\), we get that \(x_3\) cannot equal \(x_1, x_2\), and \(x_3 \in (x_1, x_2)\). Then taking \(\varepsilon = \min\{|x_1 - x_3|, |x_2 - x_3|\}\) confirms \(x_3\) is a local min. \(\square\).

Homework 3

1. Let \(f : (0,1) \rightarrow \mathbb{R}\) be differentiable and let \(c \in (0,1)\). Suppose that \(\lim_{x \rightarrow c} f'(x)\) exists and is finite. Show that this limit must be equal to \(f'(c)\).

Proof.

Suppose for contradiction that \(\lim_{x \rightarrow c} f'(x) = L \neq f'(c)\). Assume first that \(L < f'(c)\) (a similar argument can be done for \(>\)). Select \(\varepsilon = \frac{f'(c) - L}{2} > 0\). By limit definition, there exists a \(\delta >0\) such that for all \(x \in (0,1)\), if \(|x-c| < \delta\), then \(|f'(x) - L| < \frac{f'(c) - L}{2} \implies f'(x) < \frac{f'(c) + L}{2}\). Let \(x_0 < c\) such that \(|x_0 - c| < \delta\). Thus, we have for all \(x \in [x_0, c)\), \(f'(x) < \frac{f'(c) + L}{2} < f'(c)\) (1). Then consider the open interval \((f'(x_0), f'(c))\) which \(\frac{f'(c) + L}{2}\) is in. Let \(y (\frac{f'(c) + L}{2}, f'(c))\). By Intermediate Value Theorem, there exists an \(x \in (x_0, c)\) such that \(f'(x) = y\), but this contradicts (1). \(\square\)

2. Show that the function \[ f(x) = \sum_{n=0}^\infty \frac{1}{2^n} \sin (3^nx)\] is a) continuous in \(\mathbb{R}\), and b) not differentiable anywhere in \(\mathbb{R}\).

Proof.

1. Let \(c \in \mathbb{R}\). Let \((x_k) \subseteq \mathbb{R}\) such that \(x_k \rightarrow c\). Define the partial sums \(f_N(x) = \sum_{n=0}^N \frac{1}{2^n} \sin (3^nx)\). Since \(\sin\) is continuous, and linear combinations of continuous functions are continuous, we have \(f_N\) continuous for all \(N\). (Remark: If one has the tools of functional analysis, one can prove \(f_N\) is Cauchy in \((C(\mathbb{R}), d_\infty)\), and thus uniformly converges to \(f\), so \(f\) is continuous.)

   For now, let \(\varepsilon > 0\). First note: \[ |f(x) - f_N(x)| = |\sum_{n=N+1}^\infty \frac{1}{2^n} \underbrace{\sin (3^nx)}_{\leq 1}| \leq \sum_{n=N+1}^\infty \frac{1}{2^n} < \infty \] and by Cauchy, there exists an \(N \geq 1\) such that \(\sum_{n=N+1}^\infty \frac{1}{2^n} < \varepsilon/3\). Also, since \(f_N\) is continuous, we have that \(f_N(x_k) \rightarrow_{k \rightarrow \infty} f_N(c)\), and so there exists some \(k_\varepsilon \in \mathbb{N}\) such that for all \(k \geq k_\varepsilon\), \(|f_N(x_k) - f_N(c)| < \varepsilon/3\). By triangle inequality, we see for all \(k \geq k_\varepsilon\), \[ |f(x_k) - f(c)| \leq \underbrace{|f(x_k) - f_N(x_k)|}_{\text{Cauchy convergence}} + \underbrace{|f_N(x_k) - f_N(c)|}_{\text{continuity}} + \underbrace{|f_N(c) - f_(c)|}_{\text{Cauchy convergence}} < \varepsilon \] and so \(f(x_k) \rightarrow_{k \rightarrow \infty} f(c)\), showing continuity of \(f\) at any \(c \in \mathbb{R}\). \(\square\)

   2. This one makes me sad. I’ll see if I have the motivation to write up a proper solution to this one sometime. \(\square\)

3. Suppose \(f: [0,1] \rightarrow \mathbb{R}\) is continuous with \(f(0) = f(1) = 0\), and \(f\) is differentiable on \((0,1)\). Show that for any \(\lambda > 0\), there is an \(x \in (0,1)\) such that \(f'(x) = \lambda f(x)\).

Proof.

Let \(\lambda > 0\). We define \(g(t) = e^{-\lambda t}f(t)\), and so we also have \(g(0) = g(1) = 0\). By Mean Value Theorem, there is an \(x \in (0,1)\) such that \(g'(x)\cdot (1-0) = g(1) - g(0)\), and so \(g'(x) = 0\). By product rule, we thus have: \[g'(x) = e^{-\lambda t}f'(x) - \lambda e^{-\lambda t}f(x) = \underbrace{e^{-\lambda x}}_{>0}(f'(x) - \lambda f(x)) = 0\] \[ \implies f'(x) - \lambda f(x) = 0\] and so \(f'(x) = \lambda f(x)\). \(\square\)

4. (“Reverse MVT”) Suppose \(f: [0,1] \rightarrow \mathbb{R}\) is continuous, and \(f\) is differentiable on \((0,1)\).

1. Show that for any \(0 < c < 1\) such that \(f'(c)\) is not a max or min of \(f'\) in \((0,1)\), there are \(x_1, x_2 \in (0,1)\) such that \[ f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.\]
2. Is (a) false if \(f'(c)\) is the maximum of \(f'\) in \((0,1)\)? Give an example.

Proof.

