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Let $f$ be a continuous function such that $f(2x) = f(x)$ for all $x\in \mathbb{R}$. Prove that $f(x)$ is a constant function. I want to know whether my proof for it is correct.

I proceeded by taking : $$f(x)=f(\frac{x}{2})=f(\frac{x}{4})=....=f(\frac{x}{2^n})$$

For any real number $x^{'}\in \mathbb{R}$, we have $$f(x^{'})=f(\frac{x^{'}}{2^n})$$

We now take a limit on both sides:$$\lim_{n\to\infty}f(x) = \lim_{n\to\infty}f(\frac{x^{'}}{2^n})$$

The sequence $\frac{x^{'}}{2^n}$ converges to $0$. Hence we get, $f(x{'})=f(0)$ for all $x^{'}\in \mathbb{R}$, implying $f(x)$ is a constant function.

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    $\begingroup$ It should be $\lim_{n\to\infty}f(x')$, but aside from that, yes it's right. (Also, note that there's nothing special about $2$, more generally if $\lambda>0$ then an almost identical proof shows if for all $x$, $f(\lambda x) = f(x)$, then $f$ is constant) $\endgroup$ – user580918 Aug 30 at 6:54
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    $\begingroup$ Checks out to me. $\endgroup$ – C Squared Aug 30 at 6:54
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    $\begingroup$ Maybe for rigour, you can add one more step apart from the edit $x\rightarrow x'$ that $\lim_{n\to\infty} f\left(\frac{x'}{2^n}\right)= f\left(\lim_{n\to\infty}\frac{x'}{2^n}\right)$ which happens because $f$ is continuous and is a key part of why the proof works. It's sound anyway. $\endgroup$ – Fawkes4494d3 Aug 30 at 6:57
  • $\begingroup$ Please, type x’, not x^{‘} $\endgroup$ – egreg Aug 30 at 8:29
  • $\begingroup$ Noted, thank you for the correction! $\endgroup$ – Cyanide2002 Aug 30 at 9:09
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Note that $$\ f(x)=f(\frac{x}{2})=\lim_{n\to\infty}f(\frac{x}{{2}^{n}})$$ Now since $\ f(x)$ is a continuous function, it is continuous at $\ x=0$. Hence, by definition, $$\ \lim_{n\to\infty}f(\frac{x}{{2}^{n}})=f(0)$$ Therefore $$\ f(x)=f(0) \forall x\in \mathbb{R}$$ Yes, your proof is correct. But make sure that you use the definition of continuity in this step $$\ \lim_{n\to\infty}f(\frac{x}{{2}^{n}})=f(0)$$ Hope it helps!

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