Same topologies using Alaoglu's Theorem

Question

I am trying to prove that for $0 < p < r< 1$ the weak-* topologies $\tau_p:=\sigma(l_{\infty}, l_p)$ and $\tau_r:=\sigma(l_{\infty}, l_r)$ are the same when restricted to norm-bounded subsets of $\ell_\infty$.

Attempt

Let $A$ be a norm-bounded subset of $\ell_\infty$, meaning that there exists $M > 0$ such that $\|a\|_\infty < M,\forall a \in A$. Using Alaoglu's Theorem, the closed ball $\mathcal{B}_{\ell_\infty}$ is compact in both $\tau_p$ and $\tau_r$. As scalar multiplication is a homeomorphism, we know that the set $A$ is contained in some $\lambda \mathcal{B}_{\ell_\infty}, \lambda>0$, where this closed ball is still compact in both $\tau_p$ and $\tau_r$.

Here Problem 5. ( chap3. p.87, functional analysis, W.Rudin), it sais that $A$ inherits the subspace topology of $(\lambda \mathcal{B}_{\ell_\infty}, \tau_p), 0<p<1.$

(1) What does that mean exactly?

(2) This is just because $A$ is contained in some $\lambda \mathcal{B}_{\ell_\infty}$ or did we use something else to claim this?

To show that $(\lambda \mathcal{B}_{\ell_\infty}, \tau_p), 0<p<1,$ is independent of $p$, we show that $(\mathcal{B}_{\ell_\infty}, \tau_p)$ is homeomorphic to $(\mathcal{B}_{\ell_\infty}, \tau_r)$. We know that the identity map $$i: (\mathcal{B}_{\ell_\infty}, \tau_r) \to (\mathcal{B}_{\ell_\infty}, \tau_p)$$ is bijective and continuous because the topology $\tau_r$ is finer that $\tau_p$. We also know that both spaces are compact in their corresponding topologies. In the same page, it is used the fact that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

(3) Is this the easier way to show this?

(4) How do we know that $(\mathcal{B}_{\ell_\infty}, \tau_p)$ is Hausdorff? Is it a consequence of Hahn-Banach Theorem? I know that if $X$ is a locally convex topological vector space then $X^*$ separates points of $X$, but this does not apply in this case.

Any help is welcome. Thank you.

Answer 1

To answer your first question, recall that for any subset $A\subseteq X$ of a topological space $X$, the subspace topology on $A$ inherited from $X$ is defined by declaring all sets of the form $U\cap A$ open whenever $U$ is open in $X$. Thus, if $A\subseteq B\subseteq X$, the open sets in the subspace topology of $B$ inherited from $X$ are all sets of the form $B\cap U$ for $U$ open in $X$. Therefore the subspace topology on $A$ inherited by the topological space $B$ is defined by declaring open sets to be those of the form $A\cap (B\cap U)$ for $U$ open in $X$. However, $A\cap(B\cap U)=A\cap U$ because $A\subseteq B$, so the subspace topology on $A$ inherited from $X$ is exactly the same as the subspace topology on $A$ inherited from $B$. Apply this to your sets to see why the claim (2) is true.

We now set $A$ to be the norm bounded subset of $\ell^\infty(\mathbb N)$ as in your question, and $B=\lambda B_{\ell^\infty}$. To answer your question (3), I do believe that that proof method is most probably the simplest. I am assuming that you have already proved that $\ell^p(\mathbb N)^*=\ell^\infty(\mathbb N)$ for any $0<p\leq 1$, else talking about the weak* topology on $\ell^\infty(\mathbb N)$ wouldn't make much sense. With that in mind, fix some arbitrary $x\neq y\in \ell^\infty(\mathbb N)$. Thus $\|x-y\|_\infty>0$, so there must exist a $n\in \mathbb N$ such that $x(n)\neq y(n)$. Now define $z$ by $z(n)=1$ and $z(m)=0$ if $m\neq n$. Clearly $z\in \ell^p(\mathbb N)$, and $$\langle x,z\rangle=x(n)\neq y(n)=\langle y, z\rangle. $$ Therefore $\ell^p(\mathbb N)$ separates the points of $\ell^\infty(\mathbb N)$, so indeed $\ell^\infty(\mathbb N)$ with the weak* topology is Hausdorff. This means that $B\subseteq \ell^\infty(\mathbb N)$ equipped with the subspace topology is also Hausdorff (which I leave to you to show).