A sequence $(x_n)$ in a normed linear space $X$ is said to converge weakly to $x$ if $$ \lim _{n\rightarrow\infty} \ell(x_n) = \ell(x)$$ Consider the sequence $(f_n) \in C([0,1])$ defined by $$f_n(t) = t^n.$$ Does this sequence converge weakly? Does it have weakly convergent subsequences?

I don't really see the sequences here, do we get one different sequence depending on what $t$ is? $ \lim _{n\rightarrow\infty} \ell f_n(t) = 0, $ for $t \neq 1$. So it seems that f would converge to something not continuous?

  • $\begingroup$ You did not correctly state what weak convergence means. What is $\ell$? $\endgroup$ – Jonas Meyer Nov 27 '12 at 7:29

Consider evaluation functionals $$ \mathrm{ev}_t:C([0,1])\to\mathbb{C}:f\mapsto f(t) $$ Assume $(f_n)$ weakly converges to $f\in C([0,1])$, then $$ f(t)=\lim\limits_{n\to\infty}\mathrm{ev}_t(f_n)=\lim\limits_{n\to\infty}f_n(t)= \begin{cases} 0\quad\mbox{ if }\quad t\neq 1\\ 1\quad\mbox{ if }\quad t= 1 \end{cases} $$ Thus $f$ is discontinuous at $1$. Contradiction, so there is no weak limit for $(f_n)$. The same argument shows that $(f_n)$ have no weakly convergent subsequence.

  • 3
    $\begingroup$ I'm a bit confused, for any $x \in [0,1]$ don't you have a functional $\ell_x$ given by $\ell_x(f)=f(x)$. Which would mean that weak convergence implies point-wise convergence. The sequence converges pointwise to $\chi_{\{1\}}$, so I don't see how it could have a weak limit. $\endgroup$ – JSchlather Nov 27 '12 at 7:29
  • $\begingroup$ @JacobSchlather, thanks your idea is much simpler $\endgroup$ – Norbert Nov 27 '12 at 8:25
  • $\begingroup$ @DamianSobota thank you! $\endgroup$ – Norbert Apr 23 '13 at 4:50

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