When $E[f(\alpha,X)] = f(\alpha,E[X])$, where $f$ is some convex function of the first and second variables, except when the first variable takes the value $\alpha$ in which case the equality holds, then intuitively f is a (locally) linear function of the second variable. But how do you prove this, i.e. prove that $f''(\alpha, E[X]) = 0$ where the prime denotes differentiation to the second variable? It's maybe simple to prove but I can't figure it out.
By Jenson's Inequality, $E[f(X)] >= f(E[X])$ if $f''(X) >= 0$.
When two additional constraints apply:
1) $X$ is not a constant
$E[f(X)] > f(E[X])$.
By contradiction, if $E[f(X)] = f(E[X])$, then either $X$ is a constant or $f''(X) = 0$.
Edit: @Hans is correct. I had assumed f''(x) is a constant, but that was not stated in your question. You claim is not true. You could have a discreet random variable X and and some arbitrary f(x) like the below. $E[f(X)] = 0 = f(E[X])$ but the derivatives of $f$ are undefined. Although you have told us the shape of $f$ for other values of $\alpha$ we know nothing about the shape for the $\alpha$ value of interest.