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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. –  André Nicolas Nov 28 '12 at 1:39
    
Did you really mean for all $h>0$? Because if so, $f$ is constant. –  Cameron Buie Nov 28 '12 at 1:44
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1 Answer 1

I assume the domain of $f$ is a subset of an ordered field , perhaps $\mathbb{R}$. The function $f$ is constant. To see this select distinct $x_{0}$ and $x_{1}$ in the domain of $f$. Without loss of generality we have $x_{0} < x_{1}$. We have $x_{1} - x_{0} > 0$. But then $$f(x_{0}) = f(x_{0} + (x_{1} - x_{0})) = f(x_{1}).$$

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