I know that weak convergence is equivalent to strong convergence when the space has finite dimension, but $\ell^1(\mathbb{N})$ is not, so I have no idea where to begin.
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1$\begingroup$ This is a basic theorem due to Schur. See Rudin's FA for a proof. $\endgroup$ – Kavi Rama Murthy Aug 2 '20 at 23:50
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$\begingroup$ [Maybe a stupid question] Doesn't weak convergence requires an inner product? How do you define an inner product in $\ell^1(\mathbb N)$? $\endgroup$ – Dmitry Aug 2 '20 at 23:56
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1$\begingroup$ @Dmity Weak convergence replaces inner product with duality product. The real sequence $\langle y , x_n \rangle_{L^1,L^\infty}$ converges to the real number $\langle y , x \rangle_{L^1,L^\infty}$ for every $y \in L^\infty$ means that $x_n$ converges weakly to $x$. In the case of an inner product, it is the same definition, but the dual space is isomorphic to the space itself. $\endgroup$ – rubikscube09 Aug 2 '20 at 23:58
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1$\begingroup$ Note that this is true for sequences, but fails quite badly for nets (in other words, the strong and weak topology do not coincide). $\endgroup$ – MaoWao Aug 3 '20 at 1:11
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This is a bit involved. A proof can be found here: https://people.math.osu.edu/robertson.250/math857/schur.pdf. This proof uses the Vitali-Hahn-Saks theorem, a proof of which can be found here: https://en.wikipedia.org/wiki/Vitali%E2%80%93Hahn%E2%80%93Saks_theorem.
answered Aug 3 '20 at 0:52
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1$\begingroup$ I'd say this is a comment rather than an answer. $\endgroup$ – Leo Sera Aug 3 '20 at 0:55
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1$\begingroup$ It does say to avoid answering questions in comments. So this should be an answer, not a comment. (But culturally on math.stackexchange lots of people put answers in comments.) $\endgroup$ – Stephen Montgomery-Smith Aug 3 '20 at 0:56
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1$\begingroup$ @StephenMontgomery-Smith I see your point but I disagree that this answers the OP question. They are asking for a hint on where to begin, not for links on which the problem is solved for them. $\endgroup$ – Leo Sera Aug 3 '20 at 1:08
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$\begingroup$ @LeoSera Don't upvote it then. $\endgroup$ – Stephen Montgomery-Smith Aug 3 '20 at 1:19
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Here is my attempt at a hint, but I haven't actually seen the proof, so maybe this won't work. Suppose there is a sequence that converges weakly to $0$, but not strongly to $0$. Then w.l.o.g. there is a sequence $x_n \rightharpoonup 0$ and $\|x_n\| \ge 1$. Now choose a subsequence as follows. Given $x_{n_1}, x_{n_2}, \dots, x_{n_k}$, we know that there exists $N_k > N_{k-1}$ such that the sums used to calculate the norms of these vectors is almost all concentrated on $[1,N_k]$. Next, the individual components of $x_n$ converge to $0$. So we can choose $x_{n_{k+1}}$ so that the sum used to calculate the norm is almost all concentrated on $[N_k+1,\infty)$.
Now choose $y \in \ell^\infty$ so that the components of $y$ on $[N_{k-1}+1,N_k]$ are the signs of the components of $x_{n_k}$. Then see that $\langle x_{n_k}, y\rangle \not\to 0$.
To make this into a proper proof, you are going to have to quantify what 'almost all concentrated on' means.
answered Aug 3 '20 at 1:22
