# Nice Functional Analysis Problem

If $$f: \mathbb R \rightarrow \mathbb R$$ is continuous everywhere and for every real $$x$$,

$f(x) = f(2x)$

Prove that it is indeed constant. The answer involves a bit of imagination.

Note by Romanos Molfesis
2 years, 4 months ago

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I think I have seen this question before.

$f(x)=f(2x)$ $f(\frac12x)=f(x)$ and like this, we get, $f(\frac{x}{2^n})=f(x)$ Now as $$n\to\infty$$ $f(0)=f(x)$ and thus, we get that $$f(x)$$ is a constant function.

- 2 years, 4 months ago

Correct!

- 2 years, 4 months ago

We have $$f(x)=f(2^nx)$$ for all real $$x$$ and for all integers $$n$$. Now, by continuity, $$f(0)=\lim_{n\to -\infty}f(2^nx)=f(x)$$, showing that $$f(x)$$ is constant, taking the value $$f(0)$$ for all $$x$$.

- 2 years, 4 months ago

Correct!

- 2 years, 4 months ago

it is a continuous function therfore it has to be constant

- 1 year, 3 months ago