Assignment 10
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw10/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Problem 1
Let $I$ be an interval. A function $f : I \to \mathbb{R}$ satisfies a Hölder condition with exponent $\alpha > 0$ if there exists a constant $C > 0$ such that for all $x, y \in I$, \(\vert f(x)-f(y)\vert \le C\vert x-y\vert ^\alpha.\)
(a) Prove that if $f$ satisfies a Hölder condition with $\alpha > 0$, then $f$ is uniformly continuous on $I$.
Let $\vert f(x)-f(y)\vert \le C\vert x-y\vert ^\alpha$ for all $x, y \in I$.
$\forall\varepsilon > 0$, take $\delta=(\varepsilon/C)^{1/\alpha} $. If $\vert x-y\vert < \delta$, then \(\vert f(x)-f(y)\vert \le C\vert x-y\vert ^\alpha < C \left(\frac{\varepsilon}{C}\right) = \varepsilon.\)
Thus $f$ is uniformly continuous on $I$.
(b) Prove that if $f$ satisfies a Hölder condition with $\alpha > 1$, then $f$ is constant.
For $h>0$, \(\left\vert \frac{f(x+h)-f(x)}{h}\right\vert \le \frac{C\vert h\vert ^\alpha}{\vert h\vert } = C\vert h\vert ^{\alpha-1}\)
As $h \to 0$, $C\vert h\vert ^{\alpha-1} \to 0$. Thus \(\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = 0\)
Hence $f’(x) = 0$ for all $x \in I$.
Therefore $f$ is constant on $I$.
Exercise 4.1.11
Suppose $ f : I \to \mathbb{R} $ is bounded, $ g : I \to \mathbb{R} $ is differentiable at $ c \in I $, and $ g(c) = g’(c) = 0 $. Show that $ h(x) := f(x)g(x) $ is differentiable at $ c $. Hint: You cannot apply the product rule.
$\exists M \in \mathbb{R}, \text{ such that } \forall x \in I, f(x) < M$,
\[\left\vert \frac{f(x)g(x) - f(c)g(c)}{x-c} \right\vert < M \cdot \left\vert \frac{g(x)-g(c)}{x-c} \right\vert \to 0\] \[\implies \lim_{x \to c} \frac{h(x)-h(c)}{x-c} = 0\]
Problem 3
Let $ f : \mathbb{R} \to \mathbb{R} $ be a differentiable function. Prove that $ f $ is Lipschitz continuous if and only if $ f’ $ is a bounded function.
$(\Rightarrow)$ $\forall x, y \in \mathbb{R}, \ \exists L \ \text{s.t.} \ \vert f(x) - f(y)\vert \le L \vert x - y\vert $.
\[\left\vert \frac{f(x) - f(y)}{x - y} \right\vert \le L \implies f'(x) = \lim_{y \to x} \left\vert \frac{f(x) - f(y)}{x - y} \right\vert \le L\]$(\Leftarrow)$ $\forall t \in \mathbb{R}, \ \exists M > 0 \ \text{s.t.} \ \vert f’(t)\vert \le M.$
By the Mean Value Theorem, $\exists c \in (x, y), f(y) - f(x) = f’(c)(y - x)$.
\[\left\vert \frac{f(y) - f(x)}{y - x} \right\vert = \vert f'(c)\vert \le M\]$\implies \vert f(y) - f(x)\vert \le M \vert y - x\vert , \ f \ \text{is Lipschitz continuous.}$
Exercise 4.2.9
Prove the following version of L’Hôpital’s rule. Suppose $ f : (a,b) \to \mathbb{R} $ and $ g : (a,b) \to \mathbb{R} $ are differentiable functions and $ c \in (a,b) $. Suppose that $ f(c) = 0 $, $ g(c) = 0 $, $ g’(x) \neq 0 $ when $ x \neq c $, and that the limit of $ f’(x)/g’(x) $ as $ x $ goes to $ c $ exists. Show that
\(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}.\)
$\exists$ a sequence ${x_n}_{n=1}^\infty$, such that $x_n \to c.$
By Cauchy’s mean value theorem, let $\xi_n \in (x_n,c)$ if $x_n < c, (f(x_n) - f(c)) g’(\xi_n) = (g(x_n) - g(c)) f’(\xi_n)$.
\[\frac{f(x_n) - f(c)}{g(x_n) - g(c)} = \frac{f(x_n)}{g(x_n)} = \frac{f'(\xi_n)}{g'(\xi_n)}\]As $n \to \infty, x_n \to c, \xi_n \to c,$
\(\lim_{n \to \infty} \frac{f(x_n)}{g(x_n)} = \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\)