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We denote by $A^2$ the space of analytic functions on $B_1=\{z=x+iy\in \mathbb{C}, x,y\in \mathbb{R}||z|<1\}$, such that $$\left(\int\int_{B_1}|f(z)|^2 dx \, dy\right)^{1/2}<+\infty$$ In $A^2$, we define the scalar product $$\langle f\mid g\rangle_{A^2}=\int\int_{B_1}f(z) \overline{g(z)} \, dx \, dy$$ where $\overline{g(z)}$ is conjugate of $g(z)$. Prove that $A^2$, thus defined, is an Hilbert Space.

I believe that it is not a trivial demonstration. Any suggestions?

Thank you very much.

Michael Hardy
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Mark
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  • Which parts of the Hilbert Space property do you deem trivial? – Roland Feb 18 '14 at 21:26
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    I changed $<f|g>$ to $\langle f\mid g\rangle$ (coded as \langle f\mid g\rangle). That is standard. Note that using \mid instead of | automatically provides proper spacing, and proper angle brackets do NOT result in spacing seen before and after $<$ and $>$. I also put some space before $dx$ and $dy$ so that instead of $f(\cdot)dxdy$ it says $f(\cdot),dx,dy$. ${}\qquad{}$ – Michael Hardy Feb 18 '14 at 21:28
  • @MichaelHardy Thanks! – Mark Feb 18 '14 at 21:30
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    So, how to prove a space is a Hilbert space? – Mhenni Benghorbal Feb 18 '14 at 21:30
  • @MhenniBenghorbal I do not know how to proceed with this example! – Mark Feb 18 '14 at 21:36
  • You probably know that $L^2(B_1)$ is a Hilbert space? Then it reduces to showing that $A^2$ is a closed subspace. – Daniel Fischer Feb 18 '14 at 21:39
  • You have to show that $A^2$, considered as a subset of $L^2(B_1)$, is closed in $L^2(B_1)$. One standard way to do this is first to show that if a sequence $(f_n)\subset A^2$ is bounded with respect to the $A^2$ norm, then it is uniformly bounded on every compact subset of $B_1$ (this can be shown using Cauchy's formula). Once you know this, use Montel's theorem to show that if a sequence $(f_n)$ is convergent in $L^2$, then its limit belongs in fact to $A^2$. – Etienne Feb 18 '14 at 21:39

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The shortest proof I know runs as follows.

If $f\in A^2$ and if you write $f(z)=\sum_0^\infty a_n(f) z^n$ then, using polar coordinates and Parseval formula you find that $$\Vert f\Vert_{A^2}^2=\pi\,\sum_{n=0}^\infty \frac{\vert a_n(f)\vert^2}{n+1}\cdot$$

Conversely, if $(a_n)_{n\geq 0}$ is a sequence of complex numbers such that $\sum_0^\infty \frac{\vert a_n\vert^2}{n+1}<\infty$, then the formula $f(z)=\sum_0^\infty a_nz^n$ makes sense for every $z\in B_1$ and defines a function $f\in A^2$.

It follows that one defines a linear isometry $J:A^2\to \ell^2(\mathbb N)$ from $A^2$ onto $\ell^2(\mathbb N)$ by setting $$J(f)=\left(\sqrt{\frac\pi{n+1}} \,a_n(f) \right)_{n\geq 0} .$$ So $A^2$ is isometric to $\ell^2(\mathbb N)$, and hence is a Hilbert space.

Etienne
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  • Thanks. I know that the Parseval formula is: $||x||^2=\sum_n |\langle x\mid \varphi_n\rangle|^2$ where $... \varphi_n...$ is an orthonormal basis. What is it? – Mark Feb 21 '14 at 17:56
  • For each $r\in[0,1)$, you can apply Parseval in the space $L^2(0,2\pi)$ to get $\int_0^{2\pi} \vert f(re^{i\theta})\vert^2d\theta=2\pi, \sum_0^\infty \vert r^n, a_n(f)\vert^2$ – Etienne Feb 21 '14 at 18:41
  • How do I then get $\Vert f\Vert_{A^2}^2=\pi,\sum_{n=0}^\infty \frac{\vert a_n(f)\vert^2}{n+1}\cdot$ ? – Mark Feb 21 '14 at 20:02
  • You integrate in polar coordinates. – Etienne Feb 21 '14 at 21:32