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.

#Calculus

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

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.

Aditya Agarwal - 5 years, 5 months ago

Correct!

Romanos Molfesis - 5 years, 5 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 $x$ .

Otto Bretscher - 5 years, 5 months ago

Correct!

Romanos Molfesis - 5 years, 5 months ago

it is a continuous function therfore it has to be constant

sashank bonda - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$