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I was looking at the proof that $l^{p}$ is complete with respect to the standard metric.

Suppose $x^{(n)}$ is a Cauchy sequence in $l^{p}$. Then

Given $\epsilon > 0$, $\exists\,\, n_{0} \in \mathbb{N}$ such that

$$ \sum\limits_{j=1}^{\infty} |x^{(n)}_{j} j- x^{(m)}_{j}|^{p} < \epsilon^{p}$$ for all $m,n \geq n_{0}$ Thus $$ \sum\limits_{j=1}^{N} |x^{(n)}_{j} j- x^{(m)}_{j}|^{p} < \epsilon^{p}$$ for all $N \in \mathbb{N}$ provided $m,n \geq n_{0}$.

First of this implies that $x^{(n)}_{j}$ converges for every fixed $j$ to some real number $x_{j}$. Then they take the limit $m \rightarrow \infty$ in the second equation and conclude (for $n$ fixed)

$$ \sum\limits_{j=1}^{N} |x^{(n)}_{j} j- x_{j}|^{p} < \epsilon^{p}.$$

This is where my problem is. Could someone explain how this taking limits is justified?

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It's a finite sum so linearity of limits? Of course the inequality might no longer be strict. – Alex R. Jul 6 '13 at 7:41
There is a $j$ to the right of $x_j^{(n)}$, which I believe should not be there :) – Gustavo Sep 9 '15 at 1:38
up vote 4 down vote accepted

Note that for fixed $j \in \mathbb{N}$, we have

$$|x_j^{(n)}-x_j^{(m)}| \leq \left( \sum_{j=1}^{\infty} |x_j^{(n)}-x_j^{(m)}|^p \right)^{\frac{1}{p}} < \varepsilon$$

i.e. $(x_j^{(n)})_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$, thus $x_j^{(n)} \to x_j$ as $n \to \infty$ for suitable $x_j \in \mathbb{R}$. Consequently, for fixed $N \in \mathbb{N}$ (so it's a finite sum!),

$$\sum_{j=1}^N |x_j^{(n)}-x_j^{(m)}|^p \stackrel{m \to \infty}{\to} \sum_{j=1}^N |x_j^{(n)}-x_j|^p < \varepsilon^p$$


$$\sum_{j=1}^N |x_j^{(n)}-x_j|^p \leq \varepsilon^p.$$

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How does $$\sum_{j=1}^N |x_j^{(n)}-x_j|^p \leq \varepsilon^p.$$ for a fixed $N $ imply convergence to of the cauchy sequence in $l^p$ – user3503589 Jun 7 '15 at 18:26
@user3503589 I didn't claim that this implies convergence of the Cauchy sequence; I simply proved the inequality the OP was asking for. – saz Jun 7 '15 at 19:30
Ofcourse. I understand . Thanks – user3503589 Jun 7 '15 at 21:05

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