Is $\frac{1}{2}\|x\|^2$ the only function that is equal to its convex conjugate? Is $\frac{1}{2}\|x\|^2$ the only function that is equal to its convex conjugate?
The convex conjugate is defined as
$$
f^{*}(x) = \sup_y\{\langle x, y\rangle - f(y)\}.
$$
 A: This is an elementary result by J.-J. Moreau himself (see Proposition 9(a) of his famous "Proximité et dualité dans un espace hilbertien").
Let's proof it. So, let $X$ be a Hibert space with inner product $(x,y) \mapsto \langle x, y\rangle$ and norm $x \mapsto \|x\| := \langle x, x\rangle^{1/2}$. Let $f:X \rightarrow (-\infty,+\infty]$ an extended real-valued function. Define the convex conjugate of $f$, denoted $f^*$, by
$$f^*(y) := \sup_{x \in X}\langle x, y\rangle - f(x), \; \forall y \in X.$$
Note that $f^*$ is always convex l.s.c (being the supremum of affine functions) without any assumptions whatsoever on $f$. By the above
  definition of $f^*(y)$, it's clear that for any pair $(x,
y) \in X^2$, we have $f^*(y) \ge \langle x, y \rangle - f(x)$, and so

$$f(x) + f^*(y) \ge \langle x, y \rangle.
$$

This is just the
Fenchel-Young inequality, which holds absolutely, without any
assumption like convexity of $f$, etc. We now show
that $f^* = f$ iff $f = \frac{1}{2}\|.\|^2$. 
($\impliedby$) For any $y\in X$, we have $$(\frac{1}{2}\|.\|^2)^*(y)
:= \sup_{x \in X}\langle x, y\rangle - \frac{1}{2}\|x\|^2 = \sup_{x \in
  X}\frac{1}{2}\|y\|^2 - \frac{1}{2}\|x - y\|^2 =
\frac{1}{2}\|y\|^2.$$
($\implies$) Suppose $f^* = f$. Then evaluating the Fenchel-Young
inequality with $y = x$, we get $2f(x) \ge \langle x, x\rangle$, i.e
$f^*(x) = f(x) \ge \frac{1}{2}\|x\|^2$ for all $x \in X$. Also, because $f$ must be convex l.s.c, it equals its convex biconjugate $f^{**}$, and so we have
$$
f(x) = f^{**}(x) := \sup_{y \in X}\langle x, y\rangle - f^*(y) \le \sup_{y \in
  X}\langle x, y\rangle - \frac{1}{2}\|y\|^2 =: (\frac{1}{2}\|.\|^2)^*(x) =
\frac{1}{2}\|x\|^2.
$$
Thus, $f = \frac{1}{2}\|.\|^2.$ 
A: It is the only lower semi-continuous function with this property. I will try to find a source.
www.cs.cmu.edu/~yaoliang/mynotes/Intro2COf.pdf
So apparently, the Euclidian vector norm is self-conjugate. For matrix norms, the Frobenius norm is self-conjugate. I don't know how the proofs go, sadly.
