Let $f:\mathbb R \to \mathbb R$ be a function such that for any irrational number r, and any real number x we have $f(x)=f(x+r)$. Show that f is a constant function.

It's easy to see any constant function satisfies the original properti... But I don't see how to show this is the only solution.


If $y$ is irrational, the set $x = 0$, $r = y$ to get $f(y) = f(0)$.

If $y$ is rational, then set $x = y$, $r = \pi-y$ to get $f(y) = f(\pi)$

But by the first statement, $f(\pi) = f(0)$. Hence, $f$ is constant.

  • $\begingroup$ I think that in the last line, $\pi$ has to be $y$. $\endgroup$ – The Great Seo Aug 3 '14 at 4:20
  • $\begingroup$ The first statement is that $f(y) = f(0)$ for any irrational number $y$. In particular, $y = \pi$ is irrational, so $f(\pi) = f(0)$. $\endgroup$ – JimmyK4542 Aug 3 '14 at 4:22
  • $\begingroup$ If you have $f(\pi)=f(0)$, it doesn't imply that for all $q\in\Bbb Q\quad f(q)=f(0)$, which is necessary to prove that $f$ is constant. $\endgroup$ – The Great Seo Aug 3 '14 at 4:23
  • $\begingroup$ Oh, I mistaked. Your answer just omitted that. Sorry for that. $\endgroup$ – The Great Seo Aug 3 '14 at 4:27

Wrong question.

The general solution of $f(x)=f(x+r)$ is $f(x)=\theta(x)$ , where $\theta(x)$ is an arbitrary periodic functions with period $r$ .

Adding the condition $f:\mathbb{R}\to\mathbb{R}$ the general solution will be modified to $f(x)=\theta(x)$ , where $\theta(x)$ is an arbitrary periodic $\mathbb{R}\to\mathbb{R}$ functions with period $r$ .

For example a particular solution $f(x)=\sin\dfrac{2\pi x}{r}$ , it is a $\mathbb{R}\to\mathbb{R}$ function, a periodic function with period $r$ and satisfies $f(x)=f(x+r)$ .

  • 4
    $\begingroup$ $r$ is an arbitrary irrational number; this function is periodic with every irrational period. $\endgroup$ – Ian Aug 3 '14 at 4:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.