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For $a\in\mathbb R$, let $h_a$ be the Hilbert space of sequences defined by $$ h_a=\left\{(x_n):\sum_{n\in\mathbb Z}(1+n^2)^a|x_n|^2<\infty\right\}$$ and inner product $\langle(x_n),(y_n)\rangle_a=\sum_{n\in\mathbb Z}(1+n^2)^ax_n\overline{y_n}$.

Define the function $f:h_{-a}\to(h_a)^*$ by $$ f\big((x_n)\big)((y_n))=\sum_{n\in\mathbb Z}x_ny_n, $$ where $(x_n)\in h_{-a}$ and $(y_n)\in h_a$. Prove that

  1. $f((a_n))$ is well-defined as a function on $(h_a)^*$.
  2. $f$ is an invertible, continuous linear map and has a bounded inverse.

Attempt: I have shown already that the series $\sum x_ny_n$ is convergent: if $(x_n)\in h_{-a}$ and $(y_n)\in h_a$ then $$ \sum(1+n^2)^{-a}|x_n|^2<\infty,\qquad\sum(1+n^2)^a|y_n|^2<\infty, $$ so by Cauchy-Schwarz I have $$ \begin{aligned} \left|\sum|x_ny_n|\right|^2&=\left|\sum((1+n^2)^{-a/2}|x_n|)((1+n^2)^{a/2}|y_n|)\right|^2\\ &\leq\left(\sum\left|((1+n^2)^{-a/2}|x_n|)^2\right|\right)\left(\sum\left|((1+n^2)^{a/2}|y_n|)^2\right|\right)\\ &<\infty, \end{aligned} $$ which means $\sum x_ny_n$ is convergent.

To finish showing that $f$ is a function from $h_{-a}$ to $(h_a)^*$, I think that I need to show somehow that $f((x_n))$ is a continuous linear functional. That $f$ is linear is clear to me, but I don't know how to show it is continuous. I know that one way to show continuity is to show boundedness, but I also don't know how to proceed in that direction. As for Part (2), I am unsure how to begin as well; in particular, I'm having trouble visualizing what the inverse map from $(h_a)^*$ to $h_{-a}$ would look like.

Any help or hints on this problem would be greatly appreciated. Thank you in advance.

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You have proved that $|f((x_n)) ((y_n))| \leq \|(x_n)\|\|(y_n)\|$ where the norms are taken the appropriate spaces. This implies that $\|f((x_n))\|\leq \|(x_n))\|$ and hence $f$ is abounded operator with $\|f\|\leq 1$.

Hints for then second part: You know that $f$ is injective. To show that $f$is surjective use the fact that ant element of $h_a^{*}$ is given by an inner product with some element of $h_a$. This follows by Riesz Theorem. Completeness of the spaces $h_a$ can be proved exactly the way you prove completeness of $\ell^{p}$ spaces. Thus each $h_a$ is a Hilbert space and Riesz Theorem is applicable.

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  • $\begingroup$ Thank you sir for the help! I have solved the problem with your hints. $\endgroup$ – mahmir Nov 15 '20 at 13:34
  • $\begingroup$ Hi sir, would you have a chance to look at this related question? Thank you. $\endgroup$ – mahmir Nov 17 '20 at 2:17

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