Proving uniform continuity of $\frac{f(x)}{x^{\alpha}}$

Question

Suppose $A>0$,and let $f$ be function which is Lipschitz continuous in $[A,\infty)$. Prove that for all $\alpha \geq 1$, the function $\frac{f(x)}{x^{\alpha}}$ is uniformly continuous in $[A,\infty)$.

Could anyone help me with this problem? Thanks!

My try. When $\alpha=1$,then from the Lipschitz condition, we can get that $|f(x)|\leq |f(A)|+L|x-A|$. From this, we can claim there exists $L′$>0, such that $\frac{|f(x)|}{x} \leq L'$ and at last we get $$|\frac{f(y)}{y}-\frac{f(x)}{x}|\leq \frac {|f(y)-f(x)|}{y}+\frac{|x-y|}{y}\frac{|f(y)}{x}\leq L''|y-x|.$$

when $\alpha=1$,then from the Lipschitz condition ,we can get that$|f(x)|\leq |f(x)|+L|x-A|$;From this, we can claim there exists $L'>0$,such that $\frac{|f(x)|}{x} \leq L'$ — user510013, Dec 04 '17 at 12:15
when $\alpha=1$,then from the Lipschitz condition ,we can get that$|f(x)|\leq |f(x)|+L|x-A|$;From this, we can claim there exists $L'>0$,such that $\frac{|f(x)|}{x} \leq L'$,at last we can get $|\frac{f(y)}{y}-\frac{f(x)}{x}|\leq \frac {|f(y)-f(x)|}{y}+\frac{|x-y|}{y}\frac{|f(y)}{x}\leq L''|y-x|$ — user510013, Dec 04 '17 at 12:22

Robert Z · Answer 1 · 2017-12-05T07:07:55.827

1

As regards the case $\alpha=1$, you are on the right track. See also Prove that $\frac{f(x)}{x}$ is uniformly continuous in $[1, +∞)$ if $f$ is Lipschitz

For $\alpha>1$, since $f(x)/x$ is bounded, it follows that $$\lim_{x\to +\infty}\frac{f(x)}{x^{\alpha}}=\lim_{x\to +\infty}\frac{f(x)/x}{x^{\alpha-1}}=0.$$ Then use Prove that $f$ is uniform continuous

edited Dec 05 '17 at 07:07

answered Dec 04 '17 at 11:57

Robert Z

145,942

Is the conculsion of link right with the counterexample $sin(\frac{1}{x})$? – user510013 Dec 04 '17 at 12:23
Yes. Here $f$ is continuous on $[A,+\infty)$ (closed set). – Robert Z Dec 04 '17 at 12:25
but how to explain the counterexample $sin(\frac{1}{x})$ – user510013 Dec 04 '17 at 12:27
This is not a counterexample... – Robert Z Dec 04 '17 at 12:36
Thank you ,I am sorry for ingoring the fact that $sin(\frac{1}{x})$ is not uniformly continuously in open set$(A,\infty)$ – user510013 Dec 04 '17 at 12:39
Fine. I included your try under your question. It's better to show what you did. Welcome to MSE and please take a tour before asking new questions: https://math.stackexchange.com/tour – Robert Z Dec 04 '17 at 12:44
Thanks pretty much – user510013 Dec 04 '17 at 12:55

Eric Fisher · Answer 2 · 2017-12-06T14:56:29.167

Let $0 < A \le x_1 < x_2$. Then $f(x)$ is Lipschitz continuous if and only if $$ \dfrac{|f(x_1)-f(x_2)|}{|x_1-x_2|} \le K $$ for some constant $K$. Now consider $$ \dfrac{|f(x_1)/x_1^\alpha-f(x_2)/x_2^\alpha|}{|x_1-x_2|} $$ Since $\alpha \ge 1$, we may write $$ \dfrac{|f(x_1)/x_1^\alpha-f(x_2)/x_2^\alpha|}{|x_1-x_2|} < \dfrac{|f(x_1)-f(x_2)|/A^\alpha}{|x_1-x_2|} $$ Hence if $g(x)=f(x)/x^\alpha$, then $$ \dfrac{|g(x_1)-g(x_2)|}{|x_1-x_2|} \le KA^\alpha $$ So $g(x)$ is Lipschitz continuous, and every such function is uniformly continuous.

Proving uniform continuity of $\frac{f(x)}{x^{\alpha}}$

2 Answers2