A continuous, convex functional on a Banach space is weakly lower semicontinuous Let $I:X \rightarrow \mathbb{R}$ be a continuous, convex functional on a Banach space $X$ (or Hilbert for instance). Then how to prove that $I$ is weakly lower semicontinuous. i.e $ \forall u_n \rightarrow u$ weakly in $X$, $I(u) \le \liminf I(u_n)$.
Can I have some hints without using the proof that a lower semicontinuous, convex function is weakly lower semicontinous? I need another proof.
 A: I think it is difficult to prove that a lower semicontinuous, convex function is weakly lower semicontinous without “using its proof”. Of course there are different proofs, by the usual and easiest one is to pick a sub level set of the function, which is closed by continuity, it is convex by convexity, and by Mazur’s theorem it is weakly closed. Hence the function is weakly lowersemicintinuous.
If you want to deal with sequences only, then try to show Mazur’s theorem with sequences, instead of giving it for a granted result.
A: If $I$ is convex then
$$I(y) \geq I(x) + \langle g_x,y-x\rangle$$
for some subgradient $g_x \in X$. So,
$$I(u_n) \geq I(u) + \langle g_{u}, u_n-u\rangle,$$
which implies
$$\liminf_{n}I(u_n) \geq I(u) + \liminf_{n}\langle g_{u}, u_n-u\rangle = I(u) .$$
The reverse implication might fail because $\langle g_{u_n}, u_n-u\rangle $ has $g_{u_n}$ change with $n$.
I imagine that such a subgradient exists in Hilbert spaces but this needs to be checked. For the Banach space case, you would need to argue that such a linear functional exists (rather than the sub gradient).