1. Let \(c \in (0,1)\) such that \(f'(c)\) is not a max or min of \(f'\) in \((0,1)\). We define \(g(x) := f(x) - f'(c)\cdot x\), and thus, it suffices to find \(x_1 \neq x_2 \in (0,1)\) such that \(g(x_1) = g(x_2)\). Suppose for contradiction that for all \(x_1 \neq x_2\), we have \(g(x_1) \neq g(x_2)\), so namely, \(g\) is injective. Note also that \(g\) is continuous, as \(f\) is continuous. Thus, we have that \(g\) is monotone, and so we only have one of the following hold for all \(x \in (0,1)\): \(g'(x) \geq 0\) xor \(g'(x) \leq 0\). Furthermore, because \(c\) was chosen so that is was not an extremum of \(f'\), we have the there exist \(a,b \in (0,1)\) such that \(f'(a) < f'(c) < f'(b)\). But then note: \[ g'(a) = f'(a) - f'(c) < 0 \text{ and } g'(b) = f'(b) - f'(a) > 0 \] which contradicts monotonicty. \(\square\)

   2. Yes, and the proof in (a) breaks down when we try to find \(a,b\) as before. For an example, consider \(f(x) = -(x-0.5)^3\), \(f:[0,1] \rightarrow \mathbb{R}\). We have \(f\) continuous and differentiable (\(f' = -3(x-0.5)^2\)). But then consider \(x = 0.5\), which indeed, maximizes \(f'\) on \((0,1)\). However, we are unable to find \(x_1 \neq x_2\) such that \(f'(0.5) = 0 = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\), or equivalently, \(f(x_1) = f(x_2)\), due to the fact that \(f\) is injective.

5. Let \(F: \mathbb{R} \rightarrow \mathbb{R}\) be Lipschitz with constant \(L > 0\). Let \(f, g : [0, \infty) \rightarrow \mathbb{R}\) be continuous on \([0, \infty)\) and differentiable in \((0, \infty)\), and satisfy: \[f'(t) = F(f(t)), \quad g'(t) \leq F(g(t)) \quad \text{ for } t \in (0, \infty).\] Show that if \(g(0) \leq f(0)\), then \(g(t) \leq f(t)\) for all \(t >0\).

Proof.

Define \(h(t) := g(t) - f(t)\). Suppose for contradiction there is a \(t_1 \in (0, \infty)\) such that \(g(t_1) > f(t_1)\), so \(h(t_1)> 0\). If \(f(0) = g(0)\), we have \(h(0) = 0\), and if we have \(f(0) > g(0)\), we have \(h(0) > 0\). Note also that \(h\) continuous, so by IVT, there is a \(t \in (0, t_1)\) such that \(h(t) = 0\). Since there exists some \(t < t_1\) such that \(h(t) = 0\), choose \(t_0\) to be the largest of these values. In particular, we have for all \(t in (t_0, t_1)\), \(h(t) > 0\) (1). By Lipschitz, we have for all \(t \in [t_0, t_1)\), \[ |h'(t)| = |g'(t) - f'(t)| \leq |F(g(t)) - F(f(t))| \leq L(\underbrace{g(t) - f(t)}_{h(t)\geq 0}) = L\cdot h(t)\] so in particular, we have \(h'(t) \leq L\cdot h(t)\). Thus we “solve” the differential inequality: \[ h'(t) - L\cdot h(t) \leq 0\] \[\implies e^{-Lt}h'(t) - Le^{-Lt}h(t) \leq 0\] \[\implies (e^{-Lt}h(t))' \leq 0 \quad \forall t \in [t_0, t_1).\] Let \(H(t) = e^{-Lt}h(t)\), so namely, \(H(t)\) is decreasing on this interval, so we must have \(H(t_0) \geq H(t)\), and so \(h(t_0) \geq h(t)\) and thus, \(f(t) \geq g(t)\) for all \(t \in [t_0, t_1)\), which contradicts (1). \(\square\)

6. Let \(f, g\) with \(n\)-th differentiable in \((0,1)\), and suppose that for some \(c \in (0,1)\), we have \(f(c) = f'(c) = \ldots = f^{(n-1)}(c) = 0\) and \(g(c) = g'(c) = \ldots = g^{(n-1)}(c) = 0\), but that \(g^{(n)}(x)\) is never \(0\) in \((0,1)\).

1. Show that \(g^{(k)}(x)\) is not zero for \(x\) sufficiently close to \(c\) for \(0 \leq k \leq n-1\).
2. Show that \[ \lim_{x\rightarrow c} \frac{f(x)}{g(x)} = \frac{f^{(n)}(c)}{g^{(n)}(c)},\] if \(g(x) \neq 0\) for \(x \neq c\). Indicate where (a) is used.

Proof.

1. By Taylor, we have for all \(x \in (0,1)\backslash\{c\}\), there exists \(x_1 \in (x, c)\) such that \[ g^{(k)}(x) = \underbrace{\sum_{m=k}^{n-1} \frac{g^{(m)}(c)}{m!}(x-c)^m}_{=0} + \underbrace{\frac{g^{(n)}(x_1)}{n!}(x-c)^n}_{\neq 0} \] \[ \implies g^{(k)}(x) \neq 0 \] for all \(0 \leq k \leq n-1\).

   2. We can spam L’hopital’s Rule. We first check the assumptions. For all \(0 \leq k \leq n-1\),
   • (by part a) \(g^{(k)}(x) \neq 0\)
   • \(\lim_{x \rightarrow c} f^{(k)}(x) = f^{(k)}(c) = 0\)
   • \(\lim_{x \rightarrow c} g^{(k)}(x) = g^{(k)}(c) = 0\) (continuity of differentiable functions)

   and so (most of) the assumptions for L’hopital are satisfied, and we can conclude:

   \[ \lim_{x\rightarrow c} \frac{f(x)}{g(x)} = \ldots = \lim_{x\rightarrow c} \frac{f^{(k)}(x)}{g^{(k)}(x)} = \ldots = \lim_{x\rightarrow c} \frac{f^{(n-1)}(x)}{g^{(n-1)}(x)} \]

   if the last limit exists. We claim it does, and we see:

   \[ \lim_{x\rightarrow c} \frac{f^{(n-1)}(x)}{g^{(n-1)}(x)} = \lim_{x\rightarrow c} \frac{\frac{f^{(n-1)}(x)-f^{(n-1)}(c)}{x-c}}{\frac{g^{(n-1)}(x)-g^{(n-1)}(c)}{x-c}} \]

   as \(g^{(n-1)}(x) \neq 0\) (by part a) and \(f^{(n-1)}(c), g^{(n-1)}(c) = 0\). Since \(f,g\) are \(n\)-th differentiable, this evaluates to \(f^{(n)}(c)/g^{(n)}(c)\), and thus, so does \(\lim_{x\rightarrow c} \frac{f(x)}{g(x)}\). \(\square\)

7. (Walter Rudin, pg. 114, #6) Suppose \(f\) is continuous for \(x \geq 0\), \(f'(x)\) exists for \(x>0\), \(f(0) = 0\), and \(f'\) is monotonically increasing. Put \[ g(x) = f(x)/x \quad (x>0)\] and prove that \(g\) is monotonically increasing.

