If $f$ is proper, lsc, and $\frac{f(x) + f(y)}{2} = f^{**}\left(\frac{x + y}{2}\right) \implies x = y$, is $f$ necessarily convex? 
Suppose $X$ is a real Hilbert Space and $f : X \to (-\infty, \infty]$ is a lower semicontinuous, proper function. Further, suppose $f$ satisfies the following, for all $x, y \in \operatorname{dom} f$:
  $$\frac{f(x) + f(y)}{2} = f^{**}\left(\frac{x + y}{2}\right) \implies x = y.$$
  Is $f$ necessarily a convex function?

Here $^*$ refers to the Fenchel conjugate, and $\operatorname{dom} f$ is the set of points $x \in X$ such that $f(x) \neq \infty$.
I know that:


*

*$f^{**}(x) \le f(x)$ for all $x$ and $f^{**}(x) = f(x)$ for all $x$ if and only if $f$ is convex (and lsc).

*In fact, $f^{**}$ is the greatest lsc convex minorant of $f$.

*This means that
$$\frac{f(x) + f(y)}{2} \ge \frac{f^{**}(x) + f^{**}(y)}{2} \ge  f^{**}\left(\frac{x + y}{2}\right)$$
for all $x, y \in \operatorname{dom} f$.

*Therefore,
$$\frac{f(x) + f(y)}{2} = f^{**}\left(\frac{x + y}{2}\right)$$
implies that $f(x) = f^{**}(x)$ and $f(y) = f^{**}(y)$.

*Another consequence is that $f^{**}(\lambda x + (1 - \lambda y)) = \lambda f^{**}(x) + (1 - \lambda)f^{**}(y)$ for all $\lambda \in [0, 1]$.


My thoughts:


*

*Really, I just need to establish that $f^{**}(x) = f(x)$ for all $x$.

*Despite biduals showing up both in the premises and the above desired conclusion, there doesn't seem to be a direct path to manipulate one to the other, especially since not every point in $\overline{\operatorname{conv}} \operatorname{dom} f$ can be expressed as $\frac{x+y}{2}$ where $x, y \in \operatorname{dom} f$.

*The function $g(x, y) = \frac{f(x)+f(y)}{2} - f^{**}\left(\frac{x + y}{2}\right)$ is not a metric in general, even if $g(x, y) = 0 \implies x = y$.

*I get a feeling that Stegall's variational principle might help, for a variety of reasons, but one handy reason is that we may add any linear functional to $f$, without changing $g$.


Any thoughts are welcome!

The conjecture seems to be not true. Take $X=\mathbb R$,
$$
f(x)=\sqrt{|x|}.
$$
Then it holds $f^{**}\equiv 0$. Moreover, if 
$$
f(x)+f(y) = 2 f^{**}\left(\frac{x+y}2\right)=0,
$$
then necessarily $x=y=0$. But $f$ is not convex.
