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.
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 $f$.