Proof.

It suffices to show \[ g'(x) = \frac{x f'(x) - f(x)}{x^2} \geq 0 \quad \forall x >0\] or equivalently, \[ x f'(x) - f(x) \geq 0 \quad \forall x >0. \] By MVT, there is a \(c \in (0,x)\) such that \[ f(x) - \cancel{f(0)} = f'(c) (x - 0)\] \[\implies f'(c) = f(x)/x\] \[\implies f'(x) \geq f(x)/x \quad \text{(by monotonicity)} \] which gives the desired result. \(\square\)

8. (Walter Rudin, pg. 114, #15) Suppose \(a \in \mathbb{R}\), \(f\) is twice-differentiable on \((a, \infty)\), and \(M_0, M_1, M_2\) are the least upper bounds on \(|f(x)|, |f'(x)|, |f"(x)|\) respectively on \((a, \infty)\). Prove that \[ M_1^2 \leq 4M_0M_2. \]

Proof.

Assume WLOG \(M_0, M_2\) are finite. By Taylor, we have for all \(x \in (a, \infty)\), if \(h>0\), there is some \(x_1 \in (x, x+2h)\) such that \[ f'(x) = \frac{1}{2h}(f(x+2h)-f(x)) - hf"(x_1) \] which means \[ |f'(x)| \leq |\frac{1}{2h}(f(x+2h)-f(x))| - |hf"(x_1)| \leq M_0/h + hM_2 \] and by sup definition, \[ M_1 \leq M_0/h + hM_2 \quad \forall h>0.\] Then, choosing \(h = \sqrt{M_0M_2} > 0\), we see: \[ h = \frac{2\sqrt{M_0M_2} \pm \sqrt{4M_0M_2 - 4M_0M_2}}{2M_2} \] \[ \implies M_0 + M_2h^2 - 2\sqrt{M_0M_2}h = 0 \] \[ \implies M_0/h + hM_2 = 2\sqrt{M_0M_2} \] and so we get \(M_1 \leq 2\sqrt{M_0M_2}\). Squaring gives the result. \(\square\)

Homework 4

1. Let \(f: [a,b] \rightarrow \mathbb{R}\) be a function such that \(L(x) := \lim_{y\rightarrow x} f(y)\) is well defined and finite for each \(x \in [a,b]\).

1. Show that \(L\) is continuous on \([a,b]\).
2. Show that the set \(\{x \in [a,b] : f(x) \neq L(x)\}\) is at most countable.

Proof.

1. Let \(x \in [a,b], \varepsilon > 0\). Because \(L(x) = \lim_{y \rightarrow x} f(y)\) exists, we can say there exists \(\delta > 0\) such that if \(0 < |x-y| < \delta\), then \(|f(y) - L(x)| < \varepsilon/2\). Choose this \(\delta\). Note for all \(x_1 \in B_\delta(x)\), \(L(x_1)\) exists as well, and so there exists \(\delta_2(x_1) > 0\) such that \(0 < |x_1 - y| < \delta_2(x_1)\) implies \(|f(y) - L(x_1)| < \varepsilon/2\).

   Let \(x_1 \in B_\delta(x)\). Note that \((B_\delta(x) \cap B_{\delta_2(x_1)}(x_1))\backslash \{x,x_1\} \neq \emptyset\). If \(x_1 = x\), the claim follows. If \(x_1 \neq x\), because \(B_\delta(x)\) is open, there is an \(r>0\) such that \(B_r(x_1) \subseteq B_\delta(x)\), and thus, there exists a \(y \neq x, x_1\) in \(B_\delta(x) \cap B_{\delta_2}(x_1)\). Thus, we can pull a \(y\) such that \(0 < |x-y| < \delta\) and \(0< |x_1 - y| < \delta_2(x_1)\). Thus, we see: \[ |L(x) - L(x_1)| \leq |L(x) - f(y)| + |f(y) - L(x_1)| < \varepsilon/2 + \varepsilon/2 = \varepsilon \] and so \(L\) is continuous.

   2. Call \(E = \{x \in [a,b] : f(x) \neq L(x)\}\). For all \(x \in E\), define the “gap” \(G(x) = |f(x) - L(x)| > 0\). Let \(\varepsilon > 0\) and let \(D_\varepsilon = \{x \in E : G(x) > \varepsilon\}\) be the set of all discontinuities with a gap greater than \(\varepsilon\). Thus, if we consider the countable union \(\bigcup_{n\geq1} D_{1/n}\), there will exist some \(n\) such that \(G(x) > 1/n\) for all \(x \in E\), and so \(\bigcup_{n\geq1} D_{1/n} = E\). It suffices to show \(D_\varepsilon\) is countable.

   Let \(x \in D_\varepsilon \subseteq E\), so \(L(x)\) exists. Thus, there is a \(\delta_x > 0\) such that for all \(t \in P_\delta(x) = (x-\delta_x, x) \cup (x, x+\delta_x)\), \(|f(t)-L(x)| < \varepsilon/2\). By triangle inequality, we have for all \(t_1, t_2 \in P_\delta(x)\), \(|f(t_1) - f(t_2)| < \varepsilon\). Consider \(t \in P_\delta(x)\). We claim \(G(t) \leq \varepsilon\), so \(t \notin D_\varepsilon\). Indeed, since \(P_\delta(x)\) is open, consider \(B_r(t) \subseteq P_\delta(x)\) for some \(r>0\), and we have for all \(t' \in B_r(t), |f(t')-f(t)|<\varepsilon\), and sending \(t' \rightarrow t\), we have that \(|L(t)-f(t)| = G(t) \leq \varepsilon\) and so \(t \neq D_\varepsilon\).

