I'm trying to construct a sequence that converges weakly but not in the norm. I assume that in $l_\infty$, $$x_n = (0,\ldots,0,\underset{\substack{n\text{th}\\ \text{position}}}{1},1,\ldots)$$ will do the job. It does not converge to $0$ in the norm, but I do not see why it converges weakly (if it does?).
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1I'm not sure it does. See this post. However, consider the unit vectors in $\ell_2$ for an example of a weakly convergent sequence that is not norm convergent. – David Mitra Jun 08 '13 at 13:25
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1I think DavidMitra is right about your example. If we take the ultralimit $f(x)=\operatorname{\mathscr{U}-lim} x$ for some non-principal ultrafilter $\mathscr U$, then $f\in\ell_\infty^$ and $f(x_n)=1$ for each $n$. This argument wouldn't work if you took $1$ only* in the $n$-th position and zeroes on all other positions. – Martin Sleziak Jun 08 '13 at 13:30
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2Your sequence would furnish a counterexample in $c$. – David Mitra Jun 08 '13 at 13:42
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This particular sequence does not converge weakly in $\ell^\infty$: On one hand, let $\pi_k\colon\ell^\infty \to \mathbb K$ by the $k$-th coordinate projection, then as $$\pi_k(x_n) = 0, \quad n > k $$ the only possible weak limit is $0$. On the other hand, let $c \subseteq \ell^\infty$ denote the subspace of convergent sequences, let $L \in(\ell^\infty)^*$ be a Hahn-Banach extension of $\lim \colon c \to \mathbb K$. We have $L(x_n) = 1$ for all $n$, but $L(0) = 0$. So 0 isn't the weak limit either.
martini
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Thanks for the comments. I was trying to show the following: If F is a Banach space and $f_n\subsetF^∗$ is weak star converging to $f\in F^∗$ and $x\in F$ is the weak limit of $(x_ n)_n\subset F$ then $f_n(x_n)\longrightarrowf(x)$ does not hold in genearal. So I am trying to construct the corresponding sequences, without success so far. – bohem Jun 08 '13 at 13:47
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For any $F$? If you only want some $F$, take $f_n = x_n = e_n \in \ell^2$. Then $f_n(x_n) = 1 \not\to 0 = 0(0)$. – martini Jun 08 '13 at 13:53
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