0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @HennoBrandsma you are confused, this person is not showing that $\ell^1$ is a Banach space. $\endgroup$ – Tony S.F. Jun 9 at 21:57
  • $\begingroup$ @TonyS.F. I see now he's trying some alternative norm? $\endgroup$ – Henno Brandsma Jun 9 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 $|x^l_1-x^m_1|<\epsilon$. $\endgroup$ – Tony S.F. Jun 9 at 22:00
  • $\begingroup$ It holds for all $n \in \mathbb N$ so it must hold when $n=1$ $\endgroup$ – SABOY Jun 9 at 22:05
  • 1
    $\begingroup$ Did the exercise ask this, or was it "prove or disprove""??? $\endgroup$ – David C. Ullrich Jun 10 at 0:04
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.