Showing that $l^p(\mathbb{N})^* \cong l^q(\mathbb{N})$

Question

I'm reading functional analysis in the summer, and have come to this exercise, asking to show that the two spaces $l^p(\mathbb{N})^*,l^q(\mathbb{N})$ are isomorphic, that is, by showing that every $l \in l^p(\mathbb{N})^*$ can be written as $l_y(x)=\sum y_nx_n$ for some $y$ in $l^q(\mathbb N)$.

The exercise has a hint. Paraphrased: "To see $y \in l^q(\mathbb N)$ consider $x^N$ defined such that $x_ny_n=|y_n|^q$ for $n \leq N$ and $x_n=0$ for $n > N$. Now look at $|l(x^N)| \leq ||l|| ||x^N||_p$."

I can't say I understand the first part of the hint. To prove the statement I need to find a $y$ such that $l=l_y$ for some $y$. How then can I define $x$ in terms of $y$ when it is $y$ I'm supposed to find. Isn't there something circular going on?

The exercise is found on page 68 in Gerald Teschls notes at http://www.mat.univie.ac.at/~gerald/ftp/book-fa/index.html

Thanks for all answers.

Answer 1

We know that the $(e_n)$ with a $1$ at the $n$-th position and $0$s elsewhere is a Schauder basis for $\ell^p$ (which has some nice alternative equivalent definitions, I recommend Topics in Banach Space Theory by Albiac and Kalton as a nice reference about this.

So, every $x \in \ell^p$ has a unique representation by

$$x = \sum_{k = 1}^\infty y_k e_k.$$

Now consider $l \in \ell^q$. Because $l$ is bounded we also have that

$$l(x) = \sum_{k = 1}^\infty y_k l(e_k).$$

Now set $z_k = f(e_k)$. Consider the following $x_n = (y_k^{(n)})$ where

$$y_k^{(n)} = \begin{cases} \frac{|z_k|^q}{z_k} &\text{when $k \leq n$ and $z_k \neq 0$,}\\ 0 &\text{otherwise.} \end{cases}.$$

We have that $$\begin{align}l(x_n) &= \sum_{k = 1}^\infty y_k^{(n)} z_k\\ &= \sum_{k = 1}^n |z_k|^q\\ &\leq \|l\|\|x_n\|\\ &= \|l\| \left ( \sum |x_k^{(n)}|^p \right )^{\frac1p}\\ &= \|l\| \left ( \sum |x_k^{(n)}|^p \right )^{\frac1p}\\ &= \|l\| \left ( \sum |z_k|^q \right )^{\frac1q}. \end{align}$$

Hence we have that

$$\sum |z_k|^q = \|l\| \left (\sum |z_k|^q \right )^{\frac1q}.$$

Now we divide and get

$$\left ( \sum_{k = 1}^n |z_k|^q \right )^{\frac1q} \leq \|l\|.$$

Take the limit to obtain $$\left ( \sum_{k \geq 1} |z_k|^q \right )^{\frac1q} \leq \|l\|.$$

We conclude that $(z_k) \in \ell^q$.

So, now you could try doing the same for $L^p(\mathbf R^d)$ with a $\sigma$-finite measure. A small hint: Using the $\sigma$-finiteness you can reduce to the finite measure case.