Conjugates of norms How would one find the conjugate of the following :
$$f(x) = \|x\|^2 /2$$ 
The conjugate function is defined as $ f^*(y) = \max_x y^Tx - f(x)$
I am stuck at how I can derive the explicit form for $x$. 
So far, here are my steps:
To maximize I take the derivative and set to $0$. 
$$f'(x) = y - \partial\|x\| \cdot \|x\| = 0$$
$$\partial\|x\| = y/\|x\| $$
Edit : $\|x\|$ is any norm here. Not just the 2-norm. 
Where do I go from here?
 A: For norm, conjugate is indicator of unit ball for dual norm.
See Conjugate Functions at page 7-9:

By Lieven Vandenberghe.
A: Edit: I transcribed a proof from Example 3.27 (pp. 93-94) of Boyd and Vandenberghe here.

Here is a proof in the special case that $\| \cdot \|$ is the $\ell_2$-norm.
Note that $\nabla f(x) = x$.  When you set the gradient equal to $0$, you get $y - x = 0$, or $x = y$.  Thus $f^*(y) = y^T y - \|y\|^2/2 = \|y\|^2/2$.
A: My friend Hrushikesha and I think in the following way - 
$f(x)= \frac{1}{2}\parallel x \parallel ^2$
$f^*(y)=\sup_x(x^Ty-\frac{1}{2}\parallel x \parallel ^2 ) $
From Hölder's inequality ,
$x^Ty \leq \parallel x \parallel \parallel y \parallel_*$ where, $\parallel  y\parallel_*$ is dual norm.
$ \Rightarrow  \sup_x(x^Ty-\frac{1}{2}\parallel x \parallel ^2 ) \leq \sup_x (\parallel x \parallel \parallel y \parallel_* - \frac{1}{2}\parallel x \parallel ^2) = \sup_{\parallel x \parallel} (\parallel x \parallel \parallel y \parallel_* - \frac{1}{2}\parallel x \parallel ^2)$
$ = \frac{1}{2}\parallel y \parallel_* ^2$
For a particular $y$ there is $x$ for which $x^Ty = \parallel x \parallel \parallel y \parallel_*$
Thus, $f^*(y)= \frac{1}{2}\parallel y \parallel_* ^2$
A: This problem can be solved by completing the squares.
$$f^{∗}(y)=\max_x y^Tx−f(x)$$
$$f^{∗}(y)=\max_x ( y^Tx−||x||^2/2)$$
$$f^{∗}(y)=\max_x (y^Tx−||x||^2/2 - ||y||_{*}^2/2 + ||y||_{*}^2/2)$$
$$f^{∗}(y)\leq\max_x (||y||_{*}^2/2 - (||x||-||y||_{*})^2/2)$$
trivially, $f^{∗}(y)= ||y||_{*}^2/2$.
