Finately supported sequences are dense in $l^p$

1,599

Solution 1

That is actually a bit simpler. It suffices to see that since $x \in l_p$ , $\sum |x_n|^p \leq \infty$. Therefore, this series converges, which implies that $ \forall \ \epsilon > 0$, there exists $N_0$ such that $\sum_{n\geq N} |x_n|^p < \epsilon $ for all $n \geq N_0$. Then let $S_N = (x_1,\cdots,x_N,0,\cdots)$, what we have just asserted garantees that $\|x-S_N\|<\epsilon$ $\forall \ N \geq N_0$.

Solution 2

As every $x \in s$ is written $x = \left( x_1, x_2, \dots, x_n, 0, 0, \dots \right)$ and each $x_k \in \mathbb{K}$, we have $\sum_{k=1}^n \left| x_k \right|^p$ must be finite as it is the finite sum of positive numbers. From the definition of $\ell^p$, $x \in s$ satisfies the condition and indeed $s \subset \ell^p$: $$ \ell^p = \left\{ x = \left( x_1, x_2, x_3, \dots \right) \left| x_k \in \mathbb{K} \text{ and } \sum_{k \in \mathbb{N}} \left| x_k \right|^p < \infty \right. \right\} $$ with norm $|| x ||_p = \left( \sum_{k \in \mathbb{N}} \left| x_k \right|^p \right)^{1/p}$.

From the definition of closure, you already know that $\bar{s}$ is the "smallest" subset containing $s$, meaning that $s \subseteq \bar{s} \subseteq \ell^p$. In order to show that $s$ is a dense linear subspace of $\ell^p$, we must have $\bar{s} = \ell^p$, i.e. we only need to show the reverse inclusion $\ell^p \subseteq \bar{s}$.

For every $x \in \ell^p$, we have that $\sum_{k \in \mathbb{N}} \left| x_k \right|^p < \infty$ and by the Cauchy Criterion for Convergent series: given an arbitrary $\epsilon>0$, there is an $n \in \mathbb{N}$ such that $\sum_{k =n+1}^\infty \left| x_k \right|^p <\epsilon^p$. Given such an $x \in \ell^p$, we can define $y = \left( x_1, \dots, x_n, 0, 0 , \dots \right)$ and clearly $y \in s$. By computing the norm, $$|| x-y ||_p = \left( \sum_{k \in \mathbb{N}} \left| x_k - y_k \right|^p \right)^{1/p} = \left( \underbrace{\sum_{k=1}^n \left|x_k - x_k\right|^p}_{=0} + \underbrace{\sum_{k =n+1}^\infty \left| x_k \right|^p}_{<\epsilon^p} \right)^{1/p} < \epsilon,$$ we have that $x \in \bar{s}$, and more importantly $\bar{s} = \ell^p$.

Share:
1,599

Related videos on Youtube

Skortya
Author by

Skortya

Updated on June 25, 2020

Comments

  • Skortya
    Skortya over 3 years

    I want to prove that the set of finitely supported sequences $$s=\{(x_1,x_2,\dots):\exists N>0 \;st\; x_{n\geq N}=0\}$$ is dense in $l^p$, $$l^p = \{ (x_1,x_2,\dots) : \;\;x_i\in\mathbb{K} \;and\; \sum_{i=1}^\infty |x_i|^p<\infty \}$$


    I need to prove $\overline{s}=l^p$. Clearly $s\subset l^p$ and $$\overline{s} = \{x\in l^p \;st\; d(x,s)=0 \}$$ So for $s_0\in\overline{s}$, $\exists x_n\in l^p$ st $$\|x_n-s_0\|\to 0$$

    Now here I am a stuck. Obviously this should imply that $\sum_{i=1}^\infty |s_{0_i}|^p<\infty$ but I do not have a definition of a norm to work with. So how can I arrive to this conclusion?

    • Luke
      Luke over 5 years
      You ask what is the closure of $s$ inside the normed space $l^p$. So for sure it will still be a subset of $l^p$ (you take the closure with respect to the $l^p$-norm)! Therefore, the only thing you need to show is: For each $x \in l^p$, there is a sequence $(s_n)_{n \in \mathbb{N}}$ in $s$ that converges to $x$.
    • Skortya
      Skortya over 5 years
      @Luke, OK so it is just a matter of pointing out that for $x\in l^p$, there exists $(s_n)\in s$ such that for $s_r$ is the the r-tuple of the first $r$ elements in $x$. So if $x=(1, 1/4, 1/8, ...etc)$ then $s_1=(1)$ and $s_3=(1,1/4,1/8)$