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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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4 Answers 4

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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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    $\begingroup$ 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! $\endgroup$
    – MCT
    Mar 4, 2015 at 22:06
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    $\begingroup$ @MCT I always stuck with this (why) in many proofs could you please explain this little bit? why $f$ is bounded all over $\mathbb{R}$? $\endgroup$
    – Priya
    Jul 8, 2021 at 4:33
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    $\begingroup$ @Priya because for every $x \in \mathbb{R}$, there is a point $y \in [0,2]$ with $|y-x|$ being a multiple of $1$, so $f(x) = f(y)$. So if $M$ is larger than all $f(y)$ for $y \in [0,2]$, it must also be larger than all points $f(x)$ for $x \in \mathbb{R}$ $\endgroup$
    – MCT
    Jul 11, 2021 at 14:44
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A periodic continuous function is simply a function defined on a circle, $$f:S^1 \longrightarrow R.$$ Circle is a compact space. A continuous function on a compact space is both bounded and uniformly continuous.

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    $\begingroup$ But $f$ is defined on $\mathbb R$, not on the circle. $\endgroup$ Jun 24, 2018 at 18:54
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    $\begingroup$ AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle. $\endgroup$ Jun 24, 2018 at 22:38
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Here's proof for boundedness part by contradiction :

Fix $a \in \Bbb R$. Consider the domain $I=[a,a+p]$. Assume $f$ is not bounded. Then there exists a sequence $x_n \in I$ such that $|f(x_n)| \gt n$.

Since $x_n$ is bounded and consists of real numbers, it has a convergent subsequence $x_{n_k}$. But $I$ is closed hence $x_{n_k}$ converges inside $I$. i.e. $\exists x \in I$ such that $x_{n_k} \to x$.

What we have got is a sequence $x_{n_k} \to x \in I$ and $|f(x_{n_k})| \to \infty \; (\because |f(x_{n_k})| \gt n_k)$. This is contradiction for $f$ being continuous on $I$ and hence at $x$.

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    $\begingroup$ you could also argue in the following way: $f$ is bounded on $[-p,p]$ since continuous functions on compact sets attain a maximum, say at $x_0\in[-p,p]$. then there is an $M>0$ such that $|f(x)|\leq |f(x_0)|<M$ for each $x\in[-p,p]$. for any integer $n$, and for each $x\in[-p+pn,p+pn]$, we have $x\pm pn\in[-p,p]$ depending on the sign of $n$, so $|f(x)|=|f(x\pm pn)|\leq |f(x_0)|<M$ from which the conclusion follows. $\endgroup$
    – C Squared
    Jul 25, 2021 at 9:20
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It is easier to prove a more general statement.

Consider $X$ a locally compact metric space, $G$ a group of isometries such that there exists a compact subset $K\subset X$ with $G\cdot K = X$. Let $f$ be a continuous $G$-periodic function from $X$ to another metric space $Y$. Then $f$ is uniformly continuous.

Proof:

Consider $\tilde K\supset K$ a compact neighborhood (it exists since $X$ is locally compact). Then $\Delta \colon = d(K, X\backslash \tilde K) >0$. Take $\epsilon>0$ arbitrary. There exists $\delta= \delta({\epsilon})>0$ such that $\tilde x_1$, $\tilde x_2$ in $\tilde K$, $d(\tilde x_1,\tilde x_2)< \delta$ implies $d(f(\tilde x_1), f(\tilde x_2)) < \epsilon$. Let us show that $\delta'= \min(\delta, \Delta)$ works for $f$ on the whole $X$. Indeed, consider now $x_1$, $x_2$ in $X$ such that $d(x_1, x_2) < \delta'$. Let $g \in G$ such that $\tilde x_1 = g x_1 \in K$. Since $d(x_1, x_2) = d(gx_1, gx_2)$, we get $d(\tilde x_1, \tilde x_2) < \delta'\le \Delta$, so $\tilde x_2 \in \tilde K$. Now we also have $d(\tilde x_1, \tilde x_2) < \delta$, so $d(f(\tilde x_1), f(\tilde x_2)) < \epsilon$. But note that $f(x_i) = f(\tilde x_i)$. We are done.

Comment: the idea is: translate one of the points $x_i$ to the compact fundamental domain. Under the same translation, the other point will fall in the compact neighborhood of the fundamental domain. Now use the uniform compactness of $f$ on $\tilde K$. So for the original problem, consider $K=[0,p]$, $\tilde K= [-\Delta, p+\Delta]$.

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