This is Exercise 3.12 from Brezis:
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{ }\lim_{f \in E^*, \| f \| \to \infty} \{\varphi^*(f) - \langle f,\,x_0\rangle\} = \infty.\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?
After a bit of work, I was able to show now (A) $\implies$ (B). Now I got really stuck on the converse. I tried to follow the solution given in the book, but I do not understand how the solution works:
In particular, why would showing there exists constant $k > 0$ and $C$ such that S1 hold would help us to prove (A)? It does not make sense to me what Brezis meant when he said passing conjugate into S1. I am not sure how conjugate would be "passed" into inequalities?
Moreover, why can we assume after translation that $\psi(0) < \infty$? What if $\psi(0) = \infty$? How can we reduce back to the case when $\psi(0) < \infty$?
So… I can’t completely answer this question, I don’t have time to work out all the details. But … hopefully this is helpful.
To answer your last question first, proposition 1.10 of Brezis says that if $\phi$ is not identically equal to $\infty$ then also $\phi^*$ is not identically equal to $\infty$. I.e. there is some $f \in E^*$ such that $\phi^*(f) < \infty$. Thus, wlog we may assume that $f = 0$ by translating the domain of $\phi$.
To see why the inquality S1 implies the question statement will almost certainly use the fact that the Legendre transform is order reversing. I.e. if $$ \phi \leq \psi $$ then $$ \phi^* \geq \psi^*.$$ Using this fact, S1 becomes something like $$ \psi^* \leq (k \| \cdot \| - C)^* .$$
Finally, you will probably want to use the fact that $\phi^{**} = \phi$ if $\phi$ is convex.