# Show that $f$ is a constant function

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.

• 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) – user580918 Aug 30 at 6:54
• Checks out to me. – C Squared Aug 30 at 6:54
• 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. – Fawkes4494d3 Aug 30 at 6:57
• Please, type x’, not x^{‘} – egreg Aug 30 at 8:29
• Noted, thank you for the correction! – Cyanide2002 Aug 30 at 9:09

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!