Convex conjugate of the composition of a norm with a convex function

Question

Let $E$ be a normed vector space. Let $F:\mathbb R \rightarrow (-\infty, +\infty]$ be a convex lower semi continuous function such that $F(0)=0$ and $F(t)\geq 0$ $\forall t\in \mathbb{R}$. Set $\phi (x)=F(\|x\|)$.

It is easy to see that $\phi$ is convex, lower semi continuous and that $$\phi^{\ast}(f)\leq F^{\ast}(\|f\|)$$ $\forall f\in E^{\ast}$. How can show that indeed we have $$\phi^{\ast}(f)= F^{\ast}(\|f\|)?$$

Answer 1

Given $f\in E^*$, we are to maximize $f(x)-F(\|x\|)$ over $E$. Let $z\in E$ be a unit norm vector such that $f(z)=\|f\|$. Consider $x=tz$ with $t\in\mathbb{R}$; for such $x$ we have $$f(x)-F(\|x\|) = t\|f\| - F(|t|) \tag1$$ The maximum of (1) over $t\in\mathbb{R}$ is precisely $G(\|f\|)$ where $G$ is the convex conjugate of $F(|x|)$.

Unfortunately you didn't assume $F$ to be even, so an extra step is needed. Since $F\ge 0$ and $F(0)=0$, negative values of $t$ are not going to maximize (1): they lose in comparison to $t=0$. Hence, $$ \sup_{t\in\mathbb{R}} (t\|f\| - F(|t|)) = \sup_{t\ge 0} (t\|f\| - F(|t|)) = \sup_{t\ge 0} (t\|f\| - F(t)) = \sup_{t\in \mathbb{R}} (t\|f\| - F(t)) = F^*(\|f\|) $$ Here the second-to-last step is based on the same logic as above: negative values of $t$ do not matter for the supremum.

All this shows that $\phi^{\ast}(f)\ge F^{\ast}(\|f\|)$. As you said, the opposite inequality is easy: $$f(x)-F(\|x\|)\le \|f\| \|x\| - F(\|x\|) \le F^*(\|f\|)$$