Prove a step in exercise 1.19 Brezis's Functional Analysis

Question

Let $E$ be a normed vector space, and $F: \mathbb{R} \to (-\infty, \infty]$ be a convex l.s.c. function such that $F(0) = 0$ and $F(t)\ge 0, \forall t\in \mathbb{R}$. Set $\varphi(x) = F(||x||)$.

Prove that $\varphi$ is convex, l.s.c. and that $\varphi^*(f) = F^*(||f||)$.

I'm stuck at the part proving $\varphi^*(f) = F^*(||f||)$. More specifically, I don't know how to prove $$\sup_{x\in E} (f(x) - \varphi(x)) = \sup_{y\ge 0} (||f||y - F(y)).$$ Thank you very much for any hint or solution.

Answer 1

Since $f(x) \le \|f\|\cdot \|x\|$, the inequality "$\le$" follows.

Take $\epsilon>0$. Then there is $x_\epsilon$ with $\|x_\epsilon\|=1$ and $f(x) \ge \|f\|-\epsilon$. Then for $t>0$ $$ \sup_{x\in E}(f(x)-\phi(x))\ge f(tx_\epsilon) - \phi(t) \ge t (\|f\|-\epsilon) - \phi(t). $$ This holds for all $\epsilon>0$, hence $$\sup_{x\in E}(f(x)-\phi(x))\ge t \|f\| - \phi(t). $$ Taking the supremum over $t>0$ on the right hand side yields the claim.