Mean Value theorem functional equation: $f'\left(\frac{x + y}{2}\right) = \frac{f(x) - f(y)}{x -y}$

The functional equation $$ f'\left(\frac{x + y}{2}\right) = \frac{f(y) - f(x)}{y - x}\quad\text{for all real $x$ and $y$,} $$ guarantees $f'$ is continuous: Fixing $x$ and letting $y = x + 2h \to x$, we have $$ \lim_{h \to 0} f'(x + h) = \lim_{h \to 0} \frac{f(x + 2h) - f(x)}{2h} = f'(x). $$ This technical fact is needed below.

For each real $z$ and for all $h \neq 0$, the functional equation says $$ 2h\, f'(z) = f(z + h) - f(z - h). $$ Fixing $z$ and differentiating with respect to $h$ (each side is differentiable in $h$ by hypothesis), $$ 2f'(z) = f'(z + h) + f'(z - h)\quad\text{for all real $z$ and $h$.} $$

In other words, $2f'(\frac{x + y}{2}) = f'(x) + f'(y)$ for all real $x$ and $y$. Since $f'$ is continuous, it follows that $f'$ is a linear polynomial, i.e., there exist real numbers $a$ and $b$ such that $$ f'(x) = 2ax + b\quad\text{for all real $x$,} $$ and consequently that $f$ is a quadratic polynomial.

This fact is well-known (and surely answered elsewhere on site), but in the spirit of being self-contained, here's an ad hoc sketch: If $f'(x)$ and $f'(y)$ are known for some $x < y$, then

• $f'(z) = f'(x) + \dfrac{z - x}{y - x} \bigl(f'(y) - f'(x)\bigr)$ on $[x, y]$ by successive bisection and continuity;

• $f'(2y - x) = f'(y + (y - x)) = 2f'(y) - f'(x)$ and $f'(y - 2x) = f'(y) - 2f'(x)$ are uniquely determined, so $f'$ is uniquely determined (hence linear) on the larger interval $[y - 2x, 2y - x]$. Inductively, $f'$ is linear on (arbitrary closed, bounded subintervals of) $(-\infty, \infty)$.

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(Once the end result is known, more elegant arguments can doubtless be given, based on the fact that the functional equation is invariant if a quadratic polynomial is subtracted from $f$; interpolate $f$ at three points with a quadratic $ax^{2} + bx + c$, put $g(x) = f(x) - (ax^{2} + bx + c)$, and show $g \equiv 0$. I haven't carried out the details, however.)