Identify the dual of $\ell_2(\mathbb{N})$ with $\ell_2(\mathbb{N})$ itself

Question

Can someone give me a hint as to how to go about this problem?

Prove that we can identify the dual of $\ell_2(\mathbb{N})$ with $\ell_2(\mathbb{N})$ itself as follows: Let $\tau:\ell_2(\mathbb{N}) \rightarrow \mathbb{K}$ be a bounded linear functional, and prove that there exists a sequence $(y_n) \in \ell_2(\mathbb{N}$ such that $\tau((x_n))=\sum_{n=1}^\infty x_n\overline{y_n}$ for each $(x_n)\in \ell_2(\mathbb{N})$.

I don't want the solution. Just a few tips or any related links.

Answer 1

Define $\phi:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})^*$ to be that for all $x=(x_n)\in\ell^2(\mathbb{N})$ and $y=(y_n)\in\ell^2(\mathbb{N})$: $$\phi(x)(y)=\sum_{n=1}^\infty x_n\overline{y_n}.$$ It's obvious that $\phi$ is linear. Then it suffices to show that $\phi$ is an isometric isomorphism and hence every element of $\ell^2(\mathbb{N})^*$ can be represented by elements of $\ell^2(\mathbb{N})$. You may do the following steps

• Show that $\phi$ is an isometry: First use Hölder's inequality it is easy to see that $\|\phi(x)\|\leq \|x\|_2$. To show the reverse inequality, let $x\in\ell^2(\mathbb{N})$ with $\|x\|_2=1$, and define $$y_n=\begin{cases}|x_n|^2/\overline{x_n}, &x_n\neq 0\\ 0,&\text{otherwise}.\end{cases}$$ Then it is easy to see that $\|y\|_2=1$ and $$\|\phi(x)\|\geq|\phi(x)(y)|=\|x\|_2^2=1.$$ Thus for any $x\in\ell^2(\mathbb{N})\setminus\{0\}$ we have $$\|\phi(x)\|=\|x\|_2\phi\left(\frac{x}{\|x\|_2}\right)\geq\|x\|_2,$$ which means $\|\phi(x)\|\geq\|x\|_2.$ Thus $\phi$ is an isometry, hence injective.

• Secondly, show that $\phi$ is surjective. For each $n\in\mathbb{N}$, define $$e^{(n)}_m=\begin{cases}1,&n=m\\ 0,&n\neq m\end{cases}.$$ For each $\tau\in\ell^2(\mathbb{N})^*$, define $x^\tau=(x^\tau_n)$ via $x^\tau_n=\tau(e^{(n)})$. Then one can show that $x^\tau\in\ell^2$ and $\phi(x^\tau)=\tau.$

• Lastly we show that $\|\tau\|=\|x^\tau\|_2$. It is trivial that $\|\tau\|\leq\|x^\tau\|_2$. For the reverse inequality, set $$x^{\tau,(N)}_n=\begin{cases}x^\tau_n, &n\leq N\\ 0, &n>N\end{cases}.$$ For each $y\in\ell^2$, set $$z_n=\begin{cases}|y_nx^\tau_n|/x^\tau_n, &x^\tau_n\neq 0\text{ and }n\leq N\\ 0, &n>N\end{cases}.$$ Then show that $\|\phi(x^{\tau,(N)})(y)|\geq\|\tau\|\|y\|_2$, and conclude that $\|x^\tau\|_2^2=\lim_{N\to\infty}\|\phi(x^{\tau,(N)})\|^2\leq\|\tau\|^2.$

It looks long but this is not a complete solution. I left several details for you to fill in.

Answer 2

First we need to identify the isomorphism.

For any $x$ we have $\tau(x) = \tau(\sum_k x_k e_k) = \sum_k x_k \tau(e_k)$, which suggests that $t$ defined by $t_n = \overline{\tau(e_n)}$ would be the appropriate isomorphism as long as $t \in l_2$.

Since $\tau$ is bounded, we have some $K$ such that $|\tau(x)| \le K \|x\|$.

To show that $t \in l^2$, note that $|\tau(x)| = |\sum_k x_k \tau(e_k)| \le K \|x\|$. It is straightforward to prove that if $| \langle x,y \rangle | \le K \|x\|$ for all $x$, then $\|y\| \le K$. By choosing $x_k = t_k$ for $k \le n$ and zero otherwise, we can show that $\|t\| \le K$, and hence $\tau(x) = \langle t, x \rangle$.

Note that the sequence $t_n = \overline{\tau(e_n)}$ satisfies $t \in l_2$, hence $\tau(x) = \langle t, x \rangle$. This shows that $\|\tau\| \le \|t\|$, and since $\tau(\overline{t}) = \|t\|^2$, we have $\|\tau\| = \|t\|$.