# Norm closed subspace, and weak* dense

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 $\ell^1$.

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.

• 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 $k$? – Ben Apr 13 '15 at 16:44
• 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 $k\geq K$ ...". – Nicolas Apr 13 '15 at 16:49
• 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. – Ben Apr 13 '15 at 17:05
• 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). – Nicolas Apr 13 '15 at 18:07
• I added an edit of my run at it, if you would not mind taking a look still. – Ben Apr 13 '15 at 21:02