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.