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This answer leads me to below result.

Let $E := \ell^p$ with $1 < p < \infty$. Let $\pi_n: E \to \mathbb R, x \mapsto x_n$ be the canonical projection. Clearly, $\pi_n \in E'$. Let $G := \{\pi_1, \pi_2, \ldots\}$. Then $\operatorname{span} G$ is dense in $E'$.

Below is my proof for which I use the fact that $\ell^p$ space is uniformly convex and thus reflexive for all $p \in (1, \infty)$.

  • Is my conclusion that $s_n \to \sum_{m} x_m f(y^m)$ correct?

  • Does the result hold for $p=1$ or $p = \infty$?

My attempt: Define $y^n \in E$ with $y^n =(y_1^n, y_2^n, \ldots)$ by $y_m^n = \delta_{mn}$. For $x = (x_1, x_2, \ldots) \in E$, we can write $x = \sum_{m} x_m y^m$. For $x\in E$ and $f\in E'$, let $$ x^n := \sum_{m \le n} x_m y^m \in E \quad \text{and} \quad s_n := f(x^n) = \sum_{m \le n} x_m f(y^m) \in \mathbb R \quad \forall n \in \mathbb N^*. $$

We have $|x^n-x|^p_p = \sum_{m>n} |x_m|^p \xrightarrow{n \to \infty}0$ because $|x|_p^p = \sum_{m} |x_m|^p < \infty$. This implies $x^n \to x$. Because $f$ is linear continuous, we have $s_n \to f(x)$. On the other hand, $\color{blue}{s_n \to \sum_{m} x_m f(y^m)}$, so $f(x) = \sum_{m} x_m f(y^m)$.

We have $E$ is uniformly convex and thus reflexive. Let $\varphi \in E''$ such that $\varphi_{\mid G} \equiv 0$. There is $z = (z_1, z_2, \ldots) \in E$ such that $\varphi(f) = f(z)$ for all $f\in E'$. We have $$ f(z) = \sum_m z_m f (y^m) = \sum_m \pi_m(z) f (y^m) = \sum_m \varphi(\pi_m) f (y^m) = 0. $$

It follows from Hahn-Banach theorem, that $\overline{\operatorname{span} G} = E'$. This completes the proof.

Analyst
  • 5,637

1 Answers1

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For $p=1$ we have $E'=\ell^\infty.$ The isometric isomorphism is given by $$\ell^\infty \ni x\longmapsto \sum_{n=1}^\infty x_n\pi_n\in E'.$$

The linear span $V$ of $\{\pi_n\}_{n=1}^\infty$ is a separable subspace of $E'$, so is the closure $\overline{V}.$ The entire space is not separable, hence $\overline{V}\subsetneq E'.$

Another straightforward argument: consider $\varphi =\displaystyle \sum_{n=1}^\infty \pi_n.$ We have $\varphi\in E'$ but $d(\varphi, V)=1,$ where $d(\varphi, V)=\inf\{v\in V:\, \|\varphi-v\|_{E'}\}.$

For $p=\infty$ the separability argument can be applied. Indeed, $E'$ is not separable, as the dual space of a nonseparable space.

Ryszard Szwarc
  • 30,319