Dual of $\ell_p$ for $0

Question

Let $0<p<1$. Prove that there is a one-to-one correspondence $\Lambda\leftrightarrow y$ between $(\ell^p)^*$ and $\ell^\infty$ given by

$$\Lambda (x) = \sum_{k=1}^\infty x(k)y(k)$$

Attempt

I am aware that this question has been answered in Proving that the dual of $\ell^p$ is $\ell^\infty$ for $0<p<1$., but I would like to insist on some details about it.

1. I would like to know if this different approach I am giving is correct. Let us consider the map
   $$ f: \ell_1^* \cong \ell_{\infty} \to \ell_1^* \cong \ell_{\infty} $$ given by the restriction of $\Lambda \in \ell_1^* $ to $\ell_p$, that is, $f(\Lambda) := \Lambda|_{\ell_p}$. The map $f$ is surjective as a result of Hahn-Banach Theorem. Injectivity would follow if $\ell_p$ was dense in $\ell_1$. Therefore, my first question is: Is $\ell_p$ for $0<p<1$ dense in $\ell_1$?

2. Let us now consider the proof using the expression for $\Lambda$.

It is clear that given $y \in \ell_{\infty}$, the map $\Lambda$ is linear and continuos and therefore $\Lambda \in \ell_p^*$. The issue is with the other part. Let $\Lambda \in \ell_p^*$. What I would like to prove is that for every $\Lambda$ I can find some $y \in \ell_{\infty}$ such that for every $x \in \ell_p$ we have $$\Lambda (x) = \sum_{k=1}^\infty x(k)y(k),$$ right? In the page I meantioned earlier there is a way I can do this. But, can I use somehow Hahn-Banach Theorem to prove this part? Using that $\ell_1$ is a locally convex topological vector space and the inclusion $\ell_p \subset \ell_1$ for $0<p<1$, we can deduce that there exists a map $\varphi \in \ell_1^*$ such that its restriction to $\ell_p$ is $\Lambda$, but we know as well that $$ \varphi (x) = \sum_{k=1}^\infty x(k)y(k), \ \forall x \in \ell_1,$$ for some $y \in \ell_{\infty}$. Therefore, using that $\varphi|_{\ell_p} = \Lambda$, we conclude that $$\Lambda (x) = \sum_{k=1}^\infty x(k)y(k), \ \forall x \in \ell_p,$$ for some $y \in \ell_{\infty}$. Is this correct? How do we know that every map in $\ell_p^*$ is of that form?

3. Another doubt is: Does trying to prove isometry make sense in this exercise?

Any help is appreciated. Thank you.

Answer 1

As a general remark, there are many different topologies you can put on $\ell_{p}$ to make it a topological vector space. Since $\ell_{p} \subseteq \ell_{1}$, one such topology $\tau_{1}$ is to consider the topology it induces from the usual norm on $\ell_{1}$. Another such topology $\tau_{d}$ is the one induced from the metric \begin{equation} d(x,y) = \sum_{n\in\mathbb{N}} |x(n) - y(n)|^{p}. \end{equation} This provides two different ways to consider $\ell_{p}$ as a topological vector space. The topologies are distinct as one is induced by a norm whereas the other is not. So it is not immediately clear at all that the topological dual spaces $(\ell_{p}, \tau_{1})^{*}$ and $(\ell_{p}, \tau_{d})^{*}$ would coincide as sets.

One. First of all, it's not particularly clear what the map $r$ is. However, it doesn't appear that you are using this and rather use some other map $f$ that has not been defined. As for your main question for this part, $\ell_{p}$ is indeed a dense subset of $\ell_{1}$. This follows from observing that $c_{00} \subseteq \ell_{p}$ and recalling that $c_{00}$ is dense in $\ell_{1}$.

