Assignment 9
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw9/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Exercise 3.3.11
Find an example of a continuous bounded function $f : \mathbb{R} \to \mathbb{R}$ that does not achieve an absolute minimum nor an absolute maximum on $\mathbb{R}$.
Let
\[f(x) = \frac{1}{1 + e^{-x}}\]Then $f(x)\in (0, 1)$ for all $x \in \mathbb{R}$. $f$ is continuous on $\mathbb{R}$, $f$ is bounded, since $0 < f(x) < 1$, but $f$ does not attain a global minimum or maximum because the values 0 and 1 are never reached.
Exercise 3.4.3
Show that $f : (c,\infty) \to \mathbb{R}$ for some $c > 0$ defined by $f(x) = \frac{1}{x}$ is Lipschitz continuous.
For $x, y > c > 0$, \(\left\vert \frac{1}{x} - \frac{1}{y}\right\vert = \frac{\vert x-y\vert }{\vert xy\vert }\le \frac{1}{c^2} \vert x-y\vert\)
Therefore $f$ is Lipschitz continuous on $(c,\infty)$ with Lipschitz constant $K = \frac{1}{c^2}$.
Exercise 3.4.8
Show that $f : (0,\infty) \to \mathbb{R}$ defined by $f(x) = \sin(1/x)$ is not uniformly continuous.
Suppose $f(x) = \sin(1/x)$ is uniformly continuous. Then for every $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x, y \in (0,\infty)$ with $\vert x-y\vert < \delta$, we have $\vert \sin(1/x) - \sin(1/y)\vert < \varepsilon.$
Consider sequences ${x_n}{n=1}^{\infty},{y_n}{n=1}^{\infty}$ with $x_n = \frac{1}{2n\pi + \pi/2}$, $\quad y_n = \frac{1}{2n\pi - \pi/2}$.
Let $\varepsilon = 1$, choose $M = \frac{1}{2}\sqrt{\frac{1}{\pi\delta} + \frac{\pi^2}{4}}$. For $n > M$, we have
\(\vert x_n - y_n\vert = \left\vert \frac{-\pi}{(2n\pi + \pi/2)(2n\pi - \pi/2)}\right\vert =\frac{1}{\pi (4n^2-\pi^2/4)} < \delta.\)However,
\(\vert f(x_n) - f(y_n)\vert = \vert \sin(2n\pi + \pi/2) - \sin(2n\pi - \pi/2)\vert = 2 > 1 = \varepsilon.\)This contradicts the assumption of uniform continuity. Therefore, $f(x) = \sin(1/x)$ is not uniformly continuous on $(0,\infty)$.
Problem 4
Let $S \subset \mathbb{R}$. We say that $f : S \to \mathbb{R}$ is Lipschitz continuous on $S$ if there exists $L \ge 0$ such that for all $x, y \in S$, \(\vert f(x) - f(y)\vert \le L \vert x - y\vert .\)
Prove that if $f : S \to \mathbb{R}$ is Lipschitz continuous on $S$, then $f$ is uniformly continuous on $S$.
We want to show that for every $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x, y \in S$, if $\vert x-y\vert < \delta$, then $\vert f(x)-f(y)\vert < \varepsilon$.
Since $f$ is Lipschitz continuous, $\vert f(x)-f(y)\vert \le L\vert x-y\vert $.
Choose $\delta = \varepsilon/L$. Then if $\vert x-y\vert < \delta$,
\(\vert f(x)-f(y)\vert \le L\vert x-y\vert < L \cdot \frac{\varepsilon}{L} = \varepsilon.\)Thus $f$ is uniformly continuous on $S$.
Problem 5
(a) Prove that $f(x) = \cos x$ is Lipschitz continuous on $\mathbb{R}$.
For all $x, y \in \mathbb{R}$, \(\vert \cos x - \cos y\vert = 2\left\vert \sin\!\left(\frac{x+y}{2}\right)\sin\!\left(\frac{x-y}{2}\right)\right\vert \le 2\left\vert \sin\!\left(\frac{x-y}{2}\right)\right\vert \le 2 \cdot \frac{\vert x-y\vert }{2} = \vert x-y\vert\)
Thus $f(x) = \cos x$ is Lipschitz continuous with Lipschitz constant $L = 1$.
(b) Prove that $f(x) = x^{1/3}$ is uniformly continuous on $[0,1]$ and is not Lipschitz continuous on $[0,1]$.
On the closed, bounded interval $[0,1]$, the function $f(x) = x^{1/3}$ is continuous, hence by the Heine–Cantor theorem it is uniformly continuous.
To check Lipschitz continuity, consider $y=0$ and $x \to 0^+$: \(\frac{\vert f(x) - f(0)\vert }{\vert x - 0\vert } = \frac{x^{1/3}}{x} = \frac{1}{x^{2/3}} \to \infty\)
Thus there is no finite $L$ such that $\vert f(x)-f(y)\vert \le L\vert x-y\vert $ for all $x,y\in[0,1]$. Therefore $f$ is not Lipschitz continuous on $[0,1]$.
Problem 6
Let $R \in \mathbb{R}$ and $f : [R,\infty) \to \mathbb{R}$. We say that $f(x)$ converges to $L$ as $x \to \infty$ if for every $\varepsilon > 0$ there exists $M \ge R$ such that for all $x \ge M$ we have $\vert f(x) - L\vert < \varepsilon$. We write
\(\lim_{x \to \infty} f(x) = L.\)
(a) Prove that
\(\lim_{x \to \infty} \frac{x^2}{x^2 + 1} = 1.\)
For $\varepsilon > 0$, choose $M = \sqrt{1/\varepsilon}$. For all $x > M$, \(\left\vert \frac{x^2}{x^2+1} - 1\right\vert = \frac{1}{x^2+1} < \frac{1}{M^2} \le \varepsilon\)
Thus $\frac{x^2}{x^2+1} \to 1$ as $x \to \infty$.
(b) Prove that
\(\lim_{x \to \infty} \sin x\)
does not exist.
Suppose the limit exists. Take the sequence $x_n = 2n\pi + \frac{\pi}{2}$. Then $\sin x_n = 1$. Take another sequence $y_n = 2n\pi + \frac{3\pi}{2}$. Then $\sin y_n = -1$.
Both $x_n$ and $y_n$ tend to $\infty$, but $\sin x_n \to 1$ and $\sin y_n \to -1$. Hence the limit $\lim_{x \to \infty} \sin x$ does not exist.