2
$\begingroup$

I would like to ask some questions regarding convergence in the weak* topology and subspaces.

Let $X$ be a normed space with subspace $A \subset X$. Assume $X$ is endowed with the weak* topology. Let $\{u_{m}\}_{m}$ be a sequence in $X$.

  1. If $u_{m} \rightharpoonup^{*} u$ in $A$ then since $X$ is not necessarily first-countable in the weak* topology then would I be right in stating that $u$ is not necessarily in $A$?

  2. If $u_{m} \rightharpoonup^{*} u$ in $X$ then does that imply that $u_{m} \rightharpoonup^{*} u$ in $A$ if $\{u_{m}\}_{m} \subset A$? Similarly, if $u_{m} \rightharpoonup^{*} u$ then does that imply that $u_{m} \rightharpoonup^{*} u$ in $X$?

  3. Would question 2 require that $A$ is closed in $X$?

I am trying to apply this to Sobolev space $W^{1,p}(\Omega)$ and closed subspace $W^{1,p}_{0}(\Omega)$. Any assistance or hints would be appreciated.

$\endgroup$
1
$\begingroup$
  1. No. You assume that $u_n\overset{w^*}{\to} u$ in $A$ and then asks if $u\in A$. Of course, $u\in A$, because of the assumption.

  2. Yes. See this answer.

  3. No. Carefully read the proof given in the link.

$\endgroup$
8
  • $\begingroup$ For 1. Say for example if $A$ is open in $X$ and $u_{n} \rightharpoonup^{*} u$ in $A$ with $u \in \partial A$? Would one not write $u_{n} \rightharpoonup^{*} u$ in $A$ in that case as well? $\endgroup$ – user100431 Apr 21 '14 at 8:44
  • $\begingroup$ No. For example $A=\operatorname{Ball}_{\ell_2}(0,1)\setminus\{0\}$. Consider $u_n=e_n\in A$, then $u_n\overset{w^*}{\to} 0=u\in\partial{A}=\{0\}$. But $u\notin A$. $\endgroup$ – Norbert Apr 21 '14 at 9:13
  • $\begingroup$ Okay, so it would make more sense to write $u_{n} \rightharpoonup^{*} u$ in $X = {\text{Ball}}_{l_{2}(0,1)}$ where $\{u_{n}\}_{n} \subset A$? $\endgroup$ – user100431 Apr 21 '14 at 9:23
  • $\begingroup$ No, in your notation $X$ is normed space, not a ball $\endgroup$ – Norbert Apr 21 '14 at 9:25
  • $\begingroup$ Yes, but I understand. $u_{n} \rightharpoonup^{*} u$ in $A$ would imply that the convergence is related to the subspace topology and therefore it requires that $u_{n}, u \in A$. If $u_{n} \rightharpoonup^{*} u$ with $\{u_{n}\}_{n} \subset A$ but $u \in X \setminus A$ then this would be convergence with respect to the topology endowed on $X$. Is that the point of your example? $\endgroup$ – user100431 Apr 21 '14 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy