# Proof that a periodic function is bounded and uniformly continuous. [duplicate]

I need to show that if $f:\mathbb{R}\to \mathbb{R}$ is continuous and $\forall x \in \mathbb R, f(x+1)=f(x)$, then:

• $f$ is bounded,
• $f$ is uniformly continuous,
• there exists $c\in \mathbb{R}$ such that $f(c)=f(c+\pi)$.
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## marked as duplicate by Yes, Jonas Meyer real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 17 '15 at 17:16

What are your thoughts, and what have you tried so far? – Cameron Buie Aug 2 '12 at 17:52
What can you say about $f$ restricted to $[0,1]$? – David Mitra Aug 2 '12 at 17:56
@DavidMitra If $f$ is restricted to $[0,1]$ then since $f$ is continuous on $[0,1]$, $f$ is bounded on $[0,1]$. – Kns Aug 2 '12 at 17:58
Yes, and also uniformly continuous (have you seen this theorem?). Now, can you use the periodicity of $f$ to show the first two parts of your general result? – David Mitra Aug 2 '12 at 18:00
@DavidMitra Since $x$ is any real number, Can i take $x=c+\pi$? – Kns Aug 2 '12 at 18:07

Let $$g : \begin{array}{ccc} \mathbb R & \to& \mathbb R\\ c &\mapsto &f(c+\pi) - f(c).\end{array}$$ That's a continuous function.

Now, let $c_0$ be a point where $f_{|[0,1]}$ has a minimum. Because of the periodicity, $c_0$ is a minimum for the whole of $f$. So $\forall x \in \mathbb R, f(x) \geq f(c_0)$. In particular $(x = c_0 + \pi)$, $g(c_0) \geq 0$.

The same argument with “maximum” instead of minimum gives a $c_1$ where $g(c_1) \leq0$.

The intermediate value theorem now gives a point $c$ on which $g$ vanishes, QED.

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$f$ is bounded in the compact [0,1] and as $f(x)= f(x+1), x \in \mathbb{R}, f$ is bounded.Now note that $f$ is uniformly bounde in the compacts $[0,1/2],[1/2,1]$. Then given $\epsilon>0$ there are $\delta_1, \delta_2>0$ such that $$|x-y|< \delta_1 \Rightarrow |f(x)-f(y)|< \epsilon ,x, y \in [0,1]$$ $$|x-y|< \delta_1 \Rightarrow |f(x)-f(y)|< \epsilon ,x, y \in [1/2,3/2]$$ As $f(x)= f(x+1), x \in \mathbb{R}, f$ take $\delta=\min\{\delta_1,\delta_2$. Hence, $f$ is uniformily continuous. After I do the rest.

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What about third part? – Kns Aug 2 '12 at 18:24
Why not using directly the uniform continuity on $[0,1]$? – PseudoNeo Aug 2 '12 at 19:06
Because I want include the case where $x \in [0,1]$ and $Y \in (-1/2,0]$. For example -1/4, 1/8. – user29999 Aug 2 '12 at 20:04
So, use the uniform continuity on $[-42,42]$... – PseudoNeo Aug 2 '12 at 20:12