0
$\begingroup$

Represent $\ell^1$ as the space of all real functions $x$ on $S= \{(m,n): m\geq 1, n \geq 1\}$, such that $$ \|x\|_1 = \sum |x(m,n)| < \infty. $$ Let $c_0$ be the space of all real functions $\gamma$ on $S$ such that $y(m,n) \rightarrow 0$ as $m+n \rightarrow \infty$, with norm $\|y\|_\infty = \sup |y(m,n)|$.

Let M be the subspace of $\ell^1$ consisting of all $x \in \ell^1$ that satisfy the equations $$ mx(m,1) = \sum_{n=2}^\infty x(m,n) \;\;\;\;\;\;\; (m = 1, 2, 3, \ldots) $$

Prove that $M$ is a norm-closed subspace of $\ell^1$.

Prove that $M$ is weak*-dense in $\ell^1$

I think that I need to take a Cauchy sequence from $M$ and show it converges in $\ell^1$ with respect to the norm, but I am not really sure how to go about showing this.

For density I think I need to show that for any element $x \in \ell^1$ there exists a sequence $x_n \in M$ such that for any $\gamma \in c_0$ $$ \lim \sup|\gamma x_n| \rightarrow \lim \sup |\gamma x| $$ but again I am not sure, and not sure how to do so.

If anyone would be willing to help me I would greatly appreciate it.

Edit: attempt at weak* dense

Supose we a given a sequence $x \in \ell^1$. Let $\gamma$ in $c_0$ be arbitary, consider a sequence $x_k \in M$ where $$ m x_k(m,1) = \sum_{n=2}^\infty x_k(m,n) $$ and $ \gamma = y(m,n) $ where $y(m,n) \rightarrow 0$ as $m + n \rightarrow \infty$. Observe that for given $x \in \ell^1$ and arbitary $\gamma \in c_0$ there exists a scalar $\alpha$ such that $$ \sup|\gamma x(m,n)| = \alpha $$ Then we note that there exists a sequence $x_k \in M$ such that $$ m x_k(m,1) = \sum_{n=2}^\infty x_k(m,n) = \frac{\alpha}{\|\gamma\|_\infty} $$ Since $$ \sup|\gamma x_k(m,n)| \leq \sup|\gamma| \sup|x_k(m,n)| \leq \|\gamma\|_\infty \frac{\alpha}{\|\gamma\|_\infty} $$

We have $$ \|\gamma x_k(m,n)\| - \|\gamma x(m,n)\| \leq \alpha - \alpha = 0 $$

Therefore $M$ is weak* dense in $\ell^1$.

$\endgroup$
1
$\begingroup$

To show that $M $ is norm-closed in $\ell_{1} $, we take a sequence $\left(x_{k}\right)_{k\in\mathbb{N}} $ in $M $ such that $x_{k}\rightarrow x $ and we must prove that $x\in M $. We have$$\lim_{k\rightarrow+\infty}\left\Vert x_{k}-x\right\Vert _{1}=\lim_{k\rightarrow+\infty}\sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\left|x_{k}\left(m,n\right)-x\left(m,n\right)\right|=0 $$ thus for any $m\in\mathbb{N} $ and any $\varepsilon>0 $, we have $\left|x_{k}\left(m,1\right)-x\left(m,1\right)\right|<\frac{\varepsilon}{2m} $ if $k\gg1 $. Furthermore, since $x_{k}\in S $ for any $k\in\mathbb{N} $, we have $\left|mx_{k}\left(m,1\right)-\sum_{n=2}^{+\infty}x_{k}\left(m,n\right)\right|=0 $, and also, for $k\gg1 $, we have $$\left|\sum_{n=2}^{+\infty}x_{k}\left(m,n\right)-\sum_{n=2}^{+\infty}x\left(m,n\right)\right|\leq\sum_{n=2}^{+\infty}\left|x_{k}\left(m,n\right)-x\left(m,n\right)\right|\leq\sum_{n=1}^{+\infty}\left|x_{k}\left(m,n\right)-x\left(m,n\right)\right| $$ $$\leq\sum_{m=1}^{+\infty}\sum_{n=1}^{+\infty}\left|x_{k}\left(m,n\right)-x\left(m,n\right)\right|=\left\Vert x_{k}-x\right\Vert _{1}<\frac{\varepsilon}{2} $$ where we have inverted the both sums thanks to Tonelli's theorem for series of positive terms.

