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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.

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    $\begingroup$ If $\Lambda(e_n)$ is not bounded, we can assume $n_k$ is such that $\Lambda(e_{n_k})$ diverges monotonically. We can assume it tends to infinity if we want, and even more $\Lambda(e_{n_k})>2^k$. Then consider $x=(x_n)$ defined to be zero for all $n$ except for $x_{n_k}=1/2^k$. Then $x\in\ell^p$ and $\Lambda(x)\geq1+1+1+...$. Contradiction. Hence $\Lambda(e_n)$ is bounded. $\endgroup$ – Alamos Apr 30 '15 at 3:01
  • $\begingroup$ Ahh, thank you very much! That is much simpler than what I had in mind. $\endgroup$ – Aweygan Apr 30 '15 at 3:05
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    $\begingroup$ @Alamos: may I suggest you post your answer as an answer? $\endgroup$ – Martin Argerami Apr 30 '15 at 3:18

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