# Question on dense subset of l^p space

1. Jul 16, 2015

### kostas230

Let $X$ be an infinite set. Consider the set $l^p(X)$, where $1\leq p < +\infty$, of all complex functions that satisfy the inequality
$$\sup \{\sum_{x\in E} |f(x)|^p: E \subset X, \;\; |E|<\aleph_0 \} < +\infty$$.
The function $\| \|_p: l^p(X)\rightarrow \mathbb{[0,+\infty]}$ defined by
$$\| f \|_p = \sup \{ \left( \sum_{x\in E} |f(x)|^p \right)^{1/p}: E \subset X, \; |E|<\aleph_0 \}$$
makes $l^p(X)$ a complete normed vector space.

What I'm trying to show is that there exists a dense subset of $l^p(X)$ with cardinality equal to that of $X$. For every point $a\in X$ we consider the function $\delta_a: X\rightarrow \mathbb{C}$ with $\delta_a(x) = 0$, if $x\neq 0$, and $\delta_a (a) = 0$.

Let $f\in l^p(X)$. Consider the collection $\mathcal{C}$ of all finite subsets of $X$. The relation $\subset$ on $\mathcal{C}$ makes this collection a directed set. For every $E\in\mathcal{C}$, let $g_E = \sum_{a\in E} f(a) \delta_a$. The mapping $E\rightarrow g_E$ constitutes a net, which I'm trying to show that it converges to $f$.

If $\epsilon>0$ there exists a finite subset $E$ of $X$ such that
$$\|f\|_p - \left(\sum_{x\in E}|f(x)|^p \right)^{1/p}<\epsilon$$
which in turn leads to
$$\| f \|_p - \| g_E \|_p <\epsilon.$$Therefore, if $G$ a finite subset of $X$ that contains $E$ then $\| f \|_p - \| g_G \|_p < \epsilon$. What I've been having problem proving is the inequality $\| f - g_G \|_p < \epsilon$. Any ideas on this? Thanks in advance! :)

2. Jul 16, 2015

### micromass

Staff Emeritus
How would you prove it for $X = \mathbb{N}$? Can you mimic that proof?

3. Jul 17, 2015

### mathman

There seems to be at least one serious typo. $\delta_a(a)=0$ Should be =1?

4. Jul 18, 2015

### WWGD

Together with unstated conditions:
1)What is $E$
2) I think the condition $|| ||_p$ should have range in $[0, \infty)$. I am not aware of infinity-valued normed spaces.
3) This may be obvious, but I guess your bet is defined on $l^p(X)$, and not on, some other set?

5. Jul 18, 2015

### micromass

Staff Emeritus
$E$ is finite subset of $X$.

It does since $E$ is finite.

6. Jul 18, 2015

### WWGD

Ah, yes, my bad, I missed the "$|E| < \aleph_0$"; I expected a sort of $|E| < \infty$; clearly finite sums are bounded/convergent.

7. Jul 20, 2015