Equiangular sequence in $\ell^2$ 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 $\phi_{a_n}(x) \rightarrow \phi (x) \quad \forall x\in \ell^2$?
 A: 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.
A: 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 $n<m$ 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 $j<n<a$ 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.)
