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?


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.

  • $\begingroup$ 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 '15 at 22:06

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.

  • $\begingroup$ But $f$ is defined on $\mathbb R$, not on the circle. $\endgroup$ – Al Jebr Jun 24 '18 at 18:54
  • $\begingroup$ 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. $\endgroup$ – Behnam Esmayli Jun 24 '18 at 22:38

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 $x$.


Let's handle this problem with a " trial and error" approach. And for that we here Will set some propositions and wil try to prove or disprove them by playing with all the ideas of a periodic function. So, let's start the game;


"$\mathscr{A Countinous Functionon IsBounded Everywhere ! }$"

Well, start our journey via disproving it and set the correct one . And to justify the point let's have an example ;

$\mathit{f}$:(1,$\infty$)$\rightarrow$$\mathbb{R}$ such that $\mathit{f}$(x)=x;

Now it's obviously continuous on $\mathbb{R}$, but not bounded as it doesn't have any upper bound. And investigating further we will see that for a closed and bounded interval if $\mathit{f}$ is continuous , then it's definitely bounded!


"$\mathscr{ A Periodic Function Must Have A Unbounded Domain!}$"

Well, this time it wins! the periodic nature of a function drags it's domain to exceed all boundaries. You can check it by any "popular" periodic functions( are you thinking about the trigonometric one !). Analogously you can find the similarity between the "head crumbling " google adds and a periodic function ! It comes again, and again, and again.....(well, just a bad one !).


"$\mathscr{"Every Periodic Function Is Bounded!}$"

Well, for this I don't need to work hard as there are some wonderful examples given earlier.https://math.stackexchange.com/q/1004435}.just see it ,and also the other ones and try to make yours own. So, the write proposition is ," Every continuous periodic function is bounded", which we will prove later.


"$\mathbb{R}$ $\mathscr{Is A Compact Set}$"

And I will be happy enough to tell that , NO! It's not. why? Briefly it doesn't obey the $\pmb{HEINE-BOREL}$ $\mathbf{property}$, i.e every open cover of $\mathbb{R}$ doesn't have fiite sub cover. To, justify it, we may take {$I_n$} as a open covering of $\mathbb{R}$, where $I_n$=(-n,n),n$\in$ $\mathbb{N}$.Let {$I_{n_1},I_{n_2},I_{n_3}\cdots I_{n_k}$} be a finite subcovering of it . Let's also assume that $n_g$=max{$n_1,n_2,n_3\cdots n_k$}. But as per the construction $n_g$ doesn't belongs to any $I_i$, for i=1,2....k . And that's it!

And hopefully we have come so close to our goal, but before that let's check a point: " Every function which is continuous on a compact set are also uniformly continuous on it".

Well, though it's a very powerful tool but of no use to us as we have baffled "proposition 4". So, never ever try to show the uniform continuity on $\mathbb{R}$, discussing the "compactness".

NOW, for a periodic function having primitive period$\gt$0 , we always can construt a close interval say, [o,p], on which it is bounded as it is continuous and it will also be uniformly continuous. So, we may write for any

$\epsilon$$\gt$0, $\exists$ $\delta$$\gt$0 such that if x,y$\in$ g([0,p]) obeys 0$\lt$|x-y|$\lt$$\delta$ then |$\mathit{f}$(x)-$\mathit{f}$(y)|$\lt$$\epsilon$...…(1).Now we can also write

$\forall$ x $\in$ $\mathbb{R}$, $\exists$ n$\in$$\mathbb{Z}$ such that x$\ge$np ; which means x-np and y-np $\in$ $\mathbb{g}$

And this suggests that now if we in in-euality(1), in place of x and y put x-np and y-np then as


We will be able to prove that $\mathit{f}$ is uniformly continuous on $\mathbb{R}$........

$\mathscr{ And It Hits The Goal!!!}$


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