I have a countable infinite normed, equiangular sequence $a_n \in \ell^2$, i.e $\langle a_n, a_m \rangle=\theta$ for $n\not=m$ and $\langle a_m, a_m \rangle =1$ for some $\theta <1$. It's clear that the $a_n$ does not converge. Is it still possible that their duals $\phi_{a_n}=\langle a_n,\cdot \rangle$ converge pointwise, i.e $\phi_{a_n}(x) \rightarrow \phi (x) \quad \forall x\in \ell^2$?
Equiangular sequence in $\ell^2$
2 Answers
Yes. In fact, some subsequence will converge weakly, thanks to the Banach–Alaoglu theorem. (Notice that a Hilbert space is reflexive, and what you call pointwise convergence is the same as weak, or weak* convergence, the latter two being equivalent because of reflexivity.
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0+1, I upvoted. Their is an example where $a_n(k)=\delta_{nk}$ which show that we don't need to extract a subsequence. Maybe you can provide an example which works for $\theta\in (0,1)$ and where we indeed need to extract a subsequence. – 2012-06-21
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0@DavideGiraudo: Actually, I am beginning to think that you *never* actually need a subsequence. The point being that, given two sequences of the kind under consideration, there is an isometry between the spans of the two sequences mapping one sequence to the other. I further conjecture that the weak limit must be $0$, but unfortunately I don't have the time to work out the details right now. (Basically, the weak limit would be invariant under permutations of the sequence …) – 2012-06-21
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0On second thought, no, the weak limit (call it $a_*$) will not be zero, of course. Rather, it will be uniquely determined by $\langle a_n,a_*\rangle=\theta$ for all $n$. (Silly me. What was I thinking? Gotta run … but I stand by the permutation argument.) – 2012-06-21
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1In fact by Banach-Saks theorem the norm of the weal limit is $|\theta|$. – 2012-06-21
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0@DavideGiraudo: Good catch! Uh, the square root, perhaps? Sending $n\to\infty$ in $\langle\sum_1^n a_k,\sum_1^n a_k\rangle=n+\theta\cdot(n^2-n)$? – 2012-06-21
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0So you are claiming it's possible to prove the weak convergence of the sequence itself using a permutation argument? I'd appreciate if you could elaborate on that! – 2012-06-21
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0@Julian: I can do better, now that I thought about it – all while walking about town on differet errands today. Now I only have to find some time to write it up. (Assuming I haven't made a silly mistake of course.) – 2012-06-21
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0@Julian: I posted a new and much more direct answer. – 2012-06-21
For this answer, I will assume that $0\le\theta<1$. Then $a_1$, …, $a_n$ are linearly independent for any $n$, because the $n\times n$ matrix $(\langle a_j,a_k\rangle)$ is positive definite, and in particular invertible.
Let $P_n$ be the orthogonal projection on the span $V_n$ of $a_1$, …, $a_n$. Then for any $k>n$, $$P_na_k=b_n:=\frac{\theta}{1+(n-1)\theta}\sum_{j=1}^n a_j$$ because whenever $j\le n$, $$\langle P_na_k,a_j\rangle=\langle a_k,P_na_j\rangle=\langle a_k,a_j\rangle=\theta=\langle b_n,a_j\rangle$$ where $P_na_k$ and $b_n$ belong to $V_n$, which is spanned by the $a_j$. We compute $$\lVert b_n\rVert^2=\frac{\theta^2n}{1+(n-1)\theta}$$ and note that $$\lim_{n\to\infty} \lVert b_n\rVert^2=\theta.$$
Whenever $n Whenever $j
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0looks like a very nice solution! what do you mean by $p_n$ in $\lVert b_m-b_n\rVert^2=\lVert p_m\rVert^2-\lVert p_n\rVert^2$? – 2012-06-22
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0Uh, I mean $b_n$ I suppose. Thanks for pointing that out. – 2012-06-22