Showing a certain sequence is an orthonormal basis of $H^2(\mathbb{R}_{+}^{2}).$

Question

The problem is to show $$\left\{\frac{1}{\pi^{1/2}(i+z)}\left(\frac{i-z}{i+z}\right)^n\right\}_{n=1}^{\infty}$$ is an orthonormal basis of $H^2(\mathbb{R}_{+}^{2}).$ In another exercise, I have already shown $$\left\{\frac{1}{\pi^{1/2}(i+x)}\left(\frac{i-x}{i+x}\right)^n\right\}_{n=1}^{\infty}$$ is an orthonormal basis for $L^2(\mathbb{R}).$ Now they have a hint in this problem that I don't completely understand. They say it suffices to show $F \in H^2(\mathbb{R}_{+}^{2})$ and $$\int_{-\infty}^{\infty} F(x)\frac{(x+i)^n}{(x-i)^{n+1}}dx = 0$$ for $n = 0,1,2,...$ then $F = 0.$ They also claim Cauchy integral formula can be used to prove $$(\frac{d}{dz})^n(F(z)(z+i)^n)|_{z=i} = 0$$ then $F^{n}(i) = 0$ for $n = 0,1,2,...$

The first part of this hint would show that our set is orthonormal? Would we need to do the product rule $n$ times to show the second hint? I know if $F \in H^2(\mathbb{R}_{+}^{2}),$ then by Cauchy integral formula, $$F(z) = \frac{1}{2\pi i} \int_{\mathbb{R}}\frac{f(t)}{t-z}dt,\quad z\in \mathbb{R}_{+}^{2}$$ Then $$F^n(z) = \frac{n!}{2\pi i} \int_{\mathbb{R}}\frac{f(t)}{(t-z)^{n+1}}dt,\quad z\in \mathbb{R}_{+}^{2}.$$

Answer 1

Have you tried to map the semiplane to the disk?

Actually, the function

$$z\mapsto\frac{z-i}{z+i}$$ sends the real line precisely to the unit disk. If $Im\ z>0$, then one sees that $|z-i|<|z+i|$, and the reverse inequality holds for $Im\ z>0.$

A quick way to prove the result is to map everything to the unit disk, and then you simply deal with powers and with power series.

The point $i$ maps to the origin in the unit disk, which reinforces the statements given in the statement of the problem.

You have to calculate the inverse of the conformal map given, and plug it inside $F$.

Once you do that, you will see that in the unit disk everything resembles the usual computations to obtain the Taylor coefficients around $0$.

The space $L^2({\mathbb T})$ has a Hilbert basis, i.e. $e^{int}$ for $n$ an integer. Similarly, the space $H^2$ on the disk has a Hilbert basis: since we take Fourier series $\sum a_n e^{int}$ on $[-\pi,\pi]$ and consider the Laurent series $\sum a_n z^n$, in order for the latter to be analytic on the disk one need take $a_{-n}=0$ for $n>0$, so the result is:

$$\{1,z, \cdots , z^n, \cdots \} \mbox{ is a Hilbert basis of } H^2({\mathbb D}).$$

In order to link with the Hardy space over the Poincaré half-plane, it suffices to perform the conformal transform given. However, one need add a factor, which multiplies the powers in your example (those powers correspond to the powers of $z$ above):

$$\left\{\frac{1}{\pi^{1/2}(i+x)}\left(\frac{i-x}{i+x}\right)^n\right\}_{n=1}^{\infty}.$$

Check out this link, and you will see how the argument over the Klein disk transports to the upper half plane.

http://en.wikipedia.org/wiki/Hardy_space#Hardy_spaces_for_the_upper_half_plane

Answer 2

The following is how the hint works.

Let $H=H^2(R^2_+)$.

(1) $F_n=\frac{1}{\sqrt{\pi}} \frac{(i-z)^n}{(i+z)^{n+1}}\in H$.

$$\int_R \frac{|i-z|^{2n}}{|i+z|^{2n+2}}~~dx\leq\int_R\frac{1}{1+x^2}dx<\infty.$$

(2) $F_n$ is orthonormal in $H$.

For any function $F,G$ in $H$ with boundary function $f,g\in L^2(R)$, we have $$(F,G)_H=(f,g)_{L^2(R)}.$$ It is already known that $f_n=\frac{1}{\sqrt{\pi}} \frac{(i-x)^n}{(i+x)^{n+1}}$ is orthonormal in $L^2(R)$.

(3) $F_n$ is orthonormal basis in $H$. (Now it comes to the hint.)

First note that $H$ is closed by considering the boundary function whose fourier transform are support in $R^+$.

Consider $F\in H$ with boundary function $f\in L^2(R)$ and $(F,F_n)=0=(f,f_n)$ $~~(n=0,1,2,\ldots)$, that is $$\int_R f(x)\frac{(x+i)^{n}}{(x-i)^{n+1}}dx=0.$$

It is already know $$F(z)=\frac{1}{2\pi i}\int_R \frac{f(t)}{(t-z)}dt.$$ By dominated convergent thoerem and Cauchy-Swarch inequality, we have $$F^{(n)}(z)=\frac{n!}{2\pi i}\int_R \frac{f(t)}{(t-z)^{n+1}}dt.$$

First, we have $F(i)=0=\int_R \frac{f(t)}{t-i}dt$.

Assume $F^{(k)}(i)=0$ with $k=0,1,\ldots, n-1$.

Now consider $X=\frac{d}{dz}^n (F(z)(z+i)^n)\mid_{z=i}$.

On one hand $X=(2i)^n F^{(n)}(i)$. On the other hand, $$\begin{align} X&=\sum_{k=0}^n \binom{n}{k}\cdot F^{(k)}(i) \cdot \frac{n!}{k!} (2i)^k\\ &=\sum_{k=0}^n \binom{n}{k}\cdot\frac{k!}{2\pi i}\int_R \frac{f(t)}{(t-i)^{k+1}}dt\cdot \frac{n!}{k!} (2i)^k\\ &=\frac{n!}{2\pi i} \int_R f(t)\cdot \sum_{k=0}^n\frac{\binom{n}{k} (2i)^k}{(t-i)^{k+1}}dt\\ &=\frac{n!}{2\pi i} \int_R f(t)\cdot \sum_{k=0}^n\frac{\binom{n}{k} (2i)^k(t-i)^{n-k}}{(t-i)^{n+1}}dt\\ &=\frac{n!}{2\pi i} \int_R f(t)\cdot \frac{(t+i)^n}{(t-i)^{n+1}}dt=0. \end{align}$$ Hence we have $F^{(n)}(i)=0$ for all $n\geq 0$. As $F$ is analytic in the upper half plane, it is $0$ in a neighbour of $i$, we have $F$ is the zero function.