5

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!

Brian M. Scott
  • 616,228
passenger
  • 3,793
  • 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 Answers3

5

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 $c_0(\mathbb{N})$.
Norbert
  • 56,803
  • 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? – Rudy the Reindeer 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 $y_k=(1-k^{-1},k^{-1}2^{-1},k^{-1}2^{-2},k^{-1}2^{-3},\ldots)$ – Norbert Oct 22 '12 at 13:03
  • 1
    @DavideGiraudo You are right. I'll try to salvage my proof see edits. – Norbert Oct 22 '12 at 13:04
  • @Norbert: Can you prove step 2 ? – passenger Oct 22 '12 at 13:35
  • @passanger here you can find a proof of the following fact $$\mathrm{dist}(x,\mathrm{Ker}(f))=\frac{|f(x)|}{\Vert f\Vert}$$ From this fact it follows that kernel of unbounded operator is dense in the domain space. – Norbert Oct 22 '12 at 13:42
  • @passenger Not at all, but recall that my solution is an overkill, but more enlightening. A straightforward approach is suggested by David Mitra in comments to your question. – Norbert Oct 22 '12 at 13:51
  • @Norbert: I am not sure that I undertand at all David Mitra approach. – passenger Oct 22 '12 at 13:53
  • Dear @Norbert, thank you! I'm sorry for the late reply, but I didn't get pinged (yet again). It seems that pinging on SE is somewhat buggy. – Rudy the Reindeer Oct 22 '12 at 14:16
1

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 $K=0$.

Davide Giraudo
  • 172,925
1

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 $\,Y\subset X\,$ )

DonAntonio
  • 211,718
  • 17
  • 136
  • 287