Question surrounding Exercise 3.12 of Brezis, function is convex and l.s.c. for the weak* topology. Consider Exercise 3.12 of Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations.

Let $E$ be a Banach space and let $x_0 \in E$. Let $\varphi: E \to (-\infty, +\infty]$ be a convex l.s.c. function with $\varphi \not\equiv +\infty$.

*

*Show that the following properties are equivalent:\begin{align*}\text{(A)} & \text{ There exist }R, \,M < +\infty \text{ such that }\varphi(x) \le M \text{ for all }x \in E \text{ with }\|x - x_0\| \le R. \\ \text{(B)}& \text{ }\inf_{f \in E^*} \{\varphi^*(f) - \langle f,\,x_0\rangle\} \text{ is achieved.}\end{align*}


*Assuming (A) or (B) prove that$$\inf_{f \in E^*} \{\varphi^*(f) - \langle f,\,x_0\rangle\} \text{ is achieved.}$$What is the value of this inf?

My question is a simpler one than what the exercise is asking for. How do I see that the function$$\psi(f) := \varphi^*(f) - \langle f,\, x_0\rangle$$is convex and l.s.c. for the weak* topology?
 A: Convexity:
To show $\psi$ is convex we only need to show $\varphi^*$ is convex. Let $f,g\in E^*, \lambda\in [0,1]$. We want to show 
$$
\varphi^*(\lambda f+(1-\lambda)g)\leq \lambda \varphi^*(f)+(1-\lambda)\varphi^*(g)
$$
To prove this let $x\in E$ with $\varphi(x)<+\infty$. Then 
\begin{eqnarray}
\langle \lambda f+(1-\lambda)g,x\rangle-\varphi(x) & = & \lambda \langle f,x\rangle +(1-\lambda)\langle g,x\rangle-\varphi(x)\\
& = & \lambda (\langle f,x\rangle-\varphi(x)) +(1-\lambda)(\langle g,x\rangle-\varphi(x))\\
& \le & \lambda \varphi^*(f) +(1-\lambda)\varphi^*(g)
\end{eqnarray}
which implies 
$$
\varphi^*(\lambda f+(1-\lambda)g)\leq \lambda \varphi^*(f)+(1-\lambda)\varphi^*(g)
$$
Lower semi-continuity:
To show $\psi$ is lsc in the weak* topology we only need to show $\varphi^*$ is lsc in the weak* topology. 
Edit: the remaining of this answer is somewhat modified after @EricThoma pointed out an inaccuracy with my previous argument.
We have to show that, for all $\alpha\in\Bbb R$, $S_{\alpha}=\{f:\varphi^*(f)>\alpha\} $ is open in the weak$^*$ topology. Let $f\in S_{\alpha}$. Then there is $\epsilon>0$ s.t. $\varphi^*(f)\geq\alpha+2\epsilon $, which implies there is $x\in E$ s.t $\langle f , x\rangle -\varphi(x)>\alpha+\epsilon $. Consider 
$$
U=\left\{g\in E^*:\vert \langle g,x \rangle-\langle f,x \rangle \vert<\frac{\epsilon}{2}\right\}
$$
which is an open neighbourhood of $f$ in the weak-$*$ topology. Then for all $g\in U$
$$
\varphi^*(g)\geq \langle g,x\rangle -\varphi(x)\geq \langle f , x\rangle -\varphi(x)-\epsilon/2>\alpha+\epsilon/2>\alpha
$$
