Example 18.34.12.5DA140DF64C6 On the unit disk $D\subset \mathbb C$ with Euclidean area $\mathrm{d}A$, let $K_ t$ be the Bergman kernel of the Hilbert space of holomorphic functions $f$ on $D$ with the norm $(\int _ D|f|^2\rho _ t\, \mathrm{d}A)^{1/2}$, where
\[ \rho _ t(z)=1+\frac{t^2}{(1-|z|^2)^2+t^2},\qquad t\in \mathbb R. \]
The weights $-\log \rho _ t$ are bounded and depend real analytically on $(z,t)\in D\times \mathbb R$, but
8F78E08AD393
18.34.13
\begin{equation} K_ t(0,0)= \begin{cases} [\pi (1+|t|\arctan (1/|t|))]^{-1},& t\ne 0,\\ \pi ^{-1},& t=0. \end{cases} \end{equation}
In particular, $K_ t(0,0)=\pi ^{-1}-|t|/2+O(t^2)$, so $K_ t(0,0)$ is not differentiable at $t=0$.
Proof.
The density $\rho _ t$ is radial and lies between one and two. By radial symmetry, the constant functions are orthogonal to the holomorphic functions vanishing at the origin, so the reproducing value at the center is the reciprocal of $\int _ D\rho _ t\, \mathrm{d}A$. Polar coordinates and the substitution $a=1-r^2$ give, for $t\ne 0$,
\[ \int _ D\rho _ t\, \mathrm{d}A =\pi \int _0^1\left(1+\frac{t^2}{a^2+t^2}\right)\mathrm{d}a =\pi \bigl (1+|t|\arctan (1/|t|)\bigr ). \]
For $t=0$ the integral is $\pi $. This proves (18.34.13) and the expansion. For every $t\ne 0$ we have $\| \log \rho _ t\| _{L^\infty (D)}=\log 2$, whereas $\rho _0=1$. Thus the curve $t\mapsto \log \rho _ t$ is not continuous at $t=0$ for the uniform norm, and the example does not contradict proposition 18.34.5.
$\square$
Pome
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