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Assume $f\in C[0,+\infty)$, and for all $a>0$, we have $$\lim_{x\to\infty}(f(x+a)−f(x))=0.$$ Prove that $f(x)$ is uniformly continuous.

One hint is that we can use Baire category theorem, but I still don't know how to use it. Maybe there is another way to answer this question, I'm not sure. Looking forward to your answer.

YuiTo Cheng
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ling
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1 Answers1

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Fix $\epsilon>0$. We want to find $\delta>0$ such that

$$|x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon$$

For every $N\in \Bbb N$, let $E_N=\{a\mid x\geq N\Rightarrow |f(x+a)-f(x)|\leq\epsilon/4\}$. $E_N$ is closed (by continuity of $f$) and $\bigcup_{N\in\Bbb N} E_N=[0,\infty)$. By Baire Category Theorem, at least one of them, say, $E_N$ contains a closed interval $[b,c]$. For $x,y\geq N+c$, without loss of generality, say $y\geq x$, if $|y-x|<c-b$, there always exists $z\geq N$ such that $[x,y]\subset[z+b,z+c]$. Then $|f(x)-f(y)|\le |f(x)-f(z)|+|f(y)-f(z)|=|f(z+d)-f(z)|+|f(z+e)-f(z)|\le \epsilon/2$ where $d,e\in [b,c]$. For $x,y\le N+c$, as $[0,N+c]$ is compact, $f $ restricted to $[0,N+c]$ is uniformly continuous, hence there exists $\delta'>0$ satisfing the requirements in the quote box. Let $\delta=\min(c-b,\delta')$, we are done.

YuiTo Cheng
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    Thank you very much.Now,wo can use this proposition to complete the question has been asked two years ago [With the condition $\lim_{x \to \infty} (f(x+a)−f(x))=0$, how to construct $h(x)$?](https://math.stackexchange.com/questions/2228454/with-the-condition-lim-x-to-inftyfxa-fx-0-how-to-construct-hx) – ling May 05 '19 at 02:12
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    @YuiToCheng: Very nice answer (+1) – RRL May 05 '19 at 08:28