Let $X$ be a normed space and let $\{x_n\}$ be a sequence in $X$ such that $x_n\to x$ weakly. Show that there is a sequence $\{y_n\}$ such that $y_n\in \operatorname{co}\{x_1,...,x_n\}$ and $||y_n-x||\to 0.$

My attempt: put $y_1=x_1$ and for a constant $t_0\in (0,1)$ define $y_n:=t_0x_{n-1}+(1-t_0)x_n$ for $n\geq 2$. clearly $y_n\to x$ weakly. also $||.||-\mathrm{cl}~ \mathrm{co}\{\{x_n\}_{n\in {\Bbb N}}\}= w-\mathrm{cl}~ \mathrm{co}\{\{x_n\}_{n\in {\Bbb N}}\}$ so $x\in||.||-\mathrm{cl}~ \mathrm{co}\{\{x_n\}_{n\in {\Bbb N}}\} $. can I say $y_n\to x$ by norm?

Thanks in advance

  • $\begingroup$ Since the weak closure of $co \{ \{ x_n \}_n \}$ is closed in the norm topology, you can apply the characterization of closed subsets in a metric space in terms of convergent sequences. $\endgroup$ – Crostul Jul 10 '14 at 9:20
  • $\begingroup$ Do you mean $||y_n- x||\to 0$? $\endgroup$ – niki Jul 10 '14 at 9:30
  • $\begingroup$ Exactly. It is the definition of convergence. $\endgroup$ – Crostul Jul 10 '14 at 9:54

Your Answer

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

Browse other questions tagged or ask your own question.