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A function $f:\mathbb{R}\to \mathbb{R}$ is periodic if there exits $p>0$ such that $f(x+P)=f(x)$ for all $x\in \mathbb{R}$. Show that every continuous periodic function is bounded and uniformly continuous.

For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval $[x_0,x_0+P]$. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. However, I'm not allowed to state that theorem directly. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above?

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Suppose that $f$ has period one. Since $f:[0,2]\to\Bbb R$ is continuous, it is bounded, so $f$ is bounded all over $\Bbb R$ (why?). Also, $f:[0,2]\to\Bbb R$ is uniformly continuous, being continuous on a compact set. Thus, given $\varepsilon >0$ there exists $\delta>0$ such that, whenever $|x-y|<\delta,x,y\in[0,2]$, then $|f(x)-f(y)|<\varepsilon$. Pick arbitrary $x,y$, and assume $x<y$ with $|x-y|<\delta$. We may take $\delta <1$. I claim there is an integer $n$ such that $x-n,y-n\in [0,2]$. Then $|x-y|=|x-n-(y-n)|$ and $f(x-n)=f(x)$, $f(y-n)=f(y)$. Can you continue now?

Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.

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ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that! – Michael Tong Mar 4 '15 at 22:06

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