Conicide of $w^*$ and norm topology on $S_{\ell_1}$

Question

I want to show that on $S_{\ell_1}=\{x\in \ell_1: ||x||=1\}$the $w^*$-and the norm topologies are coincide. Can any one help me . Thanks

Answer 1

Since the norm topology on $\ell_1$ is finer than the weak$^\ast$ topology $\sigma(\ell_1,c_0)$, that also holds for the subspace $S_{\ell_1}$.

It remains to see that on $S_{\ell_1}$, every norm neighbourhood of $x$ contains a weak$^\ast$ neighbourhood.

So fix $x\in S_{\ell_1}$ and $\varepsilon > 0$. Choose an $N\in\mathbb{N}$ such that

$$\sum_{k=N}^\infty \lvert x_k\rvert < \frac{\varepsilon}{4}.$$

Then

$$V = \left\{ y \in \ell_1 : \lvert y_k - x_k\rvert < \frac{\varepsilon}{4N}\text{ for all } k < N\right\}$$

is a weak$^\ast$ neighbourhood of $x$ in $\ell_1$, and

$$V\cap S_{\ell_1} \subset B_\varepsilon(x)\cap S_{\ell_1} = \{z\in S_{\ell_1} : \lVert z-x\rVert < \varepsilon\}.$$

For if $y \in V\cap S_{\ell_1}$, then

$$\sum_{k < N} \lvert y_k\rvert = \sum_{k < N} \lvert x_k - (y_k-x_k)\rvert \geqslant \sum_{k < N}\lvert x_k\rvert - \sum_{k < N} \lvert y_k - x_k\rvert > 1 - \frac{\varepsilon}{4} - N\frac{\varepsilon}{4N} = 1 - \frac{\varepsilon}{2},$$

so

$$\lVert y-x\rVert = \sum_{k < N}\lvert y_k-x_k\rvert + \sum_{k=N}^\infty \lvert y_k - x_k\rvert \leqslant \sum_{k < N} \lvert y_k-x_k\rvert + \sum_{k=N}^\infty \lvert y_k\rvert + \sum_{k=N}^\infty \lvert x_k\rvert < N\frac{\varepsilon}{4N} + \frac{\varepsilon}{2} + \frac{\varepsilon}{4} = \varepsilon.$$