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!

• What does "proper" mean? – mathworker21 Aug 7 '18 at 11:29
• Proper means $\operatorname{dom} f \neq \emptyset$. – Theo Bendit Aug 8 '18 at 4:23

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.