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?

  • $\begingroup$ What was the book? $\endgroup$
    – Mikhail
    Mar 13, 2019 at 22:45
  • $\begingroup$ I can't remember now to be honest but most likely Convex Analysis and Monotone Operator Theory in Hilbert Spaces by Bauschke and Combettes $\endgroup$ Mar 13, 2019 at 22:52
  • $\begingroup$ Cauchy-Schwarz inequality $\endgroup$
    – Kumar
    Mar 26, 2021 at 13:11

2 Answers 2


The claim that $\langle x,y\rangle\leq \|y\|\|x\|_*$ holds for all $x$ and $y$ is proved as follows: Fix $x$. If $y=0$ the result is trivial. If $y\neq0$, then $\left\|\frac{y}{\|y\|}\right\|=1\leq1$, so that $\langle x,\frac{y}{\|y\|}\rangle \leq\sup_{\|y\|\leq1}\langle x,y\rangle=\|x\|_*$. Multiplying by $\|y\|$ yields the claim.


(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.

  • $\begingroup$ Late but still useful, thanks! $\endgroup$ Oct 21, 2021 at 12:54

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