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Let $f : \mathbb{R}^n \to \mathbb{R}^m$ be a continuous function. We suppose that there exist $c \in \mathbb{R}^m$ such that $$\lim_{\|x\| \to \infty} f(x) = c.$$ Show that $f$ is uniformly continuous.

We know that if $\forall \epsilon_1 > 0$, $\exists M > 0$ such that $\forall x \in \mathbb{R}^n\setminus B(0,M) $, we have $\|f(x) - c\| < \epsilon_1$.

We want to show that $\forall \epsilon_2 > 0$, $\exists \delta > 0$ such that $\forall x,\bar{x} \in \mathbb{R^n} $, we have $\|f(x) - f(\bar{x})\| < \epsilon_2$ as long as $\|x-\bar{x}\|<\delta$.

Let $x,\bar{x} \in \mathbb{R^n} $ and $\|x-\bar{x}\|<\delta$.

Take $\epsilon_1 = \frac{\epsilon_2}{2} $

$\implies \|f(x) - c\| < \frac{\epsilon_2}{2}$ as long as $\|x\|\geq M$ and $\|f(\bar{x}) - c\| < \frac{\epsilon_2}{2}$ as long as $\|\bar{x}\|\geq M$

I am stuck on this point. Which $\delta$ could I take to prove the uniform continuity of $f$?

Taj Mohamed Bandalandabad
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  • This is almost surely a duplicate -- this question (or variant of) has been asked multiple times. Have you searched the website already? – Clement C. Feb 06 '16 at 23:39
  • @ClementC. Take your time to look for a similar question which could respond to my question; I never found it. – Taj Mohamed Bandalandabad Feb 06 '16 at 23:41
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    What does it mean $x,\overline x\in [0,1]$ as $x,\overline x\in \mathbb R^n$ ? – Svetoslav Feb 07 '16 at 00:01
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    @TajMohamedBandalandabad This is almost exactly the same argument as for the one-dimensional version: http://math.stackexchange.com/questions/75491/how-does-the-existence-of-a-limit-imply-that-a-function-is-uniformly-continuous – Clement C. Feb 07 '16 at 00:36

2 Answers2

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The thing you have shown above proves the uniform continuity of $f$ outside the ball. Actually, it shows that there exist $M>0$ such that for any two points outside the ball of radius $M$, $$\|f(x)-f(y)\|<\epsilon_2.$$ Now, we need to show that continuous function $f$ is uniformly continuous for the ball $B(0,M)$.

Since the closure of $B(0,M)$ is closed, using the Heine–Cantor theorem we can say that $f$ is uniformly continuous for the the ball $B(0,M)$.

That proves your statement.

Michael Hardy
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Kerr
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Let $\epsilon>0$ is arbitrary.

From $\lim\limits_{\|x\|\to\infty}{f(x)}\to c$ it follows that $\forall \epsilon>0,\,\exists M>0:\|x\|\ge M \Rightarrow \|f(x)-c\|\leq \epsilon$. Chose this $M$, corresponding to our $\epsilon$ devided by $2$. In the ball $\|x\|\leq M+1$ (this $+1$ is to take care of the points $x,y$ from each side of the boundary of $B(0,M)$)the function $f$ is uniformly continuous, as a continuous function over the compact set $B(0,M+1)=\{x\in\mathbb R^n:\|x\|\leq M+1\}$. It is left to show that $f$ is uniformly continuous in $\mathbb R^n\setminus B(0,M)$. Here we use that $\|f(x)-c\|\leq \frac{\epsilon}{2}$: Let $x,y\in\mathbb R^n\setminus B(0,M)$. Then $\|f(x)-f(y)\|\leq \|f(x)-c\|+\|f(y)-c\|\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

Finally, from the uniform continuity in $B(0,M+1)$ for the arbitrary $\epsilon>0$ we have $\exists\delta_1>0: x,y\in B(0,M+1):\|x-y\|\leq \delta_1\Rightarrow \|f(x)-f(y)\|\leq \epsilon$. The $\delta$ for the whole $\mathbb R^n$ we chose as $\delta:=\min{\{\delta_1,1\}}$.

Svetoslav
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