Assignment 7
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw7/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Exercise 2.6.2
Suppose both $ \sum_{n=0}^\infty a_n $ and $ \sum_{n=0}^\infty b_n $ converge absolutely. Show that the product series, $ \sum_{n=0}^\infty c_n $ where
\[c_n = a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0\]also converges absolutely.
$\vert c_n\vert = \left\vert \sum_{k=0}^n a_k b_{n-k} \right\vert \leq \sum_{k=0}^n \vert a_k\vert \vert b_{n-k}\vert $, then
\[\sum_{n=0}^\infty \vert c_n\vert \leq \sum_{n=0}^\infty \sum_{k=0}^n \vert a_k\vert \vert b_{n-k}\vert = \sum_{k=0}^\infty \sum_{n=k}^\infty \vert a_k\vert \vert b_{n-k}\vert = \left( \sum_{k=0}^\infty \vert a_k\vert \right) \left( \sum_{m=0}^\infty \vert b_m\vert \right)\]Since $ \sum_{n=0}^\infty a_n $, $ \sum_{n=0}^\infty b_n $ converge absolutely, $ \sum_{n=0}^\infty \vert c_n\vert $ converges by comparison test.
Problem 2
Find all real numbers $ x $ so that the series converges.
(a) $ \sum_{n=0}^\infty 2^n x^n $
\[\limsup_{n \to \infty} \vert 2^n x^n\vert ^{1/n} = \limsup_{n \to \infty} \vert 2x\vert = \vert 2x\vert < 1 \Rightarrow x \in \left(-\frac{1}{2}, \frac{1}{2} \right)\]
(b) $ \sum_{n=0}^\infty n x^n $
\[\limsup_{n \to \infty} \vert n x^n\vert ^{1/n} = \limsup_{n \to \infty} \vert n\vert ^{1/n} \vert x\vert = \vert x\vert < 1 \Rightarrow x \in (-1, 1)\]
(c) $ \sum_{n=0}^\infty \frac{1}{(2n)!} (x - 10)^n $
\[\limsup_{n \to \infty} \left\vert \frac{(x - 10)^n}{(2n)!} \right\vert ^{1/n} = \limsup_{n \to \infty} \frac{\vert x - 10\vert }{((2n)!)^{1/n}} = 0 \Rightarrow x \in \mathbb{R}\]
(d) $ \sum_{n=0}^\infty n! x^n $
\[\limsup_{n \to \infty} \vert n! x^n\vert ^{1/n} = \limsup_{n \to \infty} (n!)^{1/n} \vert x\vert\]Since $ n! = 1 \cdot 2 \cdot \ldots \cdot n > \left( \frac{n}{2} \right)^{n/2} $, so
\((n!)^{1/n} > \left[ \left( \frac{n}{2} \right)^{n/2} \right]^{1/n} = \sqrt{\frac{n}{2}}\)Because $ \sqrt{\frac{n}{2}} \to \infty $, $ (n!)^{1/n} $ diverges. Thus $ \sum n! x^n $ converges only if $ x = 0 $.
Problem 3 (Cauchy-Schwarz Inequality)
Prove that if $ \sum \vert x_n\vert ^2 $ and $ \sum \vert y_n\vert ^2 $ converge, then the series $ \sum x_n y_n $ converges absolutely and
\(\sum_{n=1}^\infty x_n y_n \leq \left( \sum_{n=1}^\infty \vert x_n\vert ^2 \right)^{1/2} \left( \sum_{n=1}^\infty \vert y_n\vert ^2 \right)^{1/2}\)
Show that
\(\left\vert \sum_{n=1}^m x_n y_n \right\vert \leq \left( \sum_{n=1}^m \vert x_n\vert ^2 \right)^{1/2} \left( \sum_{n=1}^m \vert y_n\vert ^2 \right)^{1/2}\)By induction, base case $ m=1 $, $ x_1 y_1 = x_1 y_1 $. If this inequality holds for $ m = k $, then at $ m = k+1 $,
\[\begin{aligned} \left\vert \sum_{n=1}^{k+1} x_n y_n \right\vert &= \left\vert \sum_{n=1}^k x_n y_n + x_{k+1} y_{k+1} \right\vert \\ &\leq \left( \sum_{n=1}^k \vert x_n\vert ^2 \right)^{1/2} \left( \sum_{n=1}^k \vert y_n\vert ^2 \right)^{1/2} + \left( \vert x_{k+1}\vert ^2 \right)^{1/2} \left( \vert y_{k+1}\vert ^2 \right)^{1/2} \\ &\leq \left( \sum_{n=1}^k \vert x_n\vert ^2 + \vert x_{k+1}\vert ^2 \right)^{1/2} \left( \sum_{n=1}^k \vert y_n\vert ^2 + \vert y_{k+1}\vert ^2 \right)^{1/2} \\ &= \left( \sum_{n=1}^{k+1} \vert x_n\vert ^2 \right)^{1/2} \left( \sum_{n=1}^{k+1} \vert y_n\vert ^2 \right)^{1/2} \end{aligned}\]Then take limits $ m \to \infty $, since $ \sum \vert x_n\vert ^2 $, $ \sum \vert y_n\vert ^2 $ converge, $ \sum \vert x_n y_n\vert $ converges by comparison test. Therefore,
\[\sum_{n=1}^\infty \vert x_n y_n\vert \leq \left( \sum_{n=1}^\infty \vert x_n\vert ^2 \right)^{1/2} \left( \sum_{n=1}^\infty \vert y_n\vert ^2 \right)^{1/2}\]
Problem 4
Prove that every real number is a cluster point of the set of irrational numbers.
