Fenchel Conjugate of a norm squared I was wondering if the fenchel conjugate of the $\frac{1}{2}||u||^2$, is the $\frac{1}{2}||u||_*^2$, where $||.||_*$ is the dual norm of $||.||$. This seems to be true for the $\ell_2$ norm. However, I do not seem to be able to prove it in general. 
Does anyone know if this is even true in general?
 A: For convenience, I'll transcribe (word for word) the proof given in Example 3.27 (pp. 93-94) of Boyd and Vandenberghe (which is free online). Note that in the passage below, the domain of $f$ is $\mathbb R^n$.

Now consider the function $f(x) = (1/2) \| x \|^2$, where $\| \cdot \|$ is
  a norm, with dual norm $\| \cdot \|_*$. We will show that its
  conjugate is $f^*(y) = (1/2) \|y\|_*^2$. From $y^T x \leq \| y \|_* \|x\|$, we conclude $$ y^T x - (1/2) \| x \|^2 \leq \| y \|_* \| x \| - (1/2) \|x\|^2 $$ for all $x$. The righthand side is a quadratic function of $\|x\|$, which has maximum value $(1/2)\|y\|_*^2$. Therefore for all $x$, we have $$ y^T x - (1/2) \|x\|^2 \leq (1/2) \|y\|_*^2 $$ which shows that $f^*(y) \leq (1/2) \|y\|_*^2$.
To show the other inequality, let $x$ be any vector with $y^T x = \|y\|_* \|x\|$, scaled so that $\|x\| = \|y\|_*$. Then we have, for
  this $x$, $$ y^T x - (1/2) \|x\|^2 = (1/2) \|y\|_*^2, $$ which shows
  that $f^*(y) \geq (1/2) \|y\|_*^2$.

A: Yes, this is true. Please see page 93-94 in the book "convex optimization" https://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf
