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?
 A: It is not even necessary to assume that $f$ is continuous.


*

*By letting $y = 2x$, we see that $f(x) = f(2x)$

*Letting $y = -4x$, we get $f(-x) = \frac{f(x) + f(-4x)}{2}$.  However, from (1), $f(-4x) = f(-2x) = f(-x)$, so this simplifies to $f(-x) = f(x)$

*Finally, let $y = -x$ and simplifying gives $2f(0) = f(x) + f(-x)$.  Substituting in from (2), this becomes $f(0) = f(x)$.  Since this holds for all $x \in \mathbb{R}$, we conclude that $f$ must be constant.

A: 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 $f(x)=f(0)$.
A: Unfortunately your solution is incorrect in several ways.
First of all, the formulation of the problem is not accurate because your equation does not define a function $f$. What you probably mean is: Prove that every function $f:\mathbb{R}\to \mathbb{R}$ satisfying (equation) is constant. Also you need at least the assumption that $f$ is continuous. Otherwise the statement is not true.
Your method requires the assumption that $f$ is differentiable.
But even then, you only show that $f'$ is constant, $f'(x)=f'(y)$ for all $x,y\in \mathbb{R}$. But you cannot integrate this to $f(x)=f(y)+C$. Rather, a linear function $f(x)=\alpha x+\beta$ has a constant derivative as well.
A: The statement also holds if we only suppose that $f$ is continuous (and not necessarily differentiable).
Suppose that $f(x_0) \neq f(0)$ for some $x_0 \in \Bbb R$.  We note that
$$
f \left( \frac{x_0 + x_0}{3} \right) = \frac{f(x_0) + f(x_0)}{2} \implies\\
f\left( \frac 23 x_0\right) = f(x_0)
$$
Then, consider the sequence $\left( \frac 23\right)^n x_0 \to 0$.
A: for all $x$ real
\begin{array}{l}
 f\left( { - x} \right) = f\left( {\frac{{ - 2x - x}}{3}} \right) = \frac{{f\left( { - 2x} \right) + f\left( { - x} \right)}}{2} \\ 
  \Leftrightarrow f\left( { - x} \right) =   f\left( { - 2x} \right) \cdots  \cdots \left( 1 \right) \\ 
 f\left( 0 \right) = \frac{{f\left( x \right) + f\left( { - x} \right)}}{2} \cdots  \cdots \left( 2 \right) \\ 
 f\space \text{is continuous at  0} \\and\\
\left( 1 \right)and\left( 2 \right) \Leftrightarrow f\left( 0 \right) = \frac{{f\left( x \right) + f\left( { - 2x} \right)}}{2} = f\left( { - x} \right)  \\ 
 \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 0 = \mathop {\lim }\limits_{x \to 0} f\left( {\frac{2}{3}x} \right) = \mathop {\lim }\limits_{x \to 0} f\left( {\left( {\frac{2}{3}} \right)^n x} \right) \\ 
  \Leftrightarrow f = \text{const} \\ 
 \end{array}
