# My proof on showing $(\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$ is Banach

Show that $$(\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$$ is Banach , where $$\vert \vert \vert x\vert \vert \vert:=\sup\limits_{n}\vert \sum\limits_{j=1}^{n}x_{j} \vert$$

My idea:

let $$(x^{(k)})_{k} \subseteq \ell^{1}$$ be a Cauchy sequence, thus for any $$\epsilon > 0$$ there exists $$N \in \mathbb N$$ so that for all $$l, m \geq \mathbb N: \vert \vert \vert x^{l}-x^{m}\vert \vert \vert<\epsilon$$. Note that this means that $$\forall n \in \mathbb N$$: $$\vert \sum\limits_{j=1}^{n}x_{j}^{l}-x_{j}^{m}\vert<\epsilon\Rightarrow \vert x_{1}^{l}-x_{1}^{m}\vert < \epsilon$$ but it is then certain that $$\vert x_{2}^{l}-x_{2}^{m}\vert < 2\epsilon$$. This looks like I can reduce the "cauchyness" of $$(x^{(k)})_{k}$$ to cauchy sequences $$(x_{j}^{(k)})_{k}$$ for all $$j \in \mathbb N$$ (this true, no?). Given the completeness of $$\mathbb R$$ with the euclidean norm, and the equivalence of finite dimensional norms, $$\lim\limits_{k\to \infty}x_{j}^{k}:=x_{j}$$ exists and we now show that $$x:=(x_{j})_{j}\in \ell^{1}$$

Question: Do we show that $$\vert \vert x\vert \vert_{1} <\infty$$ or $$\vert \vert \vert x\vert \vert \vert< \infty$$, it's difficult to know since $$\ell^{1}$$ is explicitly defined by the $$\vert \vert \cdot \vert \vert_{1}$$-norm, but we're looking at the space $$(\ell^{1}, \vert \vert \vert \cdot \vert \vert \vert)$$

I will assume the latter:

let $$n \in \mathbb N$$: $$\vert \sum\limits_{j=1}^{n}x_{j}\vert=\vert \sum\limits_{j=1}^{n}\lim\limits_{m\to \infty}x_{j}^{m}\vert=\lim\limits_{m\to \infty}\vert \sum\limits_{j=1}^{n}x_{j}^{m}\vert\leq \lim\limits_{m\to \infty}\vert \vert\vert x^{m}\vert \vert\vert<\lim\limits_{m\to \infty} c <\infty$$

Note: we can take the limit out given the sum is finite and $$\vert \cdot \vert$$ is continuous.

Since this holds for all $$n \in \mathbb N$$ it has to hold for $$\sup\limits_{n}\vert \sum\limits_{j=1}^{n}x_{j}\vert<\infty$$ so $$x \in \ell^{1}$$.

We then argue for convergence similarly. Is my proof ok?

• @HennoBrandsma you are confused, this person is not showing that $\ell^1$ is a Banach space. – Tony Jun 9 at 21:57
• @TonyS.F. I see now he's trying some alternative norm? – Henno Brandsma Jun 9 at 21:58
• Indeed. @SABOY I don't see how it follow that $|\sum\limits_{j=1}^n x^l_j-x^m_j|<\epsilon$ implies that $|x^l_1-x^m_1|<\epsilon$. – Tony Jun 9 at 22:00
• It holds for all $n \in \mathbb N$ so it must hold when $n=1$ – SABOY Jun 9 at 22:05
• Did the exercise ask this, or was it "prove or disprove""??? – David C. Ullrich Jun 10 at 0:04

A little functional analysis shows it's impossible unless the two norms are equivalent; thinking about examples showing they're not equivalent leads to this: If $$x_n=(1,-1,1/2,-1/2,1/3,-1/3,\dots,1/n,-1/n,0,0,0,\dots)$$ then $$(x_n)$$ is a Cauchy sequence that doesn't converge to any $$x\in\ell_1$$.
(It is true that the $$\ell_1$$ norm is equivalent to $$||||x||||=\sup_F\left|\sum_{j\in F}x_j\right|,$$where the sup runs over all finite sets $$F\subset\Bbb N$$.)
Exercise The completion of $$\ell_1$$ in the norm above is the space of all $$x=(x_1,\dots)$$ such that $$\sum x_j$$ converges.