$l^2$ with the $1$-norm is Banach? I have to show that $l^2$ is Banach when endowed with the $1$-norm. To this regard, I take a Cauchy sequence $(x_n)_{n \in \mathbb{N}}$. For each component, I have
$$
|x_{n,k}-x_{m,k}|=(|x_{n,k}-x_{m,k}|)^{\frac{1}{2}} \le (\sum\limits_{j=0}^\infty |x_{n,j}-x_{m,j}|)^{\frac{1}{2}}=||x_n-x_m||_2&lt;\infty
$$
So, $x_n \rightarrow x$ for some sequence $x$. In order to prove that $x \in l^2$, I just observe that $x_n$ converges to a unique limit in $l^2$, and this forces $x$ to be such a limit.
I have only a doubt. If $(x_n)_{n \in \mathbb{N}} \in l^2$ is Cauchy with respect to the distance induced by $1$-norm, is it also Cauchy with respect to distance induced by $2$-norm?
Is my argument correct?
The $l^1$ norm is not a norm on $l^2$, because there are elements in $l^2$ that are not in $l^1$, e.g.,
$$
(1,\frac12, \frac13,\dots).
$$
The Cauchy sequence property is not a problem since $\|x\|_{l^2} \le \|x\|_{l^1}$ for all $x\in l^1$:
$$
\sum_{i=1}^N x_i^2 \le \left(\sum_{i=1}^N |x_i| \right)^2 \le \|x\|_{l^1}^2,
$$
which is true for all $N$.
The norms are equivalent down to a constant, see more details here, and from there you have what you need, the &quot;Cauchy&quot;-ness of the sequence remains.