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Prove that

$$Y= \left\{ x=(x_n)_{n \in\mathbb{N}} \in c_{0}(\mathbb N )~ \Bigg | ~\sum_{n=1}^{\infty} x_n = 0 \right\}$$

is a dense linear subspace of $ c_0( \mathbb N)$.

where $ \displaystyle{c_0( \mathbb N) = \left\{ x=(x_n)_{n \in\mathbb{N}} \in \mathbb R ^{\mathbb N} : \lim_{n \to \infty} x_n =0 \right\}}$

I cannot prove that it is dense.

Any help?

Thank you in advance!

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3  
For denseness, you can show each unit vector is in the closure of $Y$. For example, to show $(1,0,\ldots)$ is in the closure of $Y$, consider vectors of the form $(1,\underbrace{-1/n,\ldots ,-1/n}_{n-\text{terms}},0,\ldots)$. –  David Mitra Oct 22 '12 at 13:15
    
O.K I can show that each $e_n$ is in the closure of $Y$ but I can't see how I can get density. –  passenger Oct 22 '12 at 13:26
    
@DavidMitra: Yes you are right! I think the most simple solution! Thank you for your time! –  passenger Oct 22 '12 at 14:44
1  
Sorry, there was a "typo" in my previous comment (now deleted). I meant to say just use the fact that the linear span of the set of unit vectors is dense in $c_0$. So the closure of $Y$ contains the closure of the linear span of the set of unit vectors, and hence is all of $c_0$. –  David Mitra Oct 22 '12 at 15:53
    
Yes i understand that! Thank's again! –  passenger Oct 23 '12 at 12:28

3 Answers 3

up vote 4 down vote accepted

The elegant proof is the following. Consider linear functional $$ f:c_{00}(\mathbb{N})\to\mathbb{R}:x\mapsto\sum\limits_{n=1}^\infty x_n $$ Then

  1. Show that $f$ is unbounded and $\mathrm{Ker}(f)\subset Y$.
  2. Show that that kernel of each unbounded functional is dense in the domain space.
  3. Recall that $c_{00}(\mathbb{N})$ is dense in $c_0(\mathbb{N})$.
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I'm confused. Consider $x = (1, 0 , 0 \dots) \in c_0$. If $Y$ is dense in $c_0$ then there is a sequence $y_k = (y_n)_k \in Y$ such that $\|y_k - x\| = \max_n |y_{kn} - x_n| \to 0$ as $k \to \infty$. What would be such a sequence? –  Matt N. Oct 22 '12 at 13:00
    
The problem is that $f$ is not well defined on the whole $c_0$ (take $x_n=n^{-1}$). –  Davide Giraudo Oct 22 '12 at 13:01
1  
@MattN. Consider $y_k=(1-k^{-1},k^{-1}2^{-1},k^{-1}2^{-2},k^{-1}2^{-3},\ldots)$ –  no identity Oct 22 '12 at 13:03
1  
@DavideGiraudo You are right. I'll try to salvage my proof see edits. –  no identity Oct 22 '12 at 13:04
    
@Norbert: Can you prove step 2 ? –  passenger Oct 22 '12 at 13:35

The fact that $Y$ is a subspace is quite clear. To see density, we can use a corollary of Hahn-Banach theorem: we just need to show that each linear continuous functional on $c_0(\Bbb N)$ which vanished on $Y$ vanishes on the whole space.

Let $f$ such a functional. We have $f(e_n-e_m)=0$ if $m\neq n$, where $e_n$ is the sequence whose $n$-th term is $1$, the others $0$. So $f(e_k)=:K$. As $\left\lVert\sum_{k=0}^ne_k\right\rVert_{\infty}=1$, we show have $nK\leq \lVert f\rVert$ and $K=0$.

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Can you explain a little more the last line of your solution? –  passenger Oct 22 '12 at 13:09
    
$f(e_k)$ as the same value $K$. Then I use $l(\sum_{1\leq k\leq n} e_k)=nK, and it's supposed to be $\leq$ the norm of $f$. –  Davide Giraudo Oct 22 '12 at 13:20

Taking on Norbert's answer: since the linear functional $\,f\,$ is obviously not the zero functional, we know $\,\ker f\,$ is a maximal subspace of $\,c_0(\Bbb N)\,$ , from which it follows that

$$c_0(\Bbb N)=\operatorname{Span}\{\ker f\,,\,v\,\}\;\;,\;\forall\;v\notin\ker f$$

Perhaps this now will make it simpler to find the solution ( hint: a subset $\,A\,$ of a topological space $\,X\,$is dense in it iff$\,A\cap Y\neq\emptyset\,$ for every non-empty open subset $\,Y\subset X\,$ )

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