   Thus, each \(x \in D_\varepsilon\) has some \(P_{\delta_x}(x)\) such that no points in it are also in \(D_\varepsilon\). In particular, by density of \(\mathbb{Q}\), we can pick \(q_x \in (x-\delta_x, x) \cap \mathbb{Q}\), and it holds that for \(x \neq y \in D_\varepsilon\), \(q_x \neq q_y\), and thus we have an injective map from \(D_\varepsilon \rightarrow \mathbb{Q}\), so \(D_\varepsilon\) is at most countable, and we achieve our desired conclusion. \(\square\)

2. (Walter Rudin, pg. 138, #2) Suppose \(f \geq 0\), \(f\) continuous on \([a,b]\), and \(\int_a^b f(x) dx = 0\). Prove that \(f(x) = 0\) for all \(x \in [a,b]\).

Proof.

Suppose for contradiction there exists some \(x_0 \in [a,b]\) such that \(f(x_0) > 0\). By continuity of \(f\), there exists some \(\delta > 0\) such that for all \(x \in B_\delta(x_0)\), \(f(x) > 0\) as well. In particular, consider the interval \([\underbrace{x_0 - \delta/2}_{c}, \underbrace{x_0 + \delta/2}_{d}]\). This interval is compact and thus achieves its minimum on continuous \(f\), say at \(x^*\), with \(f(x^*)> 0\). Note that \(f\) is also integrable on the restriction \([c,d]\).

Thus, for any partition \(P = \{c=x_0, \ldots, x_n = d\}\), we have: \[ L(f,P) = \sum_{i=1}^n \inf_{[x_{i-1}, x_i]} f(x) \cdot \Delta x_i \geq \sum_{i=1}^n f(x^*) \cdot \Delta x_i = f(x^*)(d-c) > 0 \] and so \(\int_c^d f(x) dx = \sup_P L(f,P) > 0\). Finally note \(\int_a^c f(x) dx, \int_d^b f(x) dx \geq 0\), and so we have that \(\int_a^b f(x) dx > 0\), contradicting \(=0\). \(\square\)

3. (Walter Rudin, pg. 138, #5) Suppose \(f\) is a bounded real function on \([a,b]\), and \(f^2 \in \mathcal{R}\) on \([a,b]\). Does it follow that \(f \in \mathcal{R}\)? Does the answer change if we assume \(f^3 \in \mathcal{R}\)?

Proof.

No, consider \(f:[a,b]\rightarrow \mathbb{R}\) defined by \[ f(x) = \begin{cases} 1 &: x \in \mathbb{Q} \\ -1 &: x \in \mathbb{R} \backslash \mathbb{Q} \end{cases} \] which is not integrable, but \(f^2 \equiv 1\) is integrable. But if \(f^3 \in \mathcal{R}\), we have that \(f\) is integrable, which follows from Theorem 6.11 of Rudin, using the uniformly continuous function \(\phi(y) = y^{1/3}\). \(\square\)

4. (Walter Rudin, pg. 138, #8) Suppose \(f \in \mathcal{R}\) on \([a,b]\) for every \(b > a\) where \(a\) is fixed. Define \[ \int_a^\infty f(x) dx = \lim_{b \rightarrow \infty} \int_a^b f(x) dx \] if this limit exists (and is finite). We say the integral converges, and if replaced by \(|f|\), we say the integral converges absolutely. Assume that \(f(x) \geq 0\) and \(f\) decreases monotonically on \([1, \infty)\). Prove that \(\int_1^\infty f(x) dx\) converges if and only if \(\sum_a^\infty f(n)\) converges (i.e. the integral test).

Proof.

(\(\impliedby\)) Assume \(\sum_a^\infty f(n)\) converges. Fix some \(N \in \mathbb{N}\) and consider \(\int_1^{N+1} f(x) dx\). Consider the partition \(P_0 = \{1, 2, \ldots, N+1\}\). Because \(f\) is monotone-decreasing, we get that \(U(f, P_0) = \sum_{n=1}^N f(n)\), which means \[ \int_1^{N+1} f(x) dx = \inf_P U(f, P) \leq \sum_{n=1}^N f(n).\] Sending \(N \rightarrow \infty\), we get that \[ \int_1^\infty f(x) dx \leq \sum_{n=1}^\infty f(n) < \infty.\]

(\(\implies\)) Assume \(\int_1^\infty f(x) dx < \infty\). Fix some \(N \in \mathbb{N}\) and consider \(\int_1^N f(x) dx\). Consider the partition \(P = \{1, 2, \ldots, N\}\) and thus \(L(f,P) = \sum_{n=1}^{N-1} f(n+1) = \sum_{n=2}^{N} f(n)\), and so \[ \sum_{n=2}^{N} f(n) = L(f, P) \leq \int_1^N f(x) dx. \] Sending \(N \rightarrow \infty\), we get that \[ \sum_{n=1}^\infty f(n) \leq \int_1^\infty f(x) dx < \infty\] and since \(f(1)\) is finite, we get that the series converges. \(\square\)

5. Let \(f,g : [a,b] \rightarrow \mathbb{R}, f,g \in \mathcal{R}\). Show that if the set $A := {x : f(x) = g(x)} is dense in \([a,b]\) (i.e. \(\bar{A} = [a,b]\)), then \[ I_1 = \int_a^b f(x) dx = \int_a^b g(x) dx = I_2.\]

Proof.

It suffices to show that for all \(\varepsilon > 0\), we have \[ | I_1 - I_2 | < \varepsilon. \] Let \(\varepsilon > 0\). Since \(f,g\) Riemann integrable, there exists \(\delta > 0\) such that for all partitions \(P = \{a=x_0, \ldots, x_n = b\}\) with mesh\((P) < \delta\), and for all \(t_i \in [x_{i-1}, x_i]\), \[ | I_1 - \sum_{i=1}^n f(t_i) \Delta x_i | < \varepsilon/2 | \sum_{i=1}^n g(t_i) \Delta x_i - I_2| < \varepsilon/2. \] Fix some partition \(P = \{a=x_0, \ldots, x_n = b\}\) with mesh\((P) < \delta\). By density of \(A in [a,b]\), we can pick \(t'_i \in [x_{i-1}, x_i] \cap A \neq \emptyset\) such that \(f(t'_i) = g(t_i)\) for all \(i = 1, \ldots, n\). Then by triangle, we see: \[ | I_1 - I_2 | \leq | I_1 - \sum_{i=1}^n f(t'_i) \Delta x_i | + \underbrace{|\sum_{i=1}^n f(t'_i) \Delta x_i - \sum_{i=1}^n g(t_i) \Delta x_i|}_{=0} + | \sum_{i=1}^n g(t_i) \Delta x_i - I_2| < \varepsilon \] and since \(\varepsilon >0\) arbitrary, we conclude. \(\square\)

6. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be strictly monotone increasing, continuous, with \(f(0) = 0\) and \(f(\mathbb{R}) = \mathbb{R}\). Show that

1. \(f^{-1}\) is continuous
2. \[ \int_0^a f(x) dx + \int_0^b f^{-1}(x) dx \geq ab \quad \text{for any $a,b \geq 0$} \]
3. Using \(b\), let \(p,q > 0\) such that \(1/p + 1/q = 1\). Show if \(u, v \geq 0\), then: \[ uv \leq u^p/p + v^q/q\] with equality if and only if \(u^p = v^q\).

Proof.

1. Suppose for contradiction there is a \(y_0 \in \mathbb{R}\) such that \(f^{-1}\) is discontinuous at \(y_0\), i.e. there is a \((y_n)\) such that \(y_n \rightarrow y_0\), but \(f^{-1}(y_n) \cancel{\rightarrow} f^{-1}(y_0)\). Call \(x_n = f^{-1}(y_n), x_0 = f^{-1}(y_0)\). Because \(f\) is strictly monotone increasing, we have that \(f^{-1}\) is as well (Assume \(y_1 < y_2\), suppose \(f^{-1}(y_1) \geq f^{-1}(y_2)\), which means \(f(f^{-1}(y_1)) = y_1 \geq f(f^{-1}(y_2)) = y_2)\). Reduce \((y_n)\) to an increasing subsequence that still converges to \(y_0\), and thus \((x_n)\) is also increasing. Since \(y_n \leq y_0\) for all \(n\), \(x_n \leq x_0\) for all \(n\), and thus by monotone convergence, \(x_n \rightarrow x \leq x_0\). Finally, since \(y_n = f(x_n) \leq f(x)\), sending \(n \rightarrow \infty\) gives \(y_0 \leq f(x) \leq f(x_0)\) which means \(x_n \rightarrow x_0\), a contradiction, so \(f^{-1}\) is continuous. \(\square\)

   2. Let \(a,b \geq 0\).

   (Case 1: \(f(a) = b\)) We show we get equality in this case. Let \(P_1 = \{0=x_0, \ldots, x_n = a\}\) be a partition of \([0,a]\), and by monotonicity, \(P_2 = \{0=y_0 = f(x_0), \ldots, y_n = f(a) = b\}\) is a partition of \([0,b]\). Thus, by monotonicity, we see: \[ \begin{align} U(P_1,f) + L(P_2,f^{-1}) &= \sum_{i=1}^n \sup_{i} f(x) \Delta x_i + \inf_i f^{-1}(y) \Delta y_i \\ &= \sum_{i=1}^n f(x_i) (x_i - x_{i-1}) + x_{i-1} \Delta f(x_i) \\ &= \sum_{i=1}^n f(x_i) x_i - f(x_{i-1}) x_{i-1} \\ &= f(a)\cdot a - f(0) \cdot 0 \quad \text{(telescopic sum)} \\ &= ab. \end{align} \] By sup, we have \(U(P_1,f) + \sup_P L(P,f^{-1}) \geq ab\), and since \(P_1\) was picked arbitrarily, we also have \(\inf_P U(P,f) + \sup_P L(P,f^{-1}) \geq ab\). Note that we also could’ve started with an arbitrary partition of \([0,b]\) and gotten the same result, so it also follows that \(\inf_P U(P,f) + \sup_P L(P,f^{-1}) \leq ab\). Since \(f, f^{-1}\) are both integrable, we thus get that \(\int_0^a f dx + \int_0^b f^{-1} dx = ab\).

   (Case 2: \(f(a) < b\). Similar arguments can show \(>\), by switching around \(a,b\) and \(f, f^{-1}\).) We consider the partition of \([0,a]\), \(P_1 = \{x_0 = 0, \ldots, x_n = a\}\), and the partition of \([0,b]\), \(P_2 = \{0=y_0=f(x_0), \ldots, y_n=f(x_n)=f(a), y_{n+1}, \ldots, y_m = b\}\). We evaluate: \[ \begin{align} U(f, P_1) + L(f^{-1}, P_2) &= \sum_{i=1}^n \sup_{i} f(x) \Delta x_i + \inf_i f^{-1}(y) \Delta y_i + \sum_{i=n+1}^m \inf_i f^{-1}(y) \Delta y_i \\ &= \sum_{i=1}^n f(x_i) (x_i - x_{i-1}) + x_{i-1} \Delta f(x_i) + \sum_{i=n+1}^m \inf_i f^{-1}(y) \Delta y_i \\ &= f(a)\cdot a + \sum_{i=n+1}^m \underbrace{\inf_i f^{-1}(y)}_{\geq f^{-1}(y_n) = a \text{ by monotonicity}} \Delta y_i \quad \text{(telescopic sum as in case 1)} \\ &\geq f(a)\cdot a + a \sum_{i=n+1}^m \Delta y_i \\ &\geq f(a)\cdot a + a(b-f(a)) = ab. \end{align} \] By sup/inf manipulation again, we see: \[ U(P_1, f) + \sup_P L(P, f^{-1}) \geq ab \] and since \(P_1\) was arbitrarily picked, we also have: \[ \inf_P U(P, f) + \sup_P L(P, f^{-1}) \geq ab \] and so \(\int_0^a f + \int_0^b f^{-1} \geq ab\). \(\square\)

