Maximality of subdifferential of a lsc convex function.

Question

I'm trying to demonstrate that given a function (defined on a Hilbert space) $\varphi$ which is convex and l.s.c, the subdifferential ($u \mapsto \partial \varphi (u)$) is a maximal monotone operator. I would like to use the theorem that states that if a monotone operator $A$ is such that the range of $A+Id$ is the entire Hilbert space, then the operator is also maximal. (the viceversa is also true).

For reference, i'm following the proof sketched out in: Monotonicity methods, Brezis, theorem 3.

The trouble arises when I try to demonstrate that $u$ is the minimum of the functional $\psi(v) = \frac{1}{2}\|v-f \|^2+\varphi(v)$ iff $u$ solves the equation $u+\partial \varphi (u) \ni f$, for $f$ fixed in the space.

Does anyone have a clue on how to prove this? Thank you in advance.

Answer 1

Here are the steps

1. If $u$ solves $f\in u+\partial \phi(u)$, then $u$ is a minimum of $\psi$.
2. If $u$ is a minimum of $\psi$, then $0\in\partial \psi(u)$, or equivalently $f\in u+\partial \phi(u)$.

You have to prove $$ \partial \psi(u) = u-f+\partial \phi(u) $$ to prove both individual claims. This equality follows from the sum-rule for the subdifferential.