Is $\frac{1}{2}\|x\|^2$ the only function that is equal to its convex conjugate?

The convex conjugate is defined as

$$ f^{*}(x) = \sup_y\{\langle x, y\rangle - f(y)\}. $$

  • 1
    $\begingroup$ @Arthur The supremum of complex numbers doesn't make much sense, and neither do complex-valued convex functions. $\endgroup$
    – user147263
    Dec 27, 2015 at 19:20

2 Answers 2


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, $f = \frac{1}{2}\|.\|^2.$


It is the only lower semi-continuous function with this property. I will try to find a source.


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.


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