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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 $\chi = S(w).$

maximumtag
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  • By what reasoning did you deduce the existence of a subsequence with $S(w_{n'}) \rightharpoonup S(w)$? That may enable you to deduce the weak convergence of the full sequence. – Daniel Fischer Mar 10 '14 at 17:45
  • @DanielFischer Please see my edited post, I have added the details. – maximumtag Mar 10 '14 at 17:54

1 Answers1

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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 $x_n \to x$.

Daniel Fischer
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  • Thanks for answering. Would not this theorem imply that $S(w_n)$ converges strongly to $S(w)$ then? – maximumtag Mar 10 '14 at 18:59
  • Also I think we need strong convergence of $Sw_{n_{k_j}}$ to $Sw$ in the space $X$ which we don't have. – maximumtag Mar 10 '14 at 19:00
  • No, a bounded sequence in e reflexive Banach space has weakly convergent subsequences, but it need not have any strongly convergent subsequence. The situation, as I gathered from your question, is that we have a compact injection $j \colon X \hookrightarrow Y$, so a weakly convergent sequence in $X$ is strongly convergent in $Y$. And any subsequence has a further subsequence with $j(S(w_{n''}) \to j(S(w))$ pointwise a.e., so $S(w)$ is the only candidate for the weak limit. – Daniel Fischer Mar 10 '14 at 19:05
  • thanks for explaining. I have to say this subsequence stuff is confusing me atm. Let me just ask one question for now: I assume I need to apply the Theorem you wrote with $x_n = S(w_n)$. If so then your theorem states that $S(w_n)$ strongly converges to $S(w)$. This is what my first comment was about. – maximumtag Mar 10 '14 at 19:15
  • Also, are we assuming implicitly the PDE has a unique solution? I.e. this argument will not work without uniqueness? If solution is not unique I guess I could not say eg. $S(w) = \text{whatever}$. – maximumtag Mar 10 '14 at 19:16
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    We apply the theorem to $x_n = S(w_n)$, but the space is $X$ endowed with its weak topology. Only in the weak topology do we know that $S(w_n)$ has any convergent subsequence at all. And whether we need uniqueness of the solution depends on $S$. You wrote "Furthermore, I know $\chi = S(w)$", and in the added part about the pointwise a.e. convergence of a subsequence to $S(w)$. If I don't misinterpret that, you get a subsequence converging pointwise a.e. to $S(w)$ from every sequence $(w_n)$ converging weakly to $w$, which ensures that the premises of the theorem are met. – Daniel Fischer Mar 10 '14 at 19:26
  • Thanks Daniel. I think I understand everything but one part: in my edit, I said that " I am assuming we can interloop the two subsequences with indices $n_{j_k}$ and $n'$ to pass to the limit here". After seeing this thread (http://math.stackexchange.com/questions/709392/two-convergent-sequences-with-different-indices-can-i-make-them-the-same-index/709409?noredirect=1#709392) I am not sure that this assumption is valid. – maximumtag Mar 12 '14 at 13:38
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    @maximumtag The trick is usually that you don't look at unrelated subsequences. You pick a subsequence $(w_{n_k})$ such that $S(w_{n_k})$ converges weakly (to whatever). Then with the compact embedding $X \hookrightarrow Y$, you get a (norm) convergent sequence $y_k = j(S(w_{n_k}))$ in $Y$. Then you get a subsequence $(y_{k_m})$ of the $y_k$ that converges a.e. [to $j(S(w))$]. So you get a sub-sub-sequence of the $w_n$ such that a) $S(w_{n_{k_m}})$ converges weakly in $X$, and pointwise a.e. to $S(w)$. Now if you had started by extracting first an arbitrary subsequence of $(w_n)$, you then – Daniel Fischer Mar 12 '14 at 13:49
  • are precisely in the situation of the theorem. Every subsequence of $(w_n)$ has a further subsequence (by the two steps) that converges weakly to $S(w)$ (the weak limit must be $S(w)$ since a further subsequence converges pointwise a.e.). – Daniel Fischer Mar 12 '14 at 14:02