Suppose X is a Seperable Hilbert space and E={$u_{n}$ : $n\in $N} is denumerable orthonormal basis.Let ($x_{n}$) be a sequence in X and $x\in X$.Show that if $\langle x_{n}$,$u_{m} \rangle$ converges to $\langle x$,$u_{m} \rangle$ for all $m\in N $ , then $\langle x_{n},y \rangle$ converges to $\langle x$,$y \rangle$ for all $y\in X$
My attempt:
$\langle x_{n}-x$,$y \rangle$ = $\lim_{m\to\infty}\langle S_{m}$,$y \rangle$ where $S_{m}=\sum_{i=1}^{m}\langle x_{n}-x$,$u_{i} \rangle u_{i} $
Now, $\lim_{n\to\infty}\langle x_{n}-x$,$y \rangle = \lim_{n\to\infty}\lim_{m\to\infty} \langle S_{m}$,$y \rangle $
If limits above can be exchanged , since $\lim_{n\to\infty} \langle S_{m}$,$y \rangle= 0 $ ,we can conclude the proof. Im not sure if this is correct or when the limits can be exchanged.Does the seperability condition comes in play anywhere?
Best Answer
This is not correct. Let $E=\ell^2(\mathbb{N})$ and $u_m(k)=\delta_{m,k}$. The sequence given by $x_n(k)=n\delta_{n,k}$ satisfies $\lim_{n\to\infty}\langle x_n,u_m\rangle=0$ for every $m\in\mathbb{N}$, but if we take $y(k)=\frac 1 k$, then $$ \langle x_n,y\rangle=\sum_{k=1}^\infty n\delta_{n,k}\frac 1k=1. $$