If $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$, then find $f(2)$

Equality $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$ is only possible for affine functions, with equation $$f(x)=ax+b \ \ \ \ (1)$$

(see explanation below)

When you impose conditions $f(0)=1$ and $f'(0)=-1$, one obtains $b=1$ and $a=-1$. Thus equation (1) becomes $f(x)=-x+1$. Therefore $f(2)=-1.$

Explanation:

• $f(\frac{x+y}{2})\leq\frac{f(x)+f(y)}{2}$ characterizes convex functions (i.e., whose curve is above any of their tangents),

• $f(\frac{x+y}{2})\geq\frac{f(x)+f(y)}{2}$ characterizes concave functions (i.e., whose curve is under any of their tangents).

In view of that, the only functions that are both convex and concave are the affine functions.

If $f$ satisfies your equation, then $g(x) = f(x) - f(0)$ satisfies the Cauchy functional equation $g(x+y) = g(x) + g(y)$. Of course, if $f'(0)$ exists, then $g$ is continuous. The only continuous solutions of the Cauchy functional equation are the linear functions $g(x) = ax$, so the only continuous solutions of your equation are the affine functions $f(x) = a x + b$. The rest is easy.

Let $y=0$ to turn the equation one with one parameter $x$. Then differentiate both sides through implicit differentiation and you will see that:

$$f'(\frac{x}{2})=f'(x)$$

Now here https://math.stackexchange.com/a/1800067 @Eric Wosfey answers about this equation with $f'(0)=-1$:

I will assume that $f:\mathbb{R}\to\mathbb{R}$ is supposed to be $C^1$, so $f'$ exists and is continuous everywhere. Now note that for any $x\in \mathbb{R}$, $f'(x)=f'(x/2)=f'(x/4)=f'(x/8)=\dots$. But $x/2^n$ converges to $0$ as $n\to\infty$, so continuity of $f'$ now implies $$-1=f'(0)=\lim_{n\to \infty}f'(x/2^n)=f'(x).$$ So $f'(x)=-1$ for all $x$, and thus $f(x)=-x+C$ for some constant $C$.

So from his answer we see that:

$$f(x)=-x+c$$

It is given that $f(0)=1$, so substituting this in we get:

$$f(0)=c=1$$

So

$$f(x)=1-x$$

And finally,

$$f(2)=1-2=-1$$

Edit:

It may not seem clear that,

$$f'(x)=f'(x/2)=f'(x/4)=f'(x/8)...$$

But this stems from the fact we can substitute $x=u/2$ into our original equation to get:

$$f'(u/2)=f'(u/4)$$

$$f'(x/2)=f'(x/4)$$

Now substitute $x=u/2$ again and again while switching the dummy variable $u$ back to $x$ to get the result so essential in his proof.