<s>[INST] Solutions of the functional equation $f(2x) = \frac{f(x)+x}{2}$ How can I solve the following functional equation?
$$f(2x) = \frac{f(x)+x}{2},$$
for $x \in \mathbb{R}$ with $f$ being a continuous function.
 [/INST] Here's an alternative proof.  Notice that by substituting in $\frac{1}{2^{n+1}}x$ for $x$, we can obtain
$$f\left(\frac{1}{2^n}x\right) = \frac{1}{2}f\left(\frac{1}{2^{n+1}}x\right) + \frac{x}{2^{n+2}}$$
Using this expression, we find that
\begin{align*}f(x) &= \frac{1}{2}f\left(\frac{1}{2}x\right) + \frac{x}{4}\\
 &= \frac{1}{2}\left(\frac{1}{2}f\left(\frac{1}{4}x\right) + \frac{x}{8}\right) + \frac{x}{4}\\
 &= \frac{1}{4}f\left(\frac{1}{4}x\right) + \frac{x}{4}+\frac{x}{16}\\
\\&\vdots\\
&= \frac{1}{2^n}f\left(\frac{1}{2^n}x\right) + \sum_{k=1}^{n}\frac{x}{4^k}
\end{align*}
Let's look at the behavior of this last term as $n\to \infty$.  We have that $f\left(\frac{1}{2^n}x\right) \to f(0)$, which is finite, so $\frac{1}{2^n}f\left(\frac{1}{2^n}x\right) \to 0$.  It follows that
$$f(x) =  \sum_{k=1}^{\infty}\frac{x}{4^k} = \frac{x}{3}$$
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