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?

  • $\begingroup$ @HennoBrandsma you are confused, this person is not showing that $\ell^1$ is a Banach space. $\endgroup$ Jun 9, 2019 at 21:57
  • $\begingroup$ @TonyS.F. I see now he's trying some alternative norm? $\endgroup$ Jun 9, 2019 at 21:58
  • $\begingroup$ Indeed. @SABOY I don't see how it follow that $|\sum\limits_{j=1}^n x^l_j-x^m_j|<\epsilon$ implies that $|x^l_1-x^m_1|<\epsilon$. $\endgroup$ Jun 9, 2019 at 22:00
  • $\begingroup$ It holds for all $n \in \mathbb N$ so it must hold when $n=1$ $\endgroup$
    – SABOY
    Jun 9, 2019 at 22:05
  • 1
    $\begingroup$ Did the exercise ask this, or was it "prove or disprove""??? $\endgroup$ Jun 10, 2019 at 0:04

1 Answer 1


It's not so.

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 $F\subset\Bbb N$.)

Another true fact, a misunderstanding of which could lead to the assertion in the question:

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.


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