Assume for contradiction that there exists $ x \in \mathbb{R} $, $ \delta > 0 $, such that: \((x - \delta, x + \delta) \cap (\mathbb{R} \setminus \{\mathbb{Q},\{x\}\}) = \emptyset\)
That is, the interval $ (x - \delta, x + \delta) $ contains no irrational numbers.
Then $(x - \delta, x + \delta) \subseteq \mathbb{Q}$, taking the complement: \((-\infty,x - \delta] \cup [x + \delta,\infty) \cup \{\mathbb{Q},\{x\}\} = \mathbb{R}\)
That is $(-\infty,x - \delta] \cup [x + \delta,\infty) \cup {x} = \mathbb{R}\setminus\mathbb{Q}$. But since $ \mathbb{R} \setminus \mathbb{Q} $ is dense in $ \mathbb{R} $ (see Assignment 3), there must exist some $ y \in \mathbb{R} \setminus \mathbb{Q} $ such that $ y \in (x - \delta, x + \delta) $, which contradicts the equation.
Exercise 3.1.13
Suppose $ S \subseteq \mathbb{R} $, and let $ c $ be a cluster point of $ S $. Suppose $ f: S \to \mathbb{R} $ is bounded. Show that there exists a sequence ${x_n}{n=1}^{\infty}$ with $ x_n \in S \setminus {c} $ and $\lim{n \to \infty} x_n = c$, for which the sequence ${f(x_n)}_{n=1}^{\infty}$ converges.
Since $ c $ is a cluster point of $ S $, there exists a sequence ${y_n}$ such that $ y_n \in S \setminus {c} $ for all $ n \in \mathbb{N} $, and $\lim_{n \to \infty} y_n = c$.
Because $ f $ is bounded, there exists a constant $ M > 0 $ such that $ \vert f(x)\vert \leq M $ for all $ x \in S $. Hence, the sequence ${f(y_n)}$ is bounded, that is, $ \vert f(y_n)\vert \leq M $ for all $ n $.
By the Bolzano–Weierstrass Theorem, the sequence ${f(y_n)}$ has a convergent subsequence ${f(y_{n_k})}$, i.e. $\lim_{k \to \infty} f(y_{n_k}) = L$ for some $ L \in \mathbb{R} $.
Define the sequence ${x_k}$ by $ x_k = y_{n_k} $. Then:
- $ x_k = y_{n_k} \in S \setminus {c} $ for all $ k \in \mathbb{N} $, since ${y_n}$ lies in $ S \setminus {c} $,
- $\lim_{k \to \infty} x_k = \lim_{k \to \infty} y_{n_k} = c $, because ${y_n} \to c$, so does any subsequence,
- The sequence ${f(x_k)} = {f(y_{n_k})}$ converges (to $ L $).
Therefore, the sequence ${x_k}$ satisfies the required properties. Renaming the index, the sequence ${x_n}$ is the desired one.
Problem 6
Let $ S \subset \mathbb{R} $, let $ c $ be a cluster point of $ S $, and let $ f : S \to \mathbb{R} $.
(a) Assume $ \lim_{x \to c} f(x) $ exists. Prove that there exist $ B \geq 0 $ and $ \delta > 0 $ such that if $ x \in S $ and $ 0 < \vert x - c\vert < \delta $ then $ \vert f(x)\vert \leq B $.
Denote $ L := \lim_{x \to c} f(x) $. For all $ \varepsilon > 0 $, there exists $ \delta > 0 $ such that if $ x \in S $ and $ 0 < \vert x - c\vert < \delta $, then $\vert f(x) - L\vert < \varepsilon$.
Let $ \varepsilon = 1 $, choose $ B = \vert L\vert + 1 $, then
\(\vert f(x)\vert = \vert f(x) - L + L\vert \leq \vert f(x) - L\vert + \vert L\vert < 1 + \vert L\vert = B\)
(b) Assume that $ \lim_{x \to c} f(x) = L > 0 $. Prove that there exists $ \delta > 0 $ such that if $ x \in S $ and $ 0 < \vert x - c\vert < \delta $, then $ f(x) > 0 $.
For all $ \varepsilon > 0 $, there exists $ \delta > 0 $ such that if $ x \in S $ and $ 0 < \vert x - c\vert < \delta $,
\(\vert f(x) - L\vert < \varepsilon \Rightarrow L - \varepsilon < f(x) < L + \varepsilon\)Since $ L > 0 $, let $ \varepsilon = \frac{L}{2} > 0 $, then
\(L - \varepsilon = \frac{L}{2} > 0 \Rightarrow f(x) > 0\)