MHB Solution: Proving $f$ is identically zero

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Here's this week's problem!

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Problem. Let $f : \Bbb R \to \Bbb C$ be a continuously differentiable, 1-periodic function such that and $f(x/2) + f((x+1)/2) = f(x)$, for all $x\in \Bbb R$. Prove that $f$ is identically zero.
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No one answered this week's problem. You can find my solution below.

Spoiler

Let $n\in \Bbb N$. For all $x\in \Bbb R$,

$$f(x) = f\left(\frac{x}{2}\right) + f\left(\frac{x+1}{2}\right) = f\left(\frac{x}{4}\right) + f\left(\frac{x+1}{4}\right) + f\left(\frac{x+2}{4}\right) + f\left(\frac{x+3}{4}\right) = \cdots = \sum_{j = 0}^{2^n-1} f\left(\frac{x+j}{2^n}\right).$$

Differentiating with respect to $x$ results in

$$f'(x) = \frac{1}{2^n}\sum_{j = 0}^{2^n-1} f\left(\frac{x+j}{2^n}\right)$$

for all $n\in \Bbb N$. For $x\in [0,1]$, the right hand side converges to $$\int_0^1 f'(t)\, dt = f(1) - f(0) = 0.$$ Thus, $f'(x) = 0$ for all $x\in [0,1]$. Since $f$ is $1$-periodic, so is $f'$. Thus $f' = 0$ on $[k, k+1]$ for all $k\in \Bbb Z$, which implies $f' = 0$ on $\Bbb R$. Consequently, $f$ is constant. If $c$ is the constant, the functional equation for $f$ implies $c = 0$. Therefore, $f$ is identically zero.