Functional equations $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ and $f(x)=\frac{f\left(\frac{2}{3}x\right)+f\left(\frac{4}{3}x\right)}{2}$.

Question

Suppose $f$ is continuous and $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$. Can we claim that $f(x)=kx$?

What if $f$ only satisfy $f(x)=\frac{f\left(\frac{2}{3}x\right)+f\left(\frac{4}{3}x\right)}{2}$?

This functional equation was called Jensen's equation on wiki, but there is no further discussion about it

Answer 1

No (for the first equation).

But we can claim $f(x)=ax+b$ (and all functions of the form satisfy the equation).

Let $f$ be any function satisfying the functional equation. Then this remains true if we replace $f$ with $x\mapsto f(x)-f(0)-x(f(1)-f(0))$, i.e., we may assume wlog. that $f(0)=f(1)=0$. Let $S=\{\,x\in \Bbb R\mid f(x)=0\,\}$. So far we have $0\in S$, $1\in S$. Also, $x\in S\iff \frac x2\in S$. Using that, if two of $x,y,x+y$ are in $S$, then so is the third. It follows that $S$ is a dense subgroup of $\Bbb R$. By continuity of $f$, $S=\Bbb R$.

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The answer is also "No" for the second question, but for different reasons: There are some solutions that are by far not of the given form. The simplest "unusual" solution is $f(x)=|x|$.

Answer 2

For the first problem, fix $x, y \in \Bbb{R}$ and define the set $S_{x,y}$ by

$$S_{x,y} = \{\lambda \in [0, 1] : f(\lambda x + (1-\lambda) y) = \lambda f(x) + (1-\lambda)f(y) \}. $$

We easily check that $0, 1 \in S_{x,y}$ and if $\alpha, \beta \in S$ then $\frac{\alpha+\beta}{2} \in S_{x,y}$. It immediately follows that all the dyadic rationals (i.e. rationals of the form $k/2^n$ for some $k \in \Bbb{Z}$ and $n \geq 0$) in $[0, 1]$ are in $S_{x,y}$. By the continuity of $f$, this implies $S_{x,y} = [0, 1]$.

This is enough to conclude that $f$ is of the form $f(x) = ax + b$.

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For the second problem, changing the parameter a bit gives a nowhere piecewise-linear example. Indeed, consider the functional equation

$$ f(x) = \frac{f(\alpha x) + f(\beta x)}{2}, \qquad \forall x \in \Bbb{R} \tag{*}$$

where $f$ is continuous, $\alpha \in (0, 1)$ and $\alpha + \beta = 2$. If we choose $\beta = \phi = \frac{1+\sqrt{5}}{2}$ (which is the golden ratio), then $\alpha = \phi^{-2}$ and the following function

$$ f(x) = |x|^{1+2\pi i/\log \phi} $$

solves the functional equation $\text{(*)}$.

I believe that this kind of solution does not appear in our case $\alpha = \frac{2}{3}$ since the set $\{\alpha^k \beta^l\}_{k, l \in \Bbb{Z}}$ is now dense in $[0,\infty)$. But I have no good idea to begin with.