   3. From previous parts, we can choose the function \(f(x) = x^{p-1}\), which is strictly increasing, continuous, and \(f(0) = 0\). Also note that \(f^{-1}(x) = x^{q-1}\), as \(f(f^{-1}(x)) = x^{pq-p-q+1} = x\) by the definition of dual. Applying part b with \(u,v \geq 0\), we have: \[ \int_0^u x^{p-1} dx + \int_0^v x^{q-1} dx \geq uv \] and by power rule/FTC, \[ \frac{u^p}{p} + \frac{v^q}{q} \geq uv. \] For the equality, note that: \[ \begin{align} q + p = pq &\implies q+p = pq\underbrace{\frac{u^{\frac{p+q}{q}}}{u^p}}_{=1} \\ &\implies qu^p + pu^p = pqu^{\frac{p+q}{q}} \\ &\implies u^p/p + u^p/q = u^{\frac{p+q}{q}}. \end{align} \] Thus, \(u^p = v^q \iff v = u^{p/q} \iff u^p/p +v^q/q = uv.\) \(\square\)

7. Let \(f, \alpha : [a,b] \rightarrow \mathbb{R}\) be monotone increasing functions, \(f \in \mathcal{R}(\alpha)\) on \([a,b]\). Show that \(\alpha \in \mathcal{R}(f)\) on \([a,b]\) and \[ \int_a^b f d\alpha + \int_a^b \alpha df = f(b)\alpha(b) - f(a)\alpha(a). \]

Proof.

Assume \(f \in \mathcal{R}(\alpha)\). Let \(\varepsilon >0\). Thus, there is a partition \(P = \{a=x_0, \ldots, x_n =b\}\) such that: \[ \sum_{i=1}^n [\sup_i f(x) - \inf_i f(x)][\alpha(x_i)-\alpha(x_{i-1})] < \varepsilon \] and by monotonicity of \(f, \alpha\), we also get: \[ \sum_{i=1}^n [f(x_i) - f(x_{i-1})][\sup_i \alpha(x)- \inf_i \alpha(x)] < \varepsilon \] and so \(U(\alpha, P, f)-L(\alpha, P, f) < \varepsilon\) and \(\alpha \in \mathcal{R}(f)\).

To show the “integration-by-parts” formula, pick your favorite partition \(P_0 = \{a=x_0, \ldots, x_n =b\}\), and it follows by monotonicity and telescoping sums, \[ \begin{align} U(f, P_0, \alpha) + L(\alpha, P_0, f) &= \sum_i [f(x_i) \Delta \alpha_i + \alpha(x_{i-1}) \Delta f_i] \\ &= \sum_i [f(x_i) \alpha(x_i) - f(x_{i-1})\alpha(x_{i-1})] \\ &= f(b)\alpha(b) = f(a)\alpha(a) \end{align} \] and so namely, this holds for any partition of \([a,b]\). Thus we see, \[ \begin{align} &\inf_P U(f, P, \alpha) + L(\alpha, P_0, f) \leq f(b)\alpha(b) - f(a)\alpha(a) \\ &\implies L(\alpha, P_0, f) \leq f(b)\alpha(b) - f(a)\alpha(a) - \int_a^b f d\alpha \\ &\implies \sup_P L(\alpha, P, f) \leq f(b)\alpha(b) - f(a)\alpha(a) - \int_a^b f d\alpha \\ &\implies \int_a^b f d\alpha + \int_a^b \alpha df \leq f(b)\alpha(b) - f(a)\alpha(a) \\ & U(f, P_0, \alpha) + \sup_P L(\alpha, P, f) \geq f(b)\alpha(b) - f(a)\alpha(a) \\ &\implies \inf_P U(f, P, \alpha) \geq f(b)\alpha(b) - f(a)\alpha(a) - \int_a^b \alpha df \\ &\implies \int_a^b f d\alpha + \int_a^b \alpha df \geq f(b)\alpha(b) - f(a)\alpha(a) \\ \end{align} \] and so we get equality. \(\square\)

Homework 5

1. Let \(\alpha(x) = 0\) if \(x < 0\) and \(= 1\) if \(x \geq 0\). Suppose \(f : [-1, 1] \rightarrow \mathbb{R}\) is bounded and \(f \in \mathcal{R}(\alpha)\) on \([-1,1]\). Show that \(f\) is left continuous at \(x = 0\).

Proof.

Suppose for contradiction \(f\) is not left continuous at \(x=0\), i.e. there exists some \(\varepsilon >0\) such that for all \(\delta >0\), there exists \(x \in (-\delta, 0)\) such that \(|f(x) - f(0)| \geq \varepsilon\). Let \(P = \{-1=x_0,\ldots,x_n=1\}\) be a partition of \([-1,1]\). Thus, we have for some \(i = 1, \ldots, n\) that \(0 \in (x_{i-1}, x_i]\), and so take \(\delta = -x_{i-1}\) and by the supposition, we have some \(x' \in (x_{i-1}, 0)\) such that \(|f(x')-f(0)| \geq \varepsilon\). Finally, note that only \(\Delta \alpha_i = 1\) is nonzero. Thus, \[ U(f, P, \alpha) - L(f, P, \alpha) = \sup_i f(x) - \inf_i f(x) \geq \varepsilon \] and since our partition choice was arbitrary, we have that \(f \notin \mathcal{R}(\alpha)\), a contradiction. \(\square\)

2. Let \(\mathcal{C}\) be the Cantor set. Define \(f:[0,1]\rightarrow \mathbb{R}\) by \(f(x) = 1\) on \(\mathcal{C}\) and \(f(x) = 0\) if \(x \notin \mathcal{C}\). Show that \(f \in \mathcal{R}\) on \([0,1]\) (this is an example of a function with uncountable discontinuities, yet is still integrable).

Proof.

(This one probably makes more sense if you draw a picture of the layers :) ) Let \(\varepsilon >0\). Define \(\mathcal{C}_k\) to be the \(k\)-th “Cantor layer”, with initial \(\mathcal{C}_0 = [0,1]\) and for all \(k \geq 1\), $k = {x/3 : x C{k-1}} {(2+x)/3 : x C_{k-1}}, and so \(\mathcal{C} = \bigcap_k \mathcal{C}_k\).

