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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
8 months, 3 weeks 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. Aditya Agarwal · 8 months, 3 weeks ago

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@Aditya Agarwal Correct! Romanos Molfesis · 8 months, 3 weeks ago

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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 · 8 months, 3 weeks ago

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@Otto Bretscher Correct! Romanos Molfesis · 8 months, 3 weeks ago

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