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This question is apparently easy.

The problem

Let $\phi$ a continuous and compact supported function in $\mathbb{R}^n.$ Find a compact ball that contains the support of $|\phi(x+h)-\phi(x)|$ for all $|h|>0$ sufficiently small, for example, for all $h$ with 0<|h|<1/2.

What a tried so far

Let $\vec{1}=(1,1,...,1)\in\mathbb{R}^n$. Put $\phi^+(x)=\phi(x+\vec{1})$ and $\phi^-(x)=\phi(x-\vec{1})$. Then the functions $\phi^+, \phi^-$ and $\phi$ have compact support. Let $B$ a compact ball centered in the origing and containing the support of these three functions. Given $0<|h|<1/2$, I claim that the support $S$ of $|\phi(x+h)-\phi(x)|$ is contained in $B$. In fact, let $x\in S$. Suppose, for the sake of contradiction, that $x\notin B$. Then $\phi^+(x)=\phi^-(x)=\phi(x)=0$. I stucked here.

The result is clear graphically. How can I proceed from here?

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    I tried too, instead of take $\vec{1}$ take $\vec{y}$ arbitrary with $|\vec{y}|=1$ and consider a the union of all balls that contains the support of $\phi(x+y)$ for all $y$ with $|y|=1$.... – 2017-02-14
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    a more interesting question is can one find a ball whose size goes to zero as $h$ goes to 0? otherwise, it is as guest points out simply consequence of the boundedness of compactly supported continuous functions – 2017-02-14
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    @MarkJoshi pretty interesting! – 2017-02-15

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Since $\phi$ is compactly supported, find $R > 0$ so that $supp(\phi) \subset B(0;R)$

Then $x+h \in B(0;R) \iff |x+h| < R \implies |x| < R + |h| < R + \frac{1}{2}$

So the ball $B(0;R + \frac{1}{2})$ will do the trick.

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    yeap, I got it! – 2017-02-15