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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 $$ |x_{n,k}-x_{m,k}|=(|x_{n,k}-x_{m,k}|)^{\frac{1}{2}} \le (\sum\limits_{j=0}^\infty |x_{n,j}-x_{m,j}|)^{\frac{1}{2}}=||x_n-x_m||_2<\infty $$ 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?

TheWanderer
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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 $N$.

daw
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  • Yeah, it's true. There are elements whose $1$-norm is not a finite real number. This implies that $|| \cdot ||_1$ is not a norm in $l^2$. – TheWanderer Jan 19 '21 at 11:47
  • Indeed, so imo (the beginning of) this answer is sufficient and should rather be a comment. – Anne Bauval Oct 16 '23 at 09:45