Prove that a certain subspace is weak-*dense in $l^{1}.$ I am self-studying the Rudin's book about functional analysis; I am currently stuck on a detail of exercise 10, chapter 3, point c.
Let $\ell^1$ be the space of all real functions $x(m,n)$ on $\Bbb N\times \Bbb N$ with $\lVert x \rVert_1 = \sum_{m,n} |x(m,n)| <\infty$, and let $M$ be the subspace of $\ell^{1}$ given by the sequences $x$ that satisfy the following equations:
$$
m x(m,1)= \sum_{n=2}^{\infty}x(m,\,n), \quad \forall m \in \mathbb{N}.  
$$
Let $c_0$ be the space of all real functions on $\Bbb N \times \Bbb N$ such that $y(m,n)\rightarrow 0$ as $m+n\rightarrow \infty$, with norm $\lVert y\rVert_\infty = \sup |y(m,n)|$. It is shown in part (a) that $\ell^1 = (c_0)^*$.
The problem in part (c) is to show that $M$ is weak*-dense in $\ell^1$ relative to the topology induced by $c_0$.
Firstly, I observed that $\ell^{1}$ is a locally convex space, and that $M$ is convex and balanced; consequently, the weak-$*$ closure of $M$, which I will indicate $\bar{M},$ also is convex and balanced. Summing up, $\bar{M}$ is a closed, balanced and convex subspace and if we assume there is $x_{0} \in \ell^{1}\setminus \bar{M}$ then, by theorem $3.7$, there has to be a functional in $ \Lambda \in (\ell^1)^*$ such that
$$ 
\vert \Lambda(g)\vert \leq 1 \quad \forall g \in \bar{M} \quad \vert \Lambda(x_{0}) \vert >1.
$$
I was hoping this could lead me to an absurd. Assume there is a $g \in \bar{M}$ such that $\Lambda(g)\neq 0.$ Since $\alpha g \in \bar{M}$ for all $\alpha$ in $\mathbb{R}$, we should have
$$
\vert \Lambda(\alpha \cdot g)\vert \leq 1 \quad \forall \alpha \in \mathbb{R},
$$
and this is absurd.
My question concerns the case in which $\Lambda=0$ on $\bar{M}.$ How should I address it? The theorem $3.7$ is a corollary of the Hahn-Banach theorem so I have tried to slightly modify the proof of the Hahn-Banach theorem to assure that the functional of theorem $3.7$ is non zero on $\bar{M},$ but I got stuck.
 A: We use Theorem 3.5 in Rudin FA, which states that if $M$ is a subspace of a locally compact space $X$ and $x_0\not\in \overline M$, then there exists $\Lambda\in X^*$ such that $\Lambda x_0 = 1$ but $\Lambda x = 0$ on $M$.
To apply the Theorem to the problem, take $X$ to be $\ell^1$ in the weak*-topology induced by $c_0$; then $X^* = c_0$. If $M$ is not weak*-dense then there exists $x_0\in \ell^1\backslash \overline M$, and the theorem implies that there is $y\in c_0$ with $(y,x_0)=1$ and $(y,x)=0$ on $M$. We show that there is no non-trivial $y$ in $c_0$ such that $(y,x)= 0$ on $M$; it follows that there can be no such $x_0$, so $M$ must be weak*-dense.
Note first that
\begin{align}
(y,x) &= \sum_m\left(y(m,1)x(m,1) + \sum_{n>1}y(m,n) x(m,n)\right) \\
      &= \sum_m\sum_{n>1}\left[\frac{y(m,1)}{m} + y(m,n)\right]\,x(m,n).
\end{align}
Take $x(m,\cdot) = (1, 0,\ldots, 0, m, 0,\ldots)$, where the $m$ is in the $n$th position,  and all other rows are zero. Then $x\in M$, and if $(y,x)$ is to equal zero, we must have
$$ y(m,n) = -\frac{y(m,1)}{m}\qquad (n>1).$$
This equality holds for all $m$ and $n>1$, since $m$ and $n>1$ were chosen
arbitrarily. Also, if $y$ is non-trivial, then at least one $y(m,1)$ must be non-zero. But if $y(m,1)\ne 0$, then $y(m,n)$ does not go to zero as we take $m+n\rightarrow\infty$ with $m$ fixed, and therefore $y\not\in c_0$.
