Here is the definition of convex conjugates in Brezis:
My question is: How do we see that $f$ maps to $f(x) - \phi(x)$ is continuous? Here is my attempt of showing it and I wonder if this is correct:
For $x \in E$, let $\varphi_x: E^* \to (-\infty, \infty]$ be $\varphi_x(f) = f(x) - \varphi(x)$. Let $\epsilon > 0$, then $$ |\varphi_x(f) - \varphi_x(g)| = |f(x) - g(x)| \leq (\| f \|_{E^*} - \| g \|_{E^*}) \| x \|_E \leq \| f - g \|_{E^*} \| x \|_E < \epsilon $$ as long as $\delta = \frac{\epsilon}{\| x \|_E}$ such that $\| f - g \|_{E^*} < \delta$.
I would reason as follows,
Note that if $x \notin \operatorname{dom} \phi$ then $\phi(x) = +\infty$ and so $\langle f, x \rangle - \phi(x) = -\infty$. Since $\operatorname{dom} \phi$ is non empty, there is at least one $x^*$ for which $\phi(x^*)$ is real and so $\phi^*(f) \ge \langle f, x^* \rangle - \phi(x^*)$ and so we have $\phi^*(f) = \sup_{x \in E} \langle f, x \rangle - \phi(x) = \sup_{x \in \operatorname{dom} \phi} \langle f, x \rangle - \phi(x)$.
For each $x \in \operatorname{dom} f$ the map $g_x(f)= \langle f, x \rangle - \phi(x)$ is continuous (in $f$), affine and hence convex. The $\operatorname{epi} g_x$ is convex and closed.
Since $\operatorname{epi} \phi^* = \cap_{x \in \operatorname{dom} f} \operatorname{epi} g_x$ we see that $\operatorname{epi} \phi^*$ is convex and closed, and hence $\phi^*$ is lsc.