Weak Limits and Subsequences – Functional Analysis

functional-analysishilbert-spacesweak-convergence

Let $S:X \to X$ be a (nonlinear) map between a Hilbert space $X$. I want to show that $S$ is weakly continuous, so if $x_n \rightharpoonup x$, then $S(x_n) \rightharpoonup S(x)$.

To do this, I have the following facts available to me:
$$Sw_{n'} \rightharpoonup \chi$$
for a subsequence $n'$. Furthermore, I know $\chi = S(w)$. So we have that
$$S_{w_n'} \rightharpoonup S(w)$$
i.e., we have the result for a subsequence. What do I need to conclude that the whole sequence $Sw_n$ converges weakly to the same limit so I can say it is weakly continuous?

Edit More details:
Firstly, I have an estimate $|S(w_n)|_X \leq C$ where $C$ does not depend on $w_n$ so that gives $S(w_{n'}) \rightharpoonup \chi$. My $S(w)$ is a solution operator of a PDE:
$$(\frac{d}{dt}S(w), v) + \int_\Omega f(w)\nabla (S(w)) \nabla v = 0$$ for all $v$. Now since $X$ is compactly embedded in a nicer space $Y$, and convergence in $Y$ implies convergence a.e. for a subsequence, we have that $w_{n_{j_k}} \to w$ a.e. (that's a subsubsequence). This allows us to pass to the limit in the term $f(w_{n_{j_k}})$ by some DCT argument. I am assuming we can interloop the two subsequences with indices ${n_{j_k}}$ and $n'$ to pass to the limit here. This finally gives $\chi = S(w).$

Best Answer

The ingredients you have are thus

  1. $S(w_n)$ is bounded, which guarantees the existence of weakly convergent subsequences, and
  2. $S$ is such that the only possible limit of a subsequence is $S(w)$ (by pointwise a.e. convergence of a further subsequence).

That means the full sequence $S(w_n)$ converges weakly to $S(w)$, because of the

Theorem: Let $(x_n)$ be a sequence in a topological space $X$, such that every subsequence $(x_{n_k})$ of $(x_n)$ has a subsequence $\left(x_{n_{k_m}}\right)$ converging to $x\in X$. Then the full sequence converges to $x$.

Proof: Assume $(x_n)$ does not converge to $x$. That means that $x$ has a neighbourhood $U$ such that $A = \{ n : x_n \notin U\}$ is infinite. Enumerating $A$ in ascending order produces a subsequence $(x_{n_k})$ that has no term in $U$. By the premises, $(x_{n_k})$ has a subsequence $\left(x_{n_{k_m}}\right)$ converging to $x$. That means $x_{n_{k_m}} \in U$ for all large enough $m$, contradicting the construction $x_{n_k} \notin U$ for all $k$. Hence the assumption was wrong, and $x_n \to x$.

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