A function that satisfies $f\left({\frac{x+y}3}\right)=\frac{f(x)+f(y)}2$ must be a constant Prove that a function $f:\mathbb{R}\to\mathbb{R}$ which satisfies
$$f\left({\frac{x+y}3}\right)=\frac{f(x)+f(y)}2$$
is a constant function.
This is my solution: constant function have derivative $0$ for any number, so I need to prove that $f'$ is always $0$. I first calculated $\frac{d}{dx}$ and then $\frac{d}{dy}$:
$$f'\left({\frac{x+y}3}\right)\frac13=\frac{f'(x)}2$$
$$f'\left({\frac{x+y}3}\right)\frac13=\frac{f'(y)}2$$
From this I can see that $\frac{f'(x)}2=\frac{f'(y)}2$. Multiplying by $2$ and integrating I got:
$$f(x)=f(y)+C$$
for some constant $C\in\mathbb{R}$. By definition of $f$ it is true for any $x,y\in\mathbb{R}$, so I can write
$$f(y)=f(x)+C$$
Adding this two equation and simplifying I got
$$C=0$$
so $f(x)=f(y)$ for all $x,y\in\mathbb{R}$. Is my solution mathematically correct. Is this complete proof, or I missed something?

With $y=x$ we have by induction
$$f(x)=f\left(\frac23x\right)=f\left(\left(\frac23\right)^nx\right)$$ so we need just the hypothesis that $f$ is continuous at $0$ to get that $f(x)=f(0)$.
