Exercise 1.19 in Brezis' Functional Analysis I would like a clarification to an exercise in Brezis' Functional Analysis. Exercise 1.19(2) says


  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 $F(t)\geq t$ $\forall t\in \mathbb{R}$. Set $\phi (x)=F(\|x\|)$. Prove that $\phi$ is convex, lower semi continuous and that $\phi^{\ast}(f)=F^{\ast}(\|f\|)$ $\forall f\in E^{\ast}$.


Now I can prove everything except for the last statement, where I can't understand something. The book defines the conjugate function of $\phi:E\rightarrow (-\infty, +\infty]$ to be $\phi^{\ast}:E^{\ast}\rightarrow (-\infty, +\infty]$ with
$$ \phi^{\ast} (f) =\sup_{x \in E}\{ \langle f , x \rangle - \phi (x)\} $$

With this in mind, the $F^{\ast}$ of the exercise should take as argument elements of $\mathbb{R}^{\ast}$, that is continuous linear functionals $g:\mathbb R\rightarrow \mathbb R$. In this case it takes $\|f\|$, where $f\in E^{\ast}$, so $f:E\rightarrow \mathbb{R}$. I can't understand how this can happen. Surely $\|f\|$ can't be just the norm of $f$ since it would not be linear. So my question is: what is $F^{\ast}(\|f\|)$ supposed to mean in this context?

Please don't offer any hints on how to solve the exercise!
Since $F$ is convex and $F(0) = 0$ is a global minimum, $F$ is monotonically decreasing on $(-\infty, 0]$ and monotonically increasing on $[0, \infty)$. Therefore, for any $x$, $y \in \mathbb{R}$ and any $\lambda \in [0, 1]$,\begin{align*} \phi(\lambda x + (1 - \lambda)y) &amp; = F(\|\lambda x + (1 - \lambda)y\|) \le F(\lambda\|x\| + (1 - \lambda)\|y\|) \\ &amp; \le \lambda F(\|x\|) + (1 - \lambda)F(\|y\|) = \lambda\phi(x) + (1 - \lambda)\phi(y).\end{align*}Let $M \in \mathbb{R}$ and let $H = \{x : F(x) &lt; M\}$. Then $\{x : \phi(x) &lt; M\} = [0, \infty) \cap (H \cup -H)$ is closed since $H$ is closed, as $F$ is lower semicontinuous. Thus, $\phi$ is lower semicontinuous.

Finally,\begin{align*}\phi^*(f) &amp; = \sup_{x \in E} (fx - F(\|x\|)) = \sup_{x \in E} (\pm\|f\|x - F(\|x\|) \\ &amp; = \sup_{x \in E} (\|f\|x - F(\|x\|)) = \sup_{x \ge 0} (\|f\|x - F(\|x\|)),\end{align*}since $\|f\|x - F(\|x\|)$ is strictly less than $0$ for $x &lt; 0$, which is less than the $0$, which is the value achieved when $x = 0$. Therefore,$$\phi^*(f) = \sup_{x \ge 0} (\|f\|x - F(x)) = F^*(\|f\|),$$where the final equality holds since $\|f\|x - F(x)$ is strictly negative when $x &lt; 0$, and hence less than the value at $0$, which is $0$.
Since $F$ is convex and $F(0) = 0$ is a global minimum, $F$ is monotonically decreasing on $(-\infty, 0]$ and monotonically increasing on $[0, \infty)$. Therefore, for any $x$, $y \in \mathbb{R}$ and any $\lambda \in [0, 1]$,\begin{align*} \phi(\lambda x + (1 - \lambda)y) &amp; = F(\|\lambda x + (1 - \lambda)y\|) \le F(\lambda\|x\| + (1 - \lambda)\|y\|) \\ &amp; \le \lambda F(\|x\|) + (1 - \lambda)F(\|y\|) = \lambda\phi(x) + (1 - \lambda)\phi(y).\end{align*}Let $M \in \mathbb{R}$ and let $H = \{x : F(x) &lt; M\}$. Then $\{x : \phi(x) &lt; M\} = [0, \infty) \cap (H \cup -H)$ is closed since $H$ is closed, as $F$ is lower semicontinuous. Thus, $\phi$ is lower semicontinuous.

Finally,\begin{align*}\phi^*(f) &amp; = \sup_{x \in E} (fx - F(\|x\|)) = \sup_{x \in E} (\pm\|f\|x - F(\|x\|) \\ &amp; = \sup_{x \in E} (\|f\|x - F(\|x\|)) = \sup_{x \ge 0} (\|f\|x - F(\|x\|)),\end{align*}since $\|f\|x - F(\|x\|)$ is strictly less than $0$ for $x &lt; 0$, which is less than the $0$, which is the value achieved when $x = 0$. Therefore,$$\phi^*(f) = \sup_{x \ge 0} (\|f\|x - F(x)) = F^*(\|f\|),$$where the final equality holds since $\|f\|x - F(x)$ is strictly negative when $x &lt; 0$, and hence less than the value at $0$, which is $0$.