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We know that in $c_0$ the standard unit vector basis $(e_i)_{i=1}^{\infty}$ is an unconditional basis. For $n\in\mathbb{N}$, let $s_n=\sum\limits_{i=1}^{n}e_i$, my question is that

How to prove that $(s_n)_{n=1}^{\infty}$ is a Schauder basis which is not unconditional basis?

I am grateful of any help

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Looking for biorthogonal functionals seems like a good place to start. We want $\Lambda_n\in c_0^*$ so that $\Lambda_n s_n=1$ and $\Lambda_n s_m=0$ for $m\ne n$. Now $\Lambda_n$ is given by some $\ell^1$ function, and it's not hard to see what the coordinates must be, turns out $$\Lambda_n x=x_n-x_{n+1}$$works.

So if we're going to write $x=\sum a_ns_n$ the only possible choice is $a_n=x_n-x_{n+1}$. We need to show that $$x=\sum_{n=1}^\infty(x_n-x_{n+1})s_n\quad(x\in c_0),$$with convergence in norm.

This is not hard. Consider the partial sum $$\sum_{n=1}^N(x_n-x_{n+1})s_n.$$

Take that partial sum and rewrite it as a linear combination of $e_1,\dots,e_n$. The coefficients come out very simple (I should leave something to you) and you see that $$\left|\left|x-\sum_{n=1}^N(x_n-x_{n+1})s_n\right|\right|\le|x_{N+1}|+\left|\left|x-\sum_{n=1}^Nx_ne_n\right|\right|.$$This tends to zero as $N\to\infty$ because $x\in c_0$.

So every $x\in c_0$ is equal to $\sum a_ns_n$. The $\Lambda_n$ above show that the coefficients are unique, so $(s_n)$ is a Schauder basis.

But it's not unconditional. Let $x=(x_n)$ be any sequence of reals such that $x_n\to0$ but $\sum|x_n-x_{n+1}|=\infty$ (for example, $$x_n=a+\sum_{j=1}^n(-1)^j\frac1j$$for a suitable value of $a$).

Now $$x=\sum(x_n-x_{n+1})s_n.$$But there exist $\epsilon_n=\pm1$ such that $$\sum\epsilon_n(x_n-x_{n+1})s_n=\sum|x_n-x_{n+1}|s_n,$$and that last sum does not converge in $c_0$, so the basis is not unconditional.

(If it's not clear that the last sum fails to converge, consider $\Lambda_1$ applied to the $N$-th partial sum.)

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  • $\begingroup$ @Dear David C. Ullrich, It was very nice, thanks, I edit $s_n$ with $x_n$ so must we replace in your proof $s_n$ by $x_n$? $\endgroup$
    – user62498
    Commented Aug 16, 2015 at 15:39
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    $\begingroup$ @user62498 WHY did you do that? Just so that if someone spent a half hour solving your problem for you the answer he posted would become invalid? Changing the question back to the original requires only two changes $x_nn\to s_n$. Editing my answer to be consistent with the new version of the question would require many changes, and hence would be much more error-prone. I've edited your question back to the original - if you insist on $x_n$ instead of $s_n$ please explain why, so I can decide whether to simply delete my answer. Thanks. $\endgroup$ Commented Aug 16, 2015 at 16:04
  • $\begingroup$ @ David C. Ullrich, thanks I realized $\endgroup$
    – user62498
    Commented Aug 16, 2015 at 16:08

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