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Let $f$ be a continuous function on $\mathbb R$ such that $\lim_{x\to 0} f(\frac{1}{x})$ exists. Show that $f$ is uniformly continuous on $\mathbb R$.

My proof is as follows:

Let $n = \frac{1}{x}$ and $\lim_{r\to \infty} f(n)$ = $l$.

There exists a $M$ such that if $n \geq M$, $|f(n) - l|<\epsilon$.

Suppose $f$ is not uniformly continuous on $\mathbb R$.

Then for all $\epsilon > 0$, there exists a $\delta$ such that for some n, u belonging to R, $|n-u|<\delta$ $\implies$ $|f(n) - f(u)| \geq \epsilon$. Pick $u = M$, then f0r any $n$ such that $|n-u|<\delta$, we have $|f(n) - l| \geq \epsilon$. Contradiction.

Hence, $f$ is uniformly continuous.

I don't really know how to prove this. Can someone help please. Thanks

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No, this really does not work. The second line of your proof makes no sense (what is $\epsilon$?). More seriously, the statement "$f$ is not uniformly continuous" is nothing at all as what you wrote. The negation of "for all $t$ exists $y$, $P(t,y)$" is not "for all $t$ exists $y$, not $P(t,y)$". – Andrés E. Caicedo Apr 24 '14 at 6:19
that is what my textbook write, battle introduction to real analysis – user136266 Apr 24 '14 at 7:18
ok is not "for any n,u", but "some n,u" – user136266 Apr 24 '14 at 7:19
@MartinSleziak: I'd argue that it's not just related, it's pretty much a duplicate (after one line of reasoning). – Najib Idrissi Apr 25 '14 at 10:54

(This is a relatively high-level answer, which may not be helpful to the OP but hopefully explains the underlying phenomenon at work.)

The condition that $\lim\limits_{x\to 0}f(1/x)$ exists means that we can extend $f$ to a continuous function on the one-point compactification of $\mathbb R$, and by compactness this extension (and hence $f$) is uniformly continuous.

Explicitly: Let $L=\lim\limits_{x\to 0}f(1/x)$. Define $g:S^1\to \mathbb R$ by $$g(\theta)=\begin{cases} f(\tan(\theta/2)) &\text{if } -\pi<\theta<\pi\\ L &\text{if } \theta=\pm \pi \end{cases}$$ Then $g$ is continuous, since $f$ and $\tan$ are continuous and as $\theta\to \pm \pi$ we have $\cot(\theta/2)\to 0$ so $$\lim\limits_{\theta\to \pm \pi}f\left(\frac{1}{\cot(\theta/2)}\right)=\lim\limits_{x\to 0}f(1/x)=L$$ and since $S^1$ is compact, it follows that $g$ is uniformly continuous. But $f = g(2\arctan(x))$ so is a composition of uniformly continuous functions, thus is uniformly continuous.

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Note: This is tthe negation of uniform continuity of a function $f:\Bbb R\rightarrow\Bbb R$:

There exists $\varepsilon>0$ such that for each $\delta>0$ there exists $x,y\in\Bbb R$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\geq\varepsilon$.

Proof: Now, let be $L=\lim_{x\rightarrow0} f(1/x)$. Let be $\varepsilon>0$. Then, there exists $\delta>0$ such that for each $x\in(-\delta,\delta)-\{0\}$ we have $|f(1/x)-L|<\varepsilon/2$.

Take a natural number $n>(1/\delta)+1$. Then, $f$ is uniformly continuous in the compact set $[-n,n]$ so there exists $\delta'\in(0,1)$ such for each $x,y\in[-n,n]$, if $|x-y|<\delta'$ then $|f(x)-f(y)|<\varepsilon$.

Now, take $x,y\in\Bbb R$ such that $|x-y|<\min(\delta,\delta')$. Let $A=(-\infty,-1/\delta)\cup(1/\delta,\infty)$. Let $B=[-n,n]$. We have tree options:

  • If $x,y\in A$, then $$|f(x)-f(y)|\leq|f(x)-L|+|f(y)-L|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$$
  • If $x,y\in B$, then $|f(x)-f(y)|<\varepsilon$, as we have already proved.
  • If, for example, $x\in A$ and $y\in B$, one of them is in $A\cap B$, so this is actually one of the former options. Indeed, suppose that $x\notin B$. Then we have that $|x|>1/\delta$ and $|x|>n$. Since $y\in B$, $|y|\leq n$. Then, $$|y|>|x|-|y-x|>n-\delta'>\frac1\delta+1-\delta'>\frac1\delta$$ so $y\in A\cap B$.
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why when we take $n > \frac{1}{\delta} +1$, then f is uniformly continuous? – user136266 Apr 25 '14 at 0:24
i can't really understand the your first case, can you explain more please thanks – user136266 Apr 25 '14 at 0:33

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