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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!

  • 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 $(1,\underbrace{-1/n,\ldots ,-1/n}_{n-\text{terms}},0,\ldots)$.2012-10-22
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    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.2012-10-22
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    @DavidMitra: Yes you are right! I think the most simple solution! Thank you for your time!2012-10-22
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    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 $c_0$.2012-10-22
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    Yes i understand that! Thank's again!2012-10-23

3 Answers 3

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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 $c_0(\mathbb{N})$.
  • 0
    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?2012-10-22
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    The problem is that $f$ is not well defined on the whole $c_0$ (take $x_n=n^{-1}$).2012-10-22
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    @MattN. Consider $y_k=(1-k^{-1},k^{-1}2^{-1},k^{-1}2^{-2},k^{-1}2^{-3},\ldots)$2012-10-22
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    @DavideGiraudo You are right. I'll try to salvage my proof see edits.2012-10-22
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    @Norbert: Can you prove step 2 ?2012-10-22
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    @passanger [here](http://math.stackexchange.com/questions/153511/space-of-mathbbr-valued-sequences-converging-to-0-some-basic-results/153533#153533) 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.2012-10-22
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    @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.2012-10-22
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    @Norbert: I am not sure that I undertand at all David Mitra approach.2012-10-22
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    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.2012-10-22
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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?2012-10-22
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    $f(e_k)$ 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 $f$.2012-10-22
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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 $\,Y\subset X\,$ )