Assume we have a real valued random variable $X$, and a twice differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $0< \ell <f'(x) < L $ for all $x \in \mathbb{R}$, where $\ell ,L \in \mathbb{R}$. Take an arbitrary point $x^* \in \mathbb{R}$, I know I can say through the mean value theorem that
$$f(X) - f(x^*) = f'(C) ( X - x^* ) $$ where $C$ is a random variable that always lies between $X$ and $x^*$. Assuming all the following expectations exist I would like to say that there exist an $a$ such that $$E[f(X)] - f(x^*) = E[f'(C) ( X - x^* )] = a ( E[X] - x^* ) $$ where $ \ell< a < L$. Can this be done?
I believe this depends on the sign of $X - x^*$ but I am confused by the fact that in the mean value theorem for real valued functions $X - x^* > 0$.
This is not necessarily true. For a counterexample, let $X$ be uniform on $[-1,1]$, let $x^*=0$, and let $f$ be the function
$$ f(x) = \begin{cases} 1+2(x+1), & x \le -1, \\ (x+2)^2, & x\in(-1,1), \\ 9 + 6(x-1), &x \ge 1. \end{cases} $$
(This definition may seem a little out of left-field, but all it is is the function $(x+2)^2$ on $[-1,1]$, extended linearly outside $[-1,1]$ in such a way that $f'$ is continuous.)
By definition of $X$, we have $E[X]=0=x^*$, so the right hand side of your equation is zero, regardless of the value of $a$. The left hand side, however, is $E[(X+2)^2]>0$. So there is no value of $a$ which makes the two sides equal.