Proving that the dual of $\ell^p$ is $\ell^\infty$ for $0

Question

This question comes from Rudin's Functional Analysis, exercise 3.5(d). It concerns the $\ell^p$ spaces (for $0<p<1$) topologized by the metric

$d(x,y)=\sum_{k=1}^\infty |x(k)-y(k)|^p\tag*{($x,y\in\ell^p$)}$

I am asked to prove that there is a one-to-one correspondence $\Lambda\leftrightarrow y$ between $(\ell^p)^*$ and $\ell^\infty$ by

$\Lambda x = \sum_{k=1}^\infty x(k)y(k)\tag*{($x\in\ell^p$)}$

So far, I have shown that for each $y\in\ell^\infty$, the map $F_y:\ell^p\to\mathbb{C}$ given by $F_y(x)=\sum_{k=1}^\infty x(k)y(k)$ is linear (this is obvious) and continuous (by showing it is bounded on the unit ball in $\ell^p$), hence $F_y\in(\ell^p)^*$. Also, to each $\Lambda\in(\ell^p)^*$ corresponds a function $y_\Lambda:\mathbb{N}\to\mathbb{C}$ given by $y_\Lambda(n)=\Lambda(e_n)$, where $e_n:\mathbb{N}\to\mathbb{C}$ maps $n$ to $1$ and $e_n(k)=0$ for all $k\neq n$.

The issue lies in proving that $y_\Lambda$ is in $\ell^\infty$. I am currently trying to prove this via the following approach:

Since $\ell^p$ is a subspace of $\ell^1$, any $\Lambda\in(\ell^p)^*$ can be extended to a linear functional $\Lambda'\in(\ell^1)^*\cong\ell^\infty$, which is possible by a corollary to the Hahn-Banach theorem (this corollary is Theorem 3.6 in Rudin's book).

This argument seems fine, however I only know that $\Lambda$ is continuous on $\ell^p$ with respect to the $\ell^p$-metric, not the metric induced by the $\ell^1$ norm. I have hit a roadblock in showing that $\Lambda$ is continuous in this metric.

So my question is this: would anyone be able to provide any hints/suggestions regarding where I should go from here? I could do the dirty work myself, I would just like to know that this method will lead to the answer. Any assistance would be greatly appreciated.

P.S. This is my first question here on stackexchange, so any criticism about my question would go a long way.

Answer 1

This is an old posting and one of the comments addresses partially the question in the OP. Here, I would like to give a simple and direct proof that $(\ell_p)^*=\ell_\infty$. I hope the OP, who is now an expert in functional analysis, does not object to my posting.

The main tool used here is the fact (see Theorem 1.18 in Rudin's FA, 2nd edition) that:

Theorem A: If $X$ is a topological linear space and $\Lambda:X\rightarrow \mathbb{C}$ is a linear map, then $\Lambda$ is continuous iff $\Lambda$ is bounded in a neighborhood $V$ of $0$ in $X$.

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Notation:

1. For any $x,y\in \ell_p$, $$d_p(x,y)=\sum_{n\geq1}|x(n)-y(n)|^p$$
2. For any $x_0\in\ell_p$ and $a>0$, define $B_p(x_0;a):=\{x\in \ell_p: d_p(x,x_0)<a\}$.
3. For any $y\in \ell_\infty$, define $\Lambda_y:x\mapsto\sum_m y(m)x(m)$ for $x\in\ell_p$.

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Claim I: $\ell_\infty\subset (\ell_p)^*$.
Suppose $y\in \ell_\infty$. Since $0<p<1$, for any $x\in B_p(0;1)$, $|x(m)|\leq |x(m)|^p<1$ for all $m\in\mathbb{N}$. Hence $$\sum_m|y(m)||x(m)|\leq\|y\|_\infty\sum_m|x(m)|^p=\|y\|_\infty d_p(x,0)<\|y\|_\infty$$ From this, it is clear that (1) the map $\Lambda_y:x\mapsto \sum_my(m)x(m)$ is well defined (i.e. the series converges) and linear, and (2) $\Lambda_y$ is bounded in a neighborhood $B_p(0,1)$ of $0$ in $\ell_p$. Consequently, $\Lambda_y\in(\ell_p)^*$. By using the sequence $e_n$ defined as $e_n(m):=\mathbb{1}_{\{n\}}(m)$, it is easy to prove that \begin{align} \sup\{|\Lambda_yx|:x\in B_p(0;1)\}=\|y\|_\infty\tag{1}\label{one} \end{align}

Claim II: For each $\Lambda\in \ell^*_p$ there exists a unique $y\in \ell_\infty$ such that $\Lambda=\Lambda_y$.
By Theorem A there is $a>0$ and $M>0$ such that $$|\Lambda(x)|\leq M \qquad\forall \,x\in B_p(0;a).$$ Define $\bar{y}\in\mathbb{C}^\mathbb{N}$ as $\bar{y}(m)=\Lambda(e_m)$. Since $$|\bar{y}(m)|=|2^{1/p}a^{-1/p}\Lambda(2^{-1/p}a^{1/p} e_m)|\leq 2^{1/p}a^{-1/p}M$$ for all $m$, we have that $\bar{y}\in \ell_\infty$. By Claim I, $\Lambda_{\bar{y}}\in(\ell_p)^*$. Notice that $$\Lambda(x)=\Lambda_{\bar{y}}(x)\qquad\forall\,x\in L:=\operatorname{span}(e_n:n\in\mathbb{N})$$ Since $L$ is dense in $\ell_p$, it follows that $\Lambda=\Lambda_{\bar{y}}$. The uniqueness of the representation $\Lambda=\Lambda_{\bar{y}}$ follows from \eqref{one}.