If $f(x)$ is a continuous function such that $f(3x)=f(x)$ and the domain of $f$ is all non-negative real numbers. Prove that $f$ is a constant function.

What I did: $$f(3x)=f(x)=f\left(\frac{x}{3}\right)=\cdots= f\left(\frac{x}{3^n}\right)$$ Now as $n$ tends to infinity, $f(\frac{x}{3^n})$ tends to $f(0)$ and hence $f(x)=f(0)$ for all $x$.

However, I think the second step is a bit dodgy. I can't quite tell how, since I lack sufficient mathematical maturity, but it just doesn't seem right. I'd be glad if someone could provide a more rigorous proof of the problem. Thanks in advance.

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    $\begingroup$ The argument is written up a bit informally, but is fully correct. $\endgroup$ – André Nicolas May 19 '14 at 16:11
  • $\begingroup$ This is rigorous enough for my taste. $\endgroup$ – anonymous May 19 '14 at 16:18
  • $\begingroup$ If continuity is only assumed for $x>0$, then it is not true. Deleted my previous comment. $\endgroup$ – Sungjin Kim May 19 '14 at 16:18
  • $\begingroup$ Where you wrote $\displaystyle f\left(\frac{x}{3}\right)\cdots f\left(\frac{x}{3^n}\right)$ I thought you meant you were multiplying those things. That is what that notation means in standard usage. But now I suspect you meant $\displaystyle f\left(\frac{x}{3}\right)=\cdots =f\left(\frac{x}{3^n}\right)$. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 19 '14 at 16:53
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    $\begingroup$ The part that feels dodgy is that you are making the argument "a thing and the limit of a thing are the same thing", which is in general not true. (For example, 3, 3.1, 3.14, 3.141, ... are all rational numbers, therefore the limit pi is also rational. That's not a good argument.) "Continuity" is the property that a thing and its limit are the same thing. So use the formal definition of continuity in your argument if you want to make it seem less dodgy. $\endgroup$ – Eric Lippert May 19 '14 at 16:54

To make this argument a bit more formal, try something along these lines:

Suppose (for contradiction) that $f(c) \neq f(0)$ for some $c \in \mathbb{R}$. By the continuity of $f$ at $0$, there exists a $\delta > 0$ such that whenever $|x| < \delta$, $|f(x) - f(0)|< |f(c) - f(0)|$. Noting that $f(c) = f\left( \frac c{3^n}\right)$, we may derive a contradiction.

That is, we may select an $n \in \mathbb{N}$ such that $|c/3^n| < \delta$. The fact that $|c/3^n| < \delta$ and $|f(c/3^n)-f(0)| = |f(c)- f(0)|$ is a contradiction of our definition of $\delta$.

By this contradiction, we are forced to conclude that $f(x) = f(0)$ for all $x \in \mathbb{R}$. That is, $f$ is constant.

  • $\begingroup$ Reductio ad absurdum. My favorite kind of proof. :) $\endgroup$ – sayantankhan May 19 '14 at 16:35

Your argument is not dodgy at all. $f(x)$ being continuous at $x = 0$ is equivalent to the statement that for all $a_1,a_2,...$ converging to $0$ one has $f(a_1), f(a_2),...$ converges to $f(0)$. So take $a_n = {x \over 3^n}$ in your example.


Sorry I misread the question. Your argument would work perfectly. For any $x\in \mathbb{R}$, we have $$f(x) = \lim_{n\rightarrow \infty} f(x) = \lim_{n\rightarrow \infty} f(x3^{-n}) = f(\lim_{n\rightarrow \infty} x3^{-n}) = f(0)$$


Rolle's theorem guarantees us that for every $c\in\mathbb{R^+}$, between $\frac{c}{3}$ and $c$, $f'$ has a zero.

Thus we can find for every $x\in\mathbb{R^+}$ a $c$ for which $f'(x)=0$.

Then $f$ is constant.


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