If $f(x) \to 0$ as $\|x\| \to \infty$ then $f$ is uniformly continuous

Question

let $f: \mathbb R^n \to \mathbb R$ be continuous such that $f(x) \to 0$ as $\|x\| \to \infty$. how do I prove that $f$ is uniformly continuous on $\mathbb R^n$?

ok I saw a question here where $n = 1$ and i think it is the same idea but there is one thing i don't understand

let $\epsilon >0$ then there exists $M >0$ such that if $\|x\| > M$ then $|f(x)| < \epsilon/2$. and because the closed ball $B'(0,M)$ is compact $f$ is uniformly continuous there so there exist $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ if $\|x - y\| < \delta$

also if $|x| > M$ and $|y| > M$ then $|f(x) - f(y)| \le |f(x)| + |f(y)| < \epsilon$

QUESTION:

what if $|x| > M$ and $|y| \le M$?

thanks for help

Answer 1

Let $\epsilon >0$. There exists $M_\epsilon$ such that $$ |z|\geq M_\epsilon\Rightarrow |f(z)|\leq \epsilon/2. $$ In particular, $$ |x|\geq M_\epsilon\,\wedge\,|y| \geq M_\epsilon \Rightarrow|f(x)-f(y)|\leq \epsilon/2+\epsilon/2=\epsilon $$ Moreover, by the uniform continuity on $\overline{B(0,M_\epsilon+1)}$, there exists $1>\delta_\epsilon>0$ such that $$ \forall x,y\in \overline{B(0,M_\epsilon+1)}\quad |x-y|\leq \delta_\epsilon\Rightarrow |f(x)-f(y)|<\epsilon. $$ This cover all cases. In fact, if $|x-y|\leq\delta_\epsilon<1$, then either $|x|,|y|\geq M_\epsilon $ or $|x|,|y|\leq M_\epsilon +1$.

Answer 2

It will always be possible to choose $\;\delta>0\;$ so that both $\;x,\,y\;$ fulfill $\;|x|,\,|y|\le M\;$ or both $\;|x|,\,|y|>M\;$ whenever $\;\left\|x-y\right\|<\delta\;$ . Can you see how?

Answer 3

Since $\|f(x)\|\to 0$ as $x\to \infty$, for any $\varepsilon>0$ there is a compact set $K_\varepsilon$ such that $\|f(x)\|<\varepsilon$ outside $K_\epsilon$. So $f(x)$ is uniformly continuous over $K_\varepsilon$, since $f$ is a continuous function and $K_\varepsilon$ is a compact set, while outside $K_\varepsilon$ we have $\|f(x)-f(y)\|\leq 2\varepsilon$ by the triangle inequality.