Solutions of the functional equation $f(2x) = \frac{f(x)+x}{2}$

Question

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.

Answer 1

Let $f(x)=ax$. Then it's easy to show that $a=1/3$.

Next prove uniqueness.

Claim: The only function $h(x)$ which is continuous at $x=0$ and satisfying $h(2x)=h(x)/2$ is $h(x)=0$.

Proof:

It's easy to show that $h(0)=0$. Suppose $h(b)\ne0$ for some non-zero $b$. Then $h(b/2^N)=2^Nh(b)$. For all $\epsilon$, $\delta$, there exists $N$ such that $|b/2^N|<\delta$ but $|h(b/2^N)|>\epsilon$, contradicting the continuity of $h(x)$ at $0$.

Now suppose there are two functions $f(x)$ and $g(x)$ both continues at $x=0$ and satisfying $f(2x)=\frac{f(x)+x}{2}$ and $g(2x)=\frac{g(x)+x}{2}$. Define $h(x)=f(x)-g(x)$ which is continuous at $x=0$. Then $h(2x)=\frac{h(x)}{2}$ and hence $f(x)=g(x)$.

Answer 2

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 $$f(x) = \sum_{k=1}^{\infty}\frac{x}{4^k} = \frac{x}{3}$$