The aim of this exercise is to prove the Riesz-Fischer theorem for Hilbert spaces that aren't separable.
Let $I$ an index set and $1\leq p \leq \infty$. Let
$\mathcal{F}=\{F\subset I: F$ is finite$\}$. Given $(a_i)_{i\in
> I}\subset \mathbb{K}$, we define:$$\Vert (a_i)_{i\in I} \Vert_p = \sup_{F\in\mathcal{F}}\left( \sum_{i\in F}
|a_i|^p\right)^{1/p} \qquad \mbox{and} \qquad \Vert (a_i)_{i\in
I}\Vert_\infty = \sup_{i\in I}|a_i| $$
Prove that $l_p(I)=\{ (a_i)_{i\in I}\} \subset \mathbb{K}:\Vert (a_i)_{i\in I} \Vert_p <\infty \}$ is a Banach space with the norm $\Vert\cdot\Vert_p$.
$l_p(I)'$ is isometrically isomorphic to $l_q(I)$ where $1/p+1/q = 1$.
Given a orthonormal basis $S=\{x_i: i\in I\}$ for a Hilbert space $H$, prove that H is isometrically isomorphic to $l_2(I)$.
Prove that every Hilbert Space $H$ is isometrically isomorphic to $H'$.
The second item is quite the same proof of $ l_p (\mathbb{N})$, and the 4 follows immediately from 3. I'm having problems to prove item 3.
The map $T:H \longrightarrow l_2(I)$ defined by $T(x)=( \langle x, x_i\rangle)_{i\in I}$ is a linear isometry. I need help to prove that T is surjective, then I can use the open map theorem and conclude the proof. :
Let $y=(a_i)_{i\in I} \in l_2(I)$, consider $\sum_{i\in I} a_ix_i$. How can I show that this sum converge to an element of $H$? (I used summability, but I'm not satisfied with my argument, there is a proof that don't requires summability?)
Best Answer
The first question is, in what sense do you want to understand convergence of the sum?
The correct notion here is unconditional convergence, i.e. we show that there is some $h \in H$ such that for every $\epsilon > 0$, there is some finite subset $J_\epsilon \subset I$ with $\Vert h - \sum_{j\in J} a_j x_j \Vert < \epsilon$ for all finite sets $J_\epsilon \subset J \subset I$.
To this end, first note $a_i = 0$ for all but countably many $i$ (because of $(a_i)_i \in \ell^2$). Hence, let $(i_n)_n$ be pairwise distinct with $\{i_n \mid n\} = \{i \mid a_i \neq 0\}$ (if there are only finitely many, the claim is trivial).
Consider the sequence $h_N := \sum_{n=1}^N a_{i_n} x_{i_n}$. We then have (for $N \geq M \geq N_0$) $$ \Vert h_N - h_M\Vert^2 = \Vert \sum_{n=M+1}^N a_{i_n} x_{i_n}\Vert ^2 = \sum_{n=N+1}^M |a_{i_n}|^2 \leq \sum_{n=N_0 + 1}^\infty |a_{i_n}|^2 \xrightarrow[N_0 \to \infty]{} 0 $$ where we used orthogonality of the $x_i$ and Pythagoras theorem in the step before the last one.
Hence, the sequence $(h_N)_N$ is Cauchy and thus convergent to some $h \in H$ by completeness of $H$.
It remains to prove that we indeed have unconditional convergence to this "candidate limit". To see this, let $\epsilon >0$ be arbitrary. Then there is a finite $J\epsilon \subset I$ with $\sum_{i \in I\setminus J_\epsilon}|a_i|^2<\epsilon$. Now let $J_\epsilon \subset J \subset I$ be finite. We get $$ \Vert h-\sum_{j \in J} a_j x_j\Vert ^2=\lim_N \Vert \sum_{n=1}^N a_{i_n}x_{i_n}-\sum_{j\in J} a_j x_j\Vert^2=\lim_N \sum_{j \in J \Delta \{i_n \mid n=1\dots N} |a_j|^2, $$ where we again used Pythagoras theorem. Here, $\Delta$ denotes the symmetric difference.
Because of $\{i_n \mid n\in \Bbb{N}\}=\{i\mid a_i \neq 0\}$, the above limit is equal to $$ \sum_{i \in I \setminus J}|a_i|^2<\epsilon, $$ which completes the proof.