Two. You cannot apply Hahn-Banach extension results directly to $(\ell_{p}, \tau_{d})$ because it is not a locally convex space. The main issue with your argument is the following. Suppose you take $\Lambda \in (\ell_{p}, \tau_{d})^{*}$. If $\Lambda \in (\ell_{p}, \tau_{1})^{*}$ then you can apply the locally convex structure of $\ell_{1}$ to conclude that there is a continuous (with respect to the norm on $\ell_{1}$) extension of $\Lambda$ to be defined on $\ell_{1}$. (Actually, you don't even need Hahn-Banach for this. Continuous linear functionals defined on a dense subspace of a normed space have a unique extension to the whole vector space via a density argument.) From there you can obtain your corresponding element of $\ell_{\infty}$ and show the element of $(\ell_{p}, \tau_{d})^{*}$ induced from that element of $\ell_{\infty}$ is equal to $\Lambda$. But as previously mentioned, it is not at all immediately clear how elements of $(\ell_{p}, \tau_{d})^{*}$ and $(\ell_{p}, \tau_{1})^{*}$ are related. This is the issue you would need to overcome in order to continue with your argument.

Three. The space $\ell_{\infty}$ has a metric structure coming from its usual norm. The question is whether $\ell_{p}^{*}$ has some reasonably natural metric structure as well in order to talk about whether or not the correspondence is an isometry. This is not immediately obvious, apart from of course directly using the correspondence to define a metric on $\ell_{p}^{*}$ that makes the correspondence an isometry. The natural topology to take on the topological dual of a topological vector space is the weak$^{*}$ topology. However, this topology is almost never induced by a metric. So you would have to try something else. I would suggest not to look too much into this as it doesn't appear to be that useful in this context.

Answer 2

The argument suggested by the OP is equivalent to showing that if $\Lambda\in(\ell_p,\tau_p)^*$, then $\Lambda\in (\ell_p,\tau_1)^*$ where

• $0<p<1$,
• $\tau_r$, $0<r\leq1$ is the topology induced by the metric $d_r(x,y)=\Big(\sum_n|x_n-y_n|^r\Big)^{\min(1,1/r)}$.

This turns out to be true since, as it is shown in the link mentioned in the OP, $$(\ell_p,\tau_p)^*=\ell_\infty$$ in the sense that for any $\Lambda\in(\ell_p,\tau_p)^*$ there is a unique $y\in(\ell_\infty,\|\;\|_\infty)$ such that $$\Lambda x=\Lambda_y x:=\sum_nx_ny_n\quad\forall x\in\ell_p$$ Further, the same links shows that if $\Lambda=\Lambda_y$, then $$\sup_{\{x\in\ell_p:d(x_p,0)\leq1\}}|\Lambda x|=\|y\|_\infty$$

In trying to apply the Hahn-Banach theorem, the OP confounds the fact that the topologies $\tau_p$ and $\tau_1$ on $\ell_p$ are not equivalent. Indeed, although the map $J:(\ell_p,\tau_p)\rightarrow(\ell_p,\tau_1)$ given by $Jx=x$ is continuous, its inverse is not; for example, the sequence $(x_n:n\in\mathbb{N})$ defined as $$x_n(m)=\frac1{m^{1/p}} \mathbb{1}_{\{n,n+1,\ldots,2n\}}(m)$$ is in $\ell_p$ and while $$\|x_n\|_1=\sum^{2n}_{m=n}\frac{1}{m^{1/p}}\xrightarrow{n\rightarrow\infty}0,$$ we have that $$d_p(x_n,\boldsymbol{0})=\sum^{2n}_{m=n}\frac{1}{m}\xrightarrow{n\rightarrow\infty}\log2>0$$

Therefore, it seems that the natural thing to do is to consider $y\in\mathbb{C}^{\mathbb{N}}$ defined as $y(m)=\Lambda e_m$ where $e_m(k)=\mathbb{1}_{\{m\}}(k)$ whence we have -at least formally- that \begin{align} \Lambda x=\sum_m x_m y_m\tag{2}\label{two} \end{align} The work resides now on showing that $y\in\ell_\infty$ and that \eqref{two} holds. It follows from this that $\Lambda=\Lambda_y$ where $\Lambda_yx=\sum_mx_my_m$. Then the argument described by the OP follows through but as an overkill.