About converging sequence of norm-one elements of $\ell^2$ I got stuck in this question for days, so any help/hint is appreciated.
Assume that $T$ is the unit ball in $\ell^2$ and $(x_n)_{n=1}^{\infty} \subset T$ and that $(x_n(i))_{n=1}^{\infty}$ converges to $y(i)$ for each $i\in \mathbb{N}$, show that if $\Vert y \Vert_2=1$; then $(x_n)_{n=1}^{\infty}$ converges to $y$ in norm.
 A: Let's write:
$$\tag {1}
\|x_n-y\|^2=\langle x_n-y,x_n-y\rangle = \|x_n\|^2+1-2\text{Re}\,\langle x_n,y\rangle
$$
Fix $\varepsilon > 0  $. Then there exists  $k_0$ such that $\sum_{k>k_0}|y (k)|^2 +\left (\sum_{k>k_0}|y (k)|^2\right)^{1/2}<\varepsilon/2$ and $|x_n (i)-y (i)|<\varepsilon /2^i $ for $i=1,\ldots,n_0$ and $n>k_0$.
Then  \begin{align}|\langle x_n,y\rangle -1|&=\left|\sum_{k=1}^{k_0}(x_n (k)-y (k))\,\overline {y (k)}+\sum_{k>k_0}^\infty x_n (k)\,\overline {y (k)}+|y (k)|^2\,\right|\\ \ \\ &\leq\frac\varepsilon 2+\sum_{k=1}^{k_0}\frac\varepsilon  {2^{k}}<\varepsilon. 
\end{align}
Going back to $(1) $, if $n>k_0$,$$\|x_n-y\|^2\leq2-2\text {Re}\,\langle x_n,y\rangle=2\text {Re}\, (1-\langle x_n,y\rangle )<2\varepsilon. 
$$
A: Since $x_n(i)\to y(i)$ for all $i$, Fatou's lemma implies that
$$ 1=||y||_2^2=\sum_i|y(i)|^2\leq \liminf_{n\to\infty}\sum_i|x_n(i)|^2=\liminf_{n\to\infty}||x_n||_2^2 $$
On the other hand, since all the $x_n$ are in the unit ball, we have
$$ \limsup_{n\to\infty}||x_n||_2^2\leq 1$$
and combining the two inequalities above shows that
$$ \lim_{n\to\infty}||x_n||_2=1$$
Next, observe that
$$ 2|y|^2+2|x_n|^2-|x_n-y|^2\geq 0$$ 
so Fatou's lemma implies that
$$ 4\leq \liminf_{n\to\infty}\sum_i[2|y(i)|^2+2|x_n(i)|^2-|x_n(i)-y(i)|^2] $$
$$ =4-\limsup_{n\to\infty}\sum_i|x_n-y|^2$$
and rearranging shows that
$$ \limsup_{n\to\infty}\sum_i|x_n(i)-y(i)|^2=0$$
which is the desired result.
