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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?

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

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up vote 3 down vote accepted

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.

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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. –  JSchlather Nov 27 '12 at 7:29
    
@JacobSchlather, thanks your idea is much simpler –  userNaN Nov 27 '12 at 8:25
    
"Thus $f$ is discontinuous at zero." -- a mistake I guess, should be: "Thus $f$ is discontinuous at 1." –  Damian Sobota Apr 22 '13 at 23:33
    
@DamianSobota thank you! –  userNaN Apr 23 '13 at 4:50

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