TITLE: Fenchel conjugate of a norm
QUESTION [6 upvotes]: I want to show that the Fenchel conjugate of a norm is the indicator function on the unit ball of the dual norm.
The Fenchel conjugate is defined for a function $f$ as,
$$f^*(y) = \sup\limits_{x} \langle y, x\rangle -f(x),$$
the dual norm $\|\cdot\|_*$ of a norm $\|\cdot\|$ is defined as,
$$\|z\|_* = \sup\limits_{\|u\|\leq 1}\langle z, u\rangle,$$
and the indicator function of a set $C$, denoted $i_C$, is defined as,
$$i_C(x) = \begin{cases} 0 & x\in C,\\ +\infty & x\not\in C\end{cases}.$$
Then the problem is to show that $\|x\|^* = i_{\|x\|_*\leq 1}(x)=\begin{cases} 0 & \|x\|_*\leq 1\\ +\infty & \|x\|_*>1\end{cases}.$
First we consider $x$ with $\|x\|_*>1$. Then, by definition of the dual norm, we have,
$$\sup\limits_{\|y\|\leq 1}\langle y,x\rangle>1.$$
So, $\exists y:\|y\|\leq 1$ such that $\langle y,x\rangle >1$, i.e. $\langle y,x\rangle - \|y\| > 0$. If we choose $z=ty$ then we have,
$$\langle z, x\rangle - \|z\| = t(\langle y,x\rangle -\|y\|)$$
and now letting $t\to\infty$ shows that $\langle z, x\rangle - \|z\|$ is unbounded, i.e. $\|x\|^* = \infty$ for $\|x\|_*>1$.
Now, I am stuck on the remaining case. For this, we consider $x$ with $\|x\|_*\leq 1$. In the book I am reading, the next claim in the proof is,
$$\langle x,y\rangle\leq \|y\|\|x\|_*,\quad\forall y$$
how do they arrive at this claim?

REPLY [2 votes]: (I know, it's a little late) For $L^p(\Omega)$ ($1<p<\infty$) it is easy
$$
\begin{align*}
f(u) &= \|u\|_p,\\
f^*(v)&=\sup_{u\in L^p(\Omega)} \left \{ \int_\Omega u vdx  -\|u\|_p  \right\}\\
&=\sup_{t\geq  0}\sup_{u\in L^p(\Omega),\|u\|=t} \left \{ \int_\Omega u vdx  -\|u\|_p  \right\}\\
&=\sup_{t\geq  0}\sup_{u\in L^p(\Omega),\|u\|=t} \left \{ t\|v\|_q  -t  \right\}\\
&=\sup_{t\geq  0} \left \{ t\|v\|_q  -t  \right\}\\
 &=\begin{cases}
0 &: \|v\|_q\leq 1\\
\infty &: \|v\|_q> 1.
\end{cases}
\end{align*}
$$
For, $L^1(\Omega)$ and $BV(\Omega)$ things get more complicated.