Consider the sequence \(a_k = \frac{2^k(5)-2}{3^{k+1}}\) which goes to \(0\) as \(k \rightarrow \infty\). Choose \(k\) such that \(a_k < \varepsilon\), and consider \(\mathcal{C}_k\). Consider the partition of \([0,1]\), \(P = \{x_i = i/3^{k+1}: i = 0, 1, \ldots, 3^{k+1}\}\), and so we compute: \[ U(f, P) - L(f, P) = \sum_i [\sup_i f(x) - \inf_i f(x)]/3^{k+1} \quad (1)\] We split the indices \(i\) as such: if \([x_{i-1}, x_i] \cap \mathcal{C}_k \neq \emptyset\), then \(i \in A\), and if \(= \emptyset\), then \(i \in B\). For \(i \in B\), we have that all \(x \in [x_{i-1}, x_i]\) are not in \(\mathcal{C}_k\), and so are not in \(\mathcal{C}\), so \(f(x) = 0\) and \(\sup_i - \inf_i = 0\) for all \(i \in B\). For \(i \in A\), we have that for \(x \in [x_{i-1}, x_i]\), \(f(x)\) could be \(0\) or \(1\), so we bound \(\sup_i f - \inf_i f \leq 1\). Thus, \((1)\) becomes \(\leq \sum_{i\in A} 1/3^{k+1} = |A|/3^{k+1}\). Finally, note that \(|A| = 2^k(5)-2\) (as in \(\mathcal{C}_k\), there are \(2^k\) intervals, and each one overlaps with five partition segments with endpoints, minus the two intervals on the end of \([0,1]\). I recommend drawing a picture to see this, and one can prove this rigorously, but tediously, with induction.) Thus, \((1) \leq \frac{2^k(5)-2}{3^{k+1}} < \varepsilon\), and so \(f \in \mathcal{R}\). \(\square\)

3. (Walter Rudin, #10) Let \(p,q > 0\) be duals. Prove the following statements.

2. If \(f,g \in \mathcal{R}(\alpha)\), \(f, g \geq 0\), and \(\int_a^b f^p d\alpha = \int_a^b g^q d\alpha = 1\), then \(\int_a^b fg d\alpha \leq 1\).
3. (H"older’s Inequality) If \(f,g\) are real in \(\mathcal{R}(\alpha)\), then \[ \left| \int_a^b fg d\alpha \right| \leq \left(\int_a^b |f|^p \right)^{1/p}\left(\int_a^b |g|^q \right)^{1/q}. \]

Proof.

2. By (6c) from HW 4, we write: \[ \begin{align} \int_a^b fg d\alpha &\leq \int_a^b \left(\frac{f^p}{p} + \frac{g^q}{q} \right) d\alpha \\ &= 1/p + 1/q = 1. \end{align} \]

   3. First, assume \(\int |f|^p d\alpha = 0\), so the RHS is \(0\). Then we also have for all \(M>0\), \(\int M|f|^p d\alpha = 0\), and since \(g\) is bounded, we have \[ \int |g||f|d\alpha = 0 \implies |\int_a^b fg d\alpha| = 0. \] The same argument holds when \(\int |g|^q d\alpha = 0\), so now assume \(\int |f|^p d\alpha = r_f, \int |g|^q d\alpha = r_g >0\). Note that we have \[ \int \left( \frac{|f|}{\left(\int |f|^p d\alpha\right)^{1/p}}\right)^p d\alpha, \int \left( \frac{|g|}{\left(\int |g|^p d\alpha\right)^{1/q}}\right)^q d\alpha = 1. \] Thus, by part b, we have \[ \int \frac{|f||g|}{r_f^{1/p}r_g^{1/q}} d\alpha \leq 1 \implies \int |fg| d\alpha \leq r_f^{1/p}r_g^{1/q}. \] Noting that \(|\int fg d\alpha| \leq \int |fg| d\alpha\) finishes the proof. \(\square\)

4. (Walter Rudin, #15) Suppose \(f\) is real, continuously differentiable, defined on \([a,b]\), with \(f(a) = f(b) = 0\), and \[ \int_a^b (f(x))^2 dx = 1.\] Prove that \[ \int_a^b xf(x)f'(x) dx = -1/2\] and that \[ \int_a^b (f'(x))^2 dx \cdot \int_a^b x^2 (f(x))^2 dx \geq 1/4.\]

Proof.

By integration by parts, we see \[ \int_a^b x f(x)f'(x) dx = \underbrace{b\cdot \frac{1}{2}(f(b))^2 - a\cdot \frac{1}{2}(f(a))^2}_{=0} - \int_a^b 1/2 (f(x))^2 dx \] which equals \(-1/2\) by assumption. Then applying H"older’s inequality (see previous question) to \(xf(x)\) and \(f'(x)\), we get: \[ \left| \int_a^b xf(x)f'(x) d\alpha \right| \leq \left(\int_a^b |xf(x)|^2 \right)^{1/2}\left(\int_a^b |f'(x)|^2 \right)^{1/2} \] and so squaring and subbing in the first equality into the LHS, we get our result.

Midterm

1. (5pt) Let \((X,d)\) be a complete metric space. Suppose that \(A_n\) is a sequence of compact subsets in \(X\) such that \(d(x,A_n) \leq 1/n\) for any \(x \in X\). Show that \(X\) is compact.

Proof.

Remark: This is similar to Homework 1, Q5. We show sequential compactness, so let \((x_n) \subseteq X\).

2. (5pt) Let \((X,d)\) be a metric space, and consider \(f: X \rightarrow Y\). Show that \(f\) is continuous in \(X\) if and only if for any open subset \(O \subseteq Y\), the set \(f^{-1}(O)\) is open.

Proof.

3. (10pt) (True or False) If false, provide a counterexample. \(X\) denotes a metric space \((X,d)\).

1. If \(A\) is bounded in \(X\), then \(\bar{A}\) is compact in \(X\). False, consider the (open or closed) unit ball in \((\ell^2, d_2)\).
2. If \(f: X\rightarrow Y\) is continuous, then \(f(X)\) is closed in \(Y\). False, consider f(x) = arctan(x).
3. If \(A \subseteq B \subseteq X\) and if \(A\) is compact in \(B\), then \(A\) is compact in \(X\). True.
4. If \(f: \mathbb{R} \rightarrow \mathbb{R}\) is a monotone increasing function, then \(f\) has at most countable discontinuities. True.
5. If \(f:[0,1] \rightarrow \mathbb{R}\) is differentiable in \((0,1)\) and Lipschitz continuous in \([0,1]\), then \(\lim_{x \rightarrow 0^+} f'(x)\) exists. False, consider f(x) = x^2 sin(1/x), with f(0) = 0.
6. If \(f:[0,1] \rightarrow \mathbb{R}\) is nonnegative and satisfies \(f^2 \in \mathcal{R}(x)\) on \([0,1]\), then \(f \in \mathcal{R}(x)\) on \([0,1]\) as well. True.