Thus, once fixed $m\in\mathbb{N} $ ($m>0 $), for any $k\gg1 $, we get $$\left|mx\left(m,1\right)-\sum_{n=2}^{+\infty}x\left(m,n\right)\right|=\left|m\left(x\left(m,1\right)-x_{k}\left(m,1\right)\right)+\left(mx_{k}\left(m,1\right)-\sum_{n=2}^{+\infty}x_{k}\left(m,n\right)\right)+\left(\sum_{n=2}^{+\infty}x_{k}\left(m,n\right)-\sum_{n=2}^{+\infty}x\left(m,n\right)\right)\right| $$ $$\leq m\left|x\left(m,1\right)-x_{k}\left(m,1\right)\right|+\left|mx_{k}\left(m,1\right)-\sum_{n=2}^{+\infty}x_{k}\left(m,n\right)\right|+\left|\sum_{n=2}^{+\infty}x_{k}\left(m,n\right)-\sum_{n=2}^{+\infty}x\left(m,n\right)\right|<\varepsilon $$ so $x\in S $.

As for the weak* density, you have exactly written what you have to do. Notice that you will certainly use the last hypothesis done on the space $c_0$ in your proof, I let you check it.

EDIT: I try to show the *-density of $c_0$.

We identify the toplogical dual of $\ell_1$ with the Banach space $\ell_\infty=\{y:S\rightarrow\mathbb{R} \arrowvert\ |y\|_\infty:=\underset{\left(m,n\right)\in\mathbb{N}^2}{\sup} |y\left(m,n\right)|<+\infty\}$. Thus, for any $\gamma\in\ell_\infty$ and any $x\in\ell_1$, we have

$$\gamma x=\sum_{m,n}y\left(m,n\right)x\left(m,n\right)$$

which is well defined by the Holder inequality. We will say that $y$ is associated to $\gamma$.

We want to prove that for any $\gamma\in\ell_\infty$, there is a sequence $\left(\gamma_{p}\right)_{p\in\mathbb{N}}$ in $c_0$ such that $\gamma_p \overset{*}{\rightarrow} \gamma$ that is, for any $x\in\ell_1$, we have $\gamma_p x \underset{p\rightarrow+\infty}{\longrightarrow}\gamma x$.

Let us define $y_{p}\left(m,n\right)=y\left(m,n\right)$ if $m+n\leq p$} and $y_{p}\left(m,n\right)=0$ otherwise, where $y$ is associated to $\gamma$. Since $y_p=0$ for $p\gg1$ but finite, the (associated) sequence $\left(\gamma_{p}\right)_{p\in\mathbb{N}}$ is in $c_{0}$. Then for any $x\in\ell_1$, we have

$$|\gamma_p x-\gamma x|= \left|\sum_{m,n}y_p\left(m,n\right)x\left(m,n\right)-\sum_{m,n}y\left(m,n\right)x\left(m,n\right)\right| =\left|\sum_{m+n>k}y\left(m,n\right)x\left(m,n\right)\right|$$

which is the rest of a convergent serie (because $\gamma x$ is finite), so we can take $k\gg1$ so that this last term is as negligible as wanted.

This completes the proof.

$\endgroup$
  • $\begingroup$ When you say $k \gg 1$ are you referring to strictly greater than, or at least the next integer value, or is it a nice notation for referring to large $k$? $\endgroup$ – Ben Apr 13 '15 at 16:44
  • $\begingroup$ It means that $k$ is sufficiently great to have a control with $\varepsilon$. I wrote it to avoid the perpetual "for all $\varepsilon >0$, there is a $K\in\mathbb{N} $ such that for any $k\geq K$ ...". $\endgroup$ – Nicolas Apr 13 '15 at 16:49
  • $\begingroup$ Makes since. I didn't know of such notation till now I appreciate the enlightenment. As for the weak* dense. It seems that since $\gamma$ is in $c_0$ it converges to $0$ both, $x_n$ and $x$ will certainly converge to zero with the weak* topology regardless of what they are. $\endgroup$ – Ben Apr 13 '15 at 17:05
  • $\begingroup$ Well I did not look this in more details. If you have some trouble, you can tell me and I will check with you. But I am pretty sure that using the triangular inequality as I did will work (it is really often how we proceed in this kind of problem). $\endgroup$ – Nicolas Apr 13 '15 at 18:07
  • $\begingroup$ I added an edit of my run at it, if you would not mind taking a look still. $\endgroup$ – Ben Apr 13 '15 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.