# Boundedness and Uniform Continuity

Prove that a continuous periodic function on $\mathbb{R}$ is bounded and uniformly continuous on $\mathbb{R}$.

Given the continuous periodic function $f:\mathbb{R} \to \mathbb{R}$ for some period $p>0$ and $I:=[0,p]$, then (a) $f(x)=f(x+p)$ for all $x\in \mathbb{R}$ and (b) according to the Boundedness Theorem, $f$ is bounded on $I$, that is, $M>0$ is such that $|f(x)|\leq M$ for all $x\in I$. How to show that when $x\in \mathbb{R}\setminus I$, $f$ is bounded?

Because function is periodic with period $T\ne0$, we can consider function on $[0,T]$. If it not bounded on R, it must not bounded on $[0,T]$, but continuous function on closed interval is bounded on it. So, we prove first part. By the Cantor's theorem, continuous function on closed interval is uniformly continuous on it. So, $$\forall \epsilon>0 \exists \delta: \forall x_1, x_2\:|x_1-x_2|<\delta => |f(x_1)-f(x_2)|=|f(x_1-[\frac{x_1}{T}*T]-f(x_2-[\frac{x_2}{T}*T]|=|f(x'_1)-f(x'_2)|<\epsilon$$ But both $x'_1$ and $x'_2$ are in $[0,T]$, so we can consider delta to corresponding delta when we write definition on $[0,T]$.

If the function is periodic, then for some $T>0$ we have $f(t) = f(t+nT)$, for all $n \in \mathbb{Z}$ and for all $t$. Hence $\sup_{t \in \mathbb{R}} |f(t)| = \sup_{t \in [0,T]} |f(t)|$. Since $[0,T]$ is compact and $f$ is continuous $f([0,T])$ is compact and so is bounded. Hence $f$ is bounded everywhere.

Since $[0,2T]$ is compact, $f$ is uniformly continuous on $[0,2T]$. It only remains to show uniform continuity on $\mathbb{R}$. Let $\epsilon>0$ and let $\delta>0$ be such that if $x,y \in [0,2T]$ and $|x-y| < \delta$ then $|f(x)-f(y)| < \epsilon$.

Now let $x,y \in \mathbb{R}$ with $|x-y| < \min(\delta, \frac{T}{2})$. Since $f$ is $T$-periodic, we must have $[x-nT,y-nT] \subset [0,2T]$ for some $n$. Since $|f(x)-f(y)| = |f(x-nT)-f(y-nT)|$, we have $|f(x)-f(y)| < \epsilon$, hence $f$ is uniformly continuous on $\mathbb{R}$.

If $T>0$ is a period of $f$, then the restricton $f|_{[0,T]}$ of $f$ to the compact set $[0,T]$ is continuous, hence is bounded and uniformly continuous. We note that $f|_{[0,2T]}$ is also uniformly continuous, by the same argument.

Any bound for $f|_{[0,T]}$ is also a bound for $f$ itself, i.e. $f$ is bounded.

For uniform continuity let $\epsilon>0$ be given. Without loss of generality, $\epsilon<T$. The uniform continuity of $f|_{[0,2T]}$ gives us $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ if $x,y\in[0,2T]$ with $|x-y|<\delta$. However, we have to consider $x,y\in\mathbb R$ with $|x-y|<\epsilon$. There are suitable integers $n,m\in\mathbb Z$ such that $x':=x-nT$ and $y':=y-mT$ are both in $[0,T]$. Then \begin{align}T\ge|x'-y'|&=|x-y-(n-m)T|\\&\ge |(n-m)T|-|x-y|\\&>|n-m|\cdot T-\epsilon\\&>(|n-m|-1)T,\end{align} hence $|n-m|<2$.

• If $n=m$, let $x''=x'=x-nT, y''=y'=y-nT$.
• If $n=m+1$, let $x''=x'+T=x-mT, y''=y'=y-mT$.
• If $n=m-1$, let $x''=x'=x-nT, y''=y'+T=y-nT$.

In all three cases we find $x'',y''\in[0,2T]$ with $|x''-y''|=|x-y|<\delta$, hence $$|f(x)-f(y)|=|f(x'')-f(y')|<\epsilon.$$

Since the function $f(x)$ is periodic, suppose its period is T. In the interval $[0,T]$, since $f(x)$ is continuous in the closed interval, $f(x)$ is bounded in that interval. Due to the periodicity, $f(x)$ is bounded on R.

Uniformly continuity...

Hint: Since a continuous function is uniformly continuous in a closed interval, $f(x)$ is uniformly continuous in $[0,T]$, which gives a '$delta$'. It's uniformly continuous also in $[T/2,3T/2]$, which gives another '$delta$'. Combine these two things.

• How do you rigorously show that $f(x)$ is bounded on $[0,T] \implies f(x)$ is bounded on $\mathbb{R}$? Apr 17, 2013 at 2:53
• For any real number x, we can write x=nT+m, T>m>=0, where n is a integer, uniquely. f(x)=f(m). Since f(m) is bounded on [0,T], f(x) is bounded on R, for $f(m)$ on [0,T] and $f(x)$ on R share the same range. Apr 17, 2013 at 4:08