# Weak lower semi continuity weakly sequentially closed

Let X be a Banach space and $$J:X \rightarrow \mathbb{R}$$ a functional. Show that J is weakly lower semi-continuous, if and only if the set $$U_\alpha$$ is weakly sequentially closed for any $$\alpha$$ $$\in \mathbb R$$ provided that it is non-empty.

$$U_\alpha := \{w ∈ X : J(w) ≤ \alpha\}$$

I am confused on how to use the definitions to proceed.

A functional $$f : X \rightarrow \mathbb{R}$$ is weakly lower semicontinuous on X if for all $$x \in X$$ and every sequence $$x_n → x$$ which converges weakly to x ∈ X, we have $$\liminf\limits_{x\rightarrow0} f(x_n) ≥ f(x)$$.

We need to use the following lemma:

Let $$C\subset X$$ be a closed and convex subset of a normed space $$X$$. Then C is weakly sequentially closed, i.e. for a sequence $$(u^n )_n$$ in $$C$$ with $$u^n\rightharpoonup u$$ in X for $$n\rightarrow \infty$$ we have that $$u \in C$$.

and also the fact that:

Let $$X$$ be a Banach space. If a functional $$J : X \rightarrow \mathbb R$$ is continuous and convex, then $$U_\alpha$$ is weakly sequentially closed for any $$\alpha \in \mathbb R$$.