# brezis- functional analysis Example 1 page 14

I'm having trouble understanding an example in Brezis' functional analysis book. It goes

Example 1. Consider $\phi(x) = ||x||$. It is easy to check that \begin{align*} \phi^*(f) = \begin{cases} 0 \hspace{1em} &\text{if } ||f|| \leq 1, \\ +\infty &\text{if } ||f|| > 1. \end{cases} \end{align*}

where $\phi^*(f)$ is the Legendre transform of $\phi$ evaluated at $f \in E^{*}$, i.e., if $E$ is a vector space, $\phi$ is a function mapping $E$ to $\mathbb{R}$, and $E^*$ is the dual space of continuous linear functionals on $E$, then \begin{align*} \phi^*(f) = \sup_{x \in E} f(x) - \phi(x) \end{align*}

I think I was able to show that $\phi^*(f) = 0$ when $||f|| \leq 1$, but I'm not totally confident in my proof, and I have no idea how to show the second case.

I said that if $||f|| \leq 1$, then by definition we have $||f|| = \sup_{x \in E, ||x|| \leq 1} |f(x)| \leq 1$, so that \begin{align*} \phi^*(f) = \sup_{x \in E} f(x) - \phi(x) = \sup_{x \in E} f(x) - ||x|| = \sup_{x \in E}||x|| f(\frac{x}{||x||}) - ||x|| \leq \sup_{x\in E}||x|| - ||x|| = 0 \end{align*} To show "$\geq$", I said we can write \begin{align*} ||f|| \leq 1 \Rightarrow \forall x, \hspace{.5em} f(\frac{-x}{||x||}) \leq 1 \Rightarrow f(\frac{x}{||x||}) \geq -1 \end{align*} so that \begin{align*} \phi^*(f) = \sup_{x \in E}f(x) - ||x|| = \sup_{x \in E}||x||f(\frac{x}{||x||}) - ||x|| \geq -2||x|| \end{align*} for all $x \in E$, so we can pick $x$ with norm arbitrarily small, so that we obtain $\phi^*(f) \geq 0$. Thus, $\phi^*(f) = 0$.

I'm not sure about the case when $||f|| > 1$. Any suggestions?

• Just a minor comment, $\phi^*(f)$ is the Legendre transform of $\phi$ evaluated at $f$, not the Legendre transform of $f$. Commented Jan 4, 2017 at 5:27

The only problem I can see with your current proof that $\phi^*(f)=0$ for $\|f\|\leq1$ is in the line \begin{align*} \phi^*(f) = \sup_{x \in E}f(x) - \|x\| = \sup_{x \in E}\|x\|f(\frac{x}{\|x\|}) - \|x\| \geq -2\|x\|. \end{align*} You're taking the supremum over $x$ in the (closed) unit ball of $E$, so after taking the supremum (or rather obtaining a lower bound), your result should not depend on $x$. Rather, this line should read \begin{align*} \phi^*(f) = \sup_{x \in E}f(x) - ||x|| = \sup_{x \in E}||x||f(\frac{x}{||x||}) - ||x|| \geq \sup_{x \in E}-2||x||=0. \end{align*} Nonetheless, it is a fine proof.

As for the case that $\|f\|>1,$ there exists some $x$ in the unit ball of $E$ with $f(x)>1$, or even better, $f(x)>1+\varepsilon$ for some given $\varepsilon>0$. Now given $M>0$, show that we can find some $y\in E$ (the obvious choice is some scalar multiple of $x$) such that $$f(y)-\|y\|>M.$$ This will imply the result.