Prove that the unitary closed ball of $l_1 (\mathbb N)$ is closed in $l_2 (\mathbb N)$ Prove that the unitary closed ball of $l_1 (\mathbb N)$ is closed in $l_2 (\mathbb N)$.
My attempt: $$\text{Let $(x_n)_{n\in\mathbb N} $ be a sequence such that }\sum_i^\infty  |x_{n_i}| \le 1 \ \text{and} \lim_{n\to \infty  } x_n = x \text { in $ l_2(\mathbb N)$}$$
If we had that the unitary ball was not closed, then we could have $x$ such that $\sum_i^\infty  |x_{i}| > 1$ and that:
$$\forall \epsilon >0 \ , \exists p \in \mathbb N:n>p \implies  \sum_i^\infty (x_{n_i}-x_i)^2 <\epsilon  $$ After playing with this expression, I couldn't find any contradiction.
 A: Observe that $\ell_1(\mathbb{N})\subset\ell_2(\mathbb{N})$, this the unit $\|\;\|_1$-ball is contained in $\ell_2$.
Suppose $(x_k, x: k\in\mathbb{N})\subset \ell_2(\mathbb{N})$ such that

*

*$\sum_n|x_k(n)|\leq 1$,

*$\|x_k-x\|^2_2=\sum_n|x_k(n)-x(n)|^2\xrightarrow{k\rightarrow\infty}0$.

Condition (2) implies
$$|x_k(n)-x(n)|\leq \|x_k-x\|_2,\qquad k,n\in\mathbb{N}$$
Hence
$$\lim_k x_k(n)=x(n),\qquad n\in\mathbb{N}$$
The conclusion can be obtained by an application of Fatou's lemma, for
$$\sum_n|x(n)|=\sum_n\lim_k|x_k(n)|\leq\liminf_k\sum_n|x_k(n)|\leq 1$$
by condition (1).
A: Step I. The closed unit ball in $\ell_1(\mathbb N)$ is a subset of the closed unit ball of $\ell_2(\mathbb N)$.
Step II. If $x^n=(x_k^n)\in\ell_1(\mathbb N)$, $\|x^n\|_1\le 1,$ and $(x^n)$ is $\ell_1-$convergent to $x=(x_k)$, then clearly $x_k^n\to x_k$, for all $k$.
Also, $(x^n)$ is $\ell_1-$Cauchy. But,
$$
\sum_{k\in\mathbb N}|x^m_k-x_k^n|^2
=\sum_{k\in\mathbb N}|x^m_k-x_k^n||x^m_k-x_k^n| 
\le 2\sum_{k\in\mathbb N}|x^m_k-x_k^n|,
$$
since $|x^m_k-x_k^n|\le 2$, for all $m,n,k$. Thus $(x^n)$ is $\ell_2-$Cauchy, and hence is $\ell_2-$convergent.
Say $\|x^n-y\|_2\to 0$, where $y=(y_k)$, then clearly $x_k^n\to y_k$, for all $k$, and hence $y=x$.
But $x$ is in the unit closed ball of $\ell^1$.
