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Assume that f is cont. and that $f(x+h)=f(x), \forall h >0$ Show f is bounded and that it attains its min and max values.

Here's an attempt. Let $x_{0}$ be an arbitrary point in f's domain. Then f is cont. on $[x_{0},x_{0}+h]$ Since f is cont. on this closed interveral, f attains its min and max by the extreme value theorem. Hence, $m\leq|f(x)|$ and $|f(x)|\leq M$ Since $f(x)=f(x+h)$, I can just replace and hence f is bounded and cont. I feel like there should be more somewhere.

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    Idea is fine. You need to say that for any $x$, there exists an $x_1$ in the interval $[x_0,x_0+h)$ and an integer $n$ such that $x=x_1+nh$. Then $f(x)=f(x_1)$, and we are finished.2012-11-28
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    Did you really mean for **all** $h>0$? Because if so, $f$ is constant.2012-11-28

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