$f(x)-f(y)=\frac{1}{2}(x-y)^T( \nabla f(x)+ \nabla f(y))$ show that the only solution is quadratic How to show that the  only convex  function satisfying the following equation
\begin{align}
f(x)-f(y)=\frac{1}{2}(x-y)^T( \nabla f(x)+ \nabla f(y))
\end{align} 
is quadratic?   Here $f$ is continuously-differentiable. 
Can this be done without taking a derivative? I know how to do this by differentiation the above equation one more time. However, I think it should hold without any second derivative arguments. 
If needed, one can assume that $f$ is strictly convex. 
 A: Here is a sketch how it could be solved.
I will ignore some details. 
First, we will assume $f(0)=0$ and $\nabla f(0)=0$
(otherwise this can be reduced to this case by the transformation 
$g(x)=f(x)-f(0)-x^T\nabla f(0)$).
Note that if we set $y=0$ in the original equation we then obtain the equation
\begin{equation}
  \tag{1}
  \label{1}
  f(x)=\frac12 x^T \nabla f(x)
\end{equation}
for all $x\in\mathbb R^n$.
Next, let $a,b,c\in\mathbb R^n$ be given with $2c=a+b$.
By using various combinations of $a,b,c$ in the original equality for $x,y$
and combining the results one can show that
\begin{equation}
  \tag{2}
  \label{2}
  f(a)-f(b) = \frac12 (a-b)^T (\nabla f(a)+\nabla f(b))
  =  (a-b)^T ( \nabla f(\frac{a+b}{2}))
\end{equation}
holds.
By using $a=x,b=-x$ in \eqref{2} we can also obtain $f(x)=f(-x)$ for all $x$.
Let us apply \eqref{2} for $a=x+y,b=x-y$.
Then we have
\begin{equation*}
  f(x+y)-f(x-y) = (2y)^T ( \nabla f(x)).
\end{equation*}
By exchanging $x$ and $y$ in the above and using $f(x)=f(-x)$ we get
\begin{equation}
  \tag{3}
  \label{3}
  y^T \nabla f(x) = x^T \nabla f(y)
\end{equation}
for all $x,y$.
Let us define the function $B: \mathbb R^n\times \mathbb R^n\to \mathbb R$
via
\begin{equation*}
  B(x,y) = x^T\nabla f(y).
\end{equation*}
By \eqref{3} one can see that $B$ is a bilinear function.
Thus there is a matrix $Q$ such that
\begin{equation*}
  x^T Q y = x^T \nabla f(y)
\end{equation*}
holds for all $x,y$.
Finally, by \eqref{1} we see that $f$ can be written as
\begin{equation*}
  f(x)=x^T Q x
\end{equation*}
and therefore $f$ is quadratic.
Note that we never needed any convexity of $f$ or any second derivatives of $f$.
