Subdifferentiablity of a convex functional On a general Banach space, for a functional $F:X\to\mathbb{R}_{\infty}$, we define the Frechet subdifferential and subdifferential by the following:
\begin{align*}
\mbox{Frechet differential: }\partial^F F(u)&=\{\xi\in X^*|\forall w\in X, F(w)\geq F(u)+\langle\xi,w-u\rangle_{X}+o(\lVert w-u\rVert)\mbox{ for }w\to u\},\\
\mbox{subdifferential: }\partial F(u)&=\{\xi\in X^*|\forall w\in X, F(w)\geq F(u)+\langle\xi,w-u\rangle_{X}\}.
\end{align*}
I guess for convex functionals, these two notions should be same but do not know how to do.
Can someone maybe help?
 A: You can start with a $\lambda-$convex function.
For such a function, claim that the Frechet differential satisfies the following global representation:
$$\partial^{F}F(u)=\{\xi\in X^*|\forall w\in X,F(w)\geq F(u)+\langle\xi,w-u\rangle+\frac{\lambda}{2}\lVert w-u\rVert^2\}.$$
Why this?
First, by obvious observation, we can see this representation is indeed Frechet differential.
Then conversely, take $\xi\in\partial^{F}F(u)$, we then need to show $\xi$ satisfies this representation.
By $\lambda-$convexity, we have that
$$F[\theta u_1+(1-\theta)u_0]\leq \theta F(u_1)+(1-\theta)F(u_0)-\frac{\lambda}{2}\theta(1-\theta)\lVert w-u\rVert^2,$$
then let $u_1=w$ and $u_0=u$, we get
$$\frac{1}{\theta}\left[F[u+\theta(w-u)]-F(u)\right]+\frac{\lambda}{2}\theta(1-\theta)\lVert w-u\rVert^2\leq F(w)-F(u).$$
Then let $\theta\to0$ and notice that
$$\liminf_{w\to u}\frac{F(w)-F(u)-\langle\xi,w-u\rangle}{\lVert w-u\rVert}\geq0,$$
so we get the inequality
$$F(w)-F(u)\geq\langle\xi,u\rangle+\frac{\lambda}{2}\lVert w-u\rVert^2.$$
This proves the claim.
Then back to your question, the usual convexity is just $0-$convexity, then take $0$ into the representation, we can easily get the answer.
A: Clearly, the Frechet subdifferential contains the convex subdifferential. Now, let $\xi \in \partial^F f(u)$.
Since for convex functions, difference quotients $t\mapsto \frac1t ( f(u+td)-f(u))$ are monotonically increasing in $t>0$, it follows for $t\in (0,1)$
$$
f(w)-f(u) \ge \frac{ f(u+t(w-u))-f(u) }t \ge
\frac1t( \langle \xi, t(w-u)\rangle + o(\|t(w-u)\|).
$$
Passing to the limit $t\searrow 0$ proves $f(w)-f(u) \ge \langle \xi, w-u\rangle$. Since $w$ is arbitrary, it follows $\xi\in \partial f(u)$.
