Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

1 Answer 1

up vote 1 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!),

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

thus,

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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