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.

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

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