4. (10pt) Let \(\alpha : [0,1] \rightarrow \mathbb{R}\) be monotone increasing and let \(f : [0,1] \rightarrow \mathbb{R}\) be bounded.

1. (5pt) Show that, if \(f \in \mathcal{R}(\alpha)\), then \(|f| \in \mathcal{R}(\alpha)\) and \[ | \int_0^1 f d\alpha | \leq \int_0^1 |f| d\alpha.\]
2. (5pt) Suppose also \(\alpha\) is differentiable at \(x_0 \in (0,1)\). Show that if \(f \in \mathcal{R}(\alpha)\) and if we let \(F(x) := \int_0^x f(x) d\alpha\) and if \(f\) is continuous at \(x_0\), then \(F\) is differentiable at \(x_0\) and \[ F'(x_0) = f(x_0)\alpha'(x_0). \]

Proof.

Homework 6

1. Construct \((f_n), (g_n)\) which converge uniformly on some set \(E\), but \((f_n \cdot g_n)\) does not converge uniformly on \(E\).

Proof.

Let \(E = \mathbb{R}\) and we define for all \(n \geq 1\), \(f_n(x) = x, g_n(x) = 1/n\). We have that \(f_n \rightarrow f\) uniformly, where \(f(x) = x\) and \(g_n \rightarrow 0\) uniformly. Note that \(f_n g_n = x/n \rightarrow 0\) pointwise on \(\mathbb{R}\) (fix some \(x\) and send \(n \rightarrow \infty\)), but not uniformly. Indeed, set \(\varepsilon = 1\) and let \(N \geq 1\). By taking \(x_0 = N\) means \(|f_n(x_0)g_n(x_0) - 0| = 1 \cancel{<} 1\), which negates uniform convergence. \(\square\)

4. Let \(f: [0,1] \rightarrow [0,1]\) be continuous with \(f(0) = 0\) and \(f(1) = 1\). Consider the sequence of functions \(f_n :[0,1] \rightarrow [0,1]\) defined by: \[ f_1 = f \text { and } f_{n+1} = f \circ f_n. \] Prove that if \((f_n)\) converges uniformly, then \(f(x) = x\) for all \(x \in [0,1]\).

Proof.

Assume that \(f_n \rightarrow g: [0,1] \rightarrow [0,1]\) uniformly. We claim first that \(f_{n+1} = f \circ f_n \rightarrow f \circ g\) uniformly. Indeed, let \(\varepsilon > 0\). First, we have \(f\) continuous on compact, so uniformly continuous, and there is a \(\delta > 0\) such that for all \(t_1, t_2\) such that \(|t_1 - t_2| < \delta, |f(t_1) - f(t_2)| < \varepsilon\). Second, by uniform convergence, we have there is an \(N_\delta \in \mathbb{N}\) such that for all \(n \geq N_\delta\), for all \(x \in [0,1]\), we have \(|f_n(x) - g(x)| < \delta\). Thus, \(|f(f_n(x)) - f(g(x))| < \varepsilon\), and so \(f_{n+1} \rightarrow f \circ g\) uniformly.

Now since we have both \(f_{n+1} \rightarrow g\) and \(\rightarrow f \circ g\) with respect to \(d_\infty\), by uniqueness of limit, we have \(f \circ g = g\). Finally, note that \(g\) is surjective. This is because \((f_n) \subseteq C([0,1])\) as compositions of continuous functions are continuous, and since \(f_n \rightarrow g\) uniformly, \(g\) is continuous as well. Since for all \(n\), \(f_n(0) = 0, f_n(1) = 1\), we have \(g(0) = 0, g(1) = 1\), and thus surjectivity is given by IVT.

With this, let \(y \in [0,1]\), so there is an \(x \in [0,1]\) such that \(g(x) = y\), which also equals \(f(g(x)) = f(y)\). \(\square\)

6. (Convolution Approximation) For each \(n \geq 1\), let \(f_n : [-1,1] \rightarrow \mathbb{R}\) be continuous, nonnegative, and assume that

1. _-1^1 f_n(y) dy = 1
2. For each \(c>0\), \((f_n)\) converges uniformly to \(0\) on \([-1, -c] \cup [c, 1]\). Show that for every \(g \in C([-1, 1])\), we have: \[ \lim_{n \rightarrow \infty} \int_-1^1 f_n(y)g(y) dy = g(0). \] (or \((f_n \star g)(0) \rightarrow g(0)\))

Proof.

Let \(g \in C([-1, 1])\). Let \(\varepsilon > 0\). Note: 1) Since \(g\) is continuous on compact domain, it is uniformly continuous and bounded (let \(|g| \leq M\)). In particular, we also have a \(\delta > 0\) such that for all \(t \in B_\delta(0)\), \(g(t) \in B_{\varepsilon/6}(g(0))\). 2) Thus, we also have \(f_n \rightarrow 0\) uniformly on \([-1, -\delta] \cup [\delta, 1]\), we have there is an \(n_\varepsilon \in \mathbb{N}\) such that for all \(n \geq n_\varepsilon\), for all \(x \in [-1, -\delta] \cup [\delta, 1]\), \(|f_n(x)| < \varepsilon/(6M)\). Thus, let \(n \geq n_\varepsilon\). We compare: \[ | \int_-1^1 f_n(y)g(y) dy - g(0) | \geq \\ \underbrace{| \int_-1^-\delta f_n(y)g(y) dy|}_{A} + \underbrace{| \int_-\delta^\delta f_n(y)g(y) dy - g(0) |}_{B} + \underbrace{| \int_\delta^1 f_n(y)g(y) dy |}_{C} \\ \]

For (A), we have: $$

$$

Homework 7

Homework 8

Homework 9

Last Updated: 6/27/2025, more updates coming (maybe) :)