Prove that which is an orthonomal basis of $L^2(R)$

Question

Prove $$\left \{\frac{1}{π^{\frac{1}{2}}}\left (\frac{i-x}{i＋x} \right )^n\frac{1}{i＋x} \right \}_{n＝-\infty }^{n＝\infty}$$is an orthonomal basis of $L^2(R)$ I tried something like taking the inner product of the bases where n is not equal but I found that it's not equal to zero, and then I tried to find a function where the inner product approaches but I couldn't find it。Who can help me?

Answer 1

This is just very simple calculation:

$$\begin{align} (U(F), U(G))&= \frac{1}{\pi}\int_{-\infty}^\infty \frac{1}{1+x^2} F(e^{2i\arctan x})\overline{G(e^{2i\arctan x})}dx \\ &=\frac{1}{2\pi}\int_{\mathbb R}F(e^{2i\arctan x})\overline{G(e^{2i\arctan x})}d(2\arctan x) \\ &= \frac{1}{\pi}\int_{-\pi}^{\pi} F(e^{i\theta})\overline{G(e^{i\theta})}d\theta \\ &=(F, G)\end{align}$$

Therefore $U$ is an isometry. It immediately follows that the set of functions from (b) is orthonormal as $\{e^{i\theta n}\}_{n=-\infty}^{\infty}$ is an orthonormal basis of $L^2(\mathbb T)$, but not necessarily a basis yet. To show $U$ is unitary, or equivalenty the system in (b) is indeed a basis, it's left to show $U$ is surjective.

Using the map $x\mapsto \frac{i-x}{i+x}$ is a homeomorphism from $\mathbb R$ to $\mathbb T\setminus \{-1\}$, for each $f\in L^2(\mathbb R)$, we can uniquely specify $F(e^{i\theta})$ such that $F(\frac{i-x}{i+x})=\sqrt{\pi} (1+x)f(x)$ by finding the unique $x$ such that $\frac{i-x}{i+x}=e^{i\theta}$ (except when $\theta=\pm\pi$, but this is not a problem as we are considering only functions that are defined a.e.). Then again from the above calculation, it can be shown that if $f\in L^2(\mathbb R)$, $F\in L^2(\mathbb T)$.

BTW, the map $x\mapsto \frac{i-x}{i+x}$ is the famous Cayley transform which has many applications in complex analysis, group theory, differential geometry and operator theory.