Suppose that $(x^{(n)})_{n=1}^\infty$ is a sequence of sequences of complex numbers that is bounded in the following sense: there exists $M>0$ such that
$$\displaystyle\sum_{j=1}^\infty j^2|x_j^{(n)}|^2 \leq M$$
for all $n$. Show that $(x^{(n)})_{n=1}^\infty$ has a convergent subsequence that converges in $\ell^2$.
So I used a diagonal argument to obtain a subsequence $(x^{(n_j)})_{j=1}^\infty$ that converges component-wise, but I need to show the subsequence converges in $\ell^2$. Of course it doesn't necessarily without the boundedness condition so somehow I have to use this. However I have tried and can't seem to use it correctly to show convergence of the sequences. I've tried with the definition $\sum_{k=1}^\infty |x_k^{(n_j)}-y_k|^2 \to \infty$ as $j \to \infty$ (where the $y_k$ are the component-wise limits) as well as trying to show $(x^{(n_j)})_{j=1}^\infty$ is Cauchy in $\ell^2$ but I couldn't make progress. Perhaps it uses the Cauchy-Schwarz Inequality but I couldn't use that in a meaningful way either. Hints, but not full solutions, would be appreciated.
Best Answer
Hint: Define
$$E=\{x\in l^2: \sum_{j=1}^\infty j^2|x_j|^2 \leq M\}.$$
You'll be done if you show that if $(y^{(n})$ is a sequence in $E$ that converges to $y$ in each component, then $y^{(n)}\to y$ in $l^2.$ To do this observe
$$\sum_{j=1}^{\infty}|y_j^{(n)}-y_j|^2 = \sum_{j=1}^{J}|y_j^{(n)}-y_j|^2 + \sum_{j=J+1}^{\infty}\frac{4M}{j^2}.$$