Basis coefficients in strong convergence of sequence in Hilbert space Let $\{v_i\}_{i\in I}\subset H$ be a basis of a Hilbert space which is not necessarily orthogonal, and a sequence $x_n\in H$ such that:
$$x_n=\sum_{i\in I}{\alpha_n^iv_i}$$
Now, assume that $x_n\to x$ (strongly convergence) and:
$$x=\sum_{i\in I}{\alpha^iv_i}$$
Is it true that $\alpha_n^i\to \alpha^i$ for every $i\in I$?
 A: Consider the Hilbert space $\ell^2$ and let $(e_n)_{n=1}^\infty$ be the canonical vectors.
Denote $e = \sum_{n=1}^\infty \frac1n e_n$ and let $B$ be an algebraic basis for $\ell^2$ containing the linearly independent set $S = \{e_n : n\in \Bbb{N}\} \cup \{e\}$.
Consider
$$x_n := \sum_{k=1}^n \frac1k e_k = \sum_{b \in B} \alpha_n^b b, \quad \alpha_n^b = \begin{cases}
\frac1k, &\text{if $b = e_k$ with $1 \le k \le n$}, \\
0, &\text{if $b = e_k$ with $k > n$}, \\
0, &\text{if $b = e$}, \\
0, &\text{if $b \in B\setminus S$}, \\
\end{cases} $$
$$x := \sum_{k=1}^\infty \frac1k e_k = e = \sum_{b \in B} \alpha^b b, \quad \alpha^b = \begin{cases}
0, &\text{if $b = e_k$ with $k\in\Bbb{N}$}, \\
1, &\text{if $b = e$}, \\
0, &\text{if $b \in B\setminus S$}, \\
\end{cases} $$
Then $x_n \to x$ strongly in $\ell^2$ but
$$\lim_{n\to\infty} \alpha^b_n = \begin{cases}
\frac1k, &\text{if $b = e_k$ with $k\in\Bbb{N}$}, \\
0, &\text{if $b = e$}, \\
0, &\text{if $b \in B\setminus S$}, \\
\end{cases} \ne \alpha_b.$$
