Proving operator is self-adjoint w.r.t. given inner product

Question

Let $s$ be a nonnegative half-integer and $\mathscr P_s$ be the space of complex polynomials $p(z)$ of degree at most $2s$ in the formal variable $z \in \Bbb C$, equipped with the sesquilinear product $$(p,q) := \frac{(2s+1)!}{\pi}\int_\Bbb C \frac{\overline{p(z)} q(z)}{(1+|z|^2)^{2s+2}} d^2z ,$$ where $d^2z$ is the Lebesgue measure on $\Bbb C\simeq \Bbb R^2$. I would like to show that the operator $$S_3 := z\frac d{dz}-s\Bbb 1 : \mathscr P_s \to \mathscr P_s$$ is Hermitian w.r.t. the given product. The $-s\Bbb 1$ part is clearly ok, but I do not understand how the derivative $d/dz$ transfers to the antilinear part of the product: with real differential operators one can usually integrate by parts, whereas here the derivative $d/dz$ should also act on a polynomial in $\bar z$ and on the denominator, which contains both $z$ and $\bar z$ – two very non-complex differentiable objects. Is there a way to show the Hermiticity of $S_3$ without e.g. expanding the polynomials as linear combinations of an orthonormal basis? Or am I fundamentally misunderstanding how to deal with area integrals on $\Bbb C$?

Answer 1

For the space $\mathscr{P}_s$, we could consider a straightforward basis like $\{ z^1, \ldots, z^{2 s} \}$. The basis is orthogonal since:

$$(z^p, z^q) \propto \int_{\mathbb{C}} \frac{\overline{z^p} z^q}{(1 + |z|^2)^{2 s + 2}} \, \mathrm{d}^2 z = \int_0^\infty \frac{r^{p + q + 1} \mathrm{d} r}{(1 + r^2)^{2 s + 2}} \cdot \int_0^{2 \pi} e^{-i (q - p) \theta} \, \mathrm{d} \theta \propto \delta_{p, q}$$

The radial integral converges since the integrand behaves like $1 / r^3$ as $r \to \infty$. The angular integral is only nonzero when $p = q$, hence the Kronecker delta $\delta_{p, q}$ in the result.

Let's examine the effect of $S_3$ on a basis element:

$$S_3 z^p = \left(z \frac{\mathrm{d}}{\mathrm{d}z} - s 1\right) z^p = (p - s) z^p$$

Finally, to test whether $S_3$ is self-adjoint, we can compute:

$$(z^p, (S_3^* - S_3) z^q) = (z^p, (p - s - (q - s)) z^q) = (p - q) (z^p, z^q) \propto (p - q) \delta_{p, q} = 0$$

Thus, $S_3$ is indeed self-adjoint.