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 $$\alpha \in \mathbb R$$.

Suppose that $$J$$ is weakly lower semi-continuous on $$X$$, that is, according to your definition, for all $$x \in X$$ and every sequence $$x_n \rightharpoonup x$$ we have $$\liminf\limits_{n\rightarrow\infty} J(x_n) \geq J(x).$$

Suppose $$U_\alpha \neq \emptyset$$. Given a sequence $$(x_n)\in U_\alpha^\mathbb{N}$$, we have $$\alpha \geq J(x_n)$$, and by w.l.s.c., $$\alpha \geq \liminf\limits_{n\rightarrow\infty} J(x_n) \geq J(x),$$ therefore $$x\in U_\alpha$$ so $$U_\alpha$$ is weakly sequentially closed.

Conversely given a sequence $$(x_n)\in X$$ with $$x_n \rightharpoonup x$$. Consider the sequence $$(J(x_n))\in\mathbb{R}^\mathbb{N}$$.

By definition, $$\liminf \limits_{n\rightarrow\infty} J(x_n) = \lim \limits_{N\rightarrow\infty} \inf\limits_{n \geq N} J(x_n).$$ Let us suppose first $$\liminf \limits_{n\rightarrow\infty} J(x_n)$$ is finite, $$\liminf \limits_{n\rightarrow\infty} J(x_n) =\alpha \in \mathbb{R}.$$

Choose $$\epsilon>0$$ (thanks @ammath). Then for every $$n\in \mathbb{N}$$, since the sequence $$\inf\limits_{n \geq N} J(x_n)$$ is monotone increasing (the sets on which the infimum is taken are nested), a subsequence ($$x_{\sigma(n)}$$) satisfies $$J(x_{\sigma(n)})\leq \alpha +\epsilon$$, so $$x_n\in U_{\alpha+\epsilon}$$ (as the weak limit of the subsequence is the weak limit of the sequence). If $$U_{\alpha+\epsilon}$$ is weakly sequentially closed then $$x\in U_{\alpha+\epsilon}$$. , which means $$\epsilon+\liminf \limits_{n\rightarrow\infty} J(x_n)=\epsilon + \alpha \geq J(x).$$ Passing to the limit as $$\epsilon\to0$$, we have obtained $$\liminf \limits_{n\rightarrow\infty} J(x_n)\geq J(x).$$ Finally, let us discuss the $$\liminf \limits_{n\rightarrow\infty}J(x_n)=-\infty$$ issue (thanks @daw below). By assumption, $$J(x)\in \mathbb{R}$$. Take $$\alpha = J(x)-1$$. A subsequence $$(x_{\sigma(n)})$$ is in $$U_{\alpha}$$. Therefore $$x\in U_{\alpha}$$, that is, $$J(x)\leq J(x)-1$$ which doesn't happen.

• If lim inf is infinite, then $J(x)=-\infty$ is a contradiction since $J(x)\in \mathbb R$. So this case cannot happen.
– daw
Jan 26 at 9:08
• @daw thanks, corrected.
– user145413
Jan 26 at 10:50
• The solution is false. What if, for example, $J(x_n) = \frac 1n$? Then $\alpha = 0$ and you won't get your subsequence with $J(x_{\sigma(n)})\le\alpha$. Feb 24 at 18:20
• @amsmath. Thanks.
– user145413
Mar 22 at 22:07