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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 $\phi_{a_n}(x) \rightarrow \phi (x) \quad \forall x\in \ell^2$?

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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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    @Julian: I posted a new and much more direct answer.2012-06-21
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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 we find $P_nb_m=b_n$ (since $P_n=P_nP_m$ – apply to $a_k$ for $k$ large), and so an application of Pythagoras yields $\lVert b_m-b_n\rVert^2=\lVert b_m\rVert^2-\lVert b_n\rVert^2$ which together with the above shows that $(b_n)$ is a Cauchy sequence. Let $b$ be its limit. Clearly, $P_nb=b_n$ for any $n$.

Whenever $j we find, as above, $\langle a_k,a_j\rangle=\theta=\langle b_n,a_j\rangle,$ and letting $b_n\to\infty$ we conclude $\langle a_k,a_j\rangle=\theta=\langle b,a_j\rangle.$ Now letting $k\to\infty$ we see that $a_k\rightharpoonup b$ (this notation means weak convergence), because the sequence $(a_k)$ is bounded and the $a_j$ span all of $\bigcup_{n=1}^\infty V_n$. (The orthogonal complement, if not trivial, is trivally taken care of.)

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    Uh, I mean $b_n$ I suppose. Thanks for pointing that out.2012-06-22