# 微积分网课代修|函数代写Function theory代考|MATH824 Calculation of the Bergman Kernel for the Disk

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## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Calculation of the Bergman Kernel for the Disk

Proposition 1.3.1. The Bergman kernel for the unit disk $D$ is
$$K(z, \zeta)=\frac{1}{\pi} \cdot \frac{1}{(1-z \bar{\zeta})^{2}} .$$
The Bergman metric for the disk is
$$g(z)=\frac{2}{\left(1-|z|^{2}\right)^{2}} .$$
This is (up to a constant multiple) the well-known Poincaré, or PoincaréBergman, metric.

This fact is so important that we now present three proofs. Some interesting function theory will occur along the way.

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Construction of the Bergman Kernel for the Disk by Conformal Invariance

Let $D \subseteq \mathbb{C}$ be the unit disk. First we notice that, if either $f \in A^{2}(D)$ or $\bar{f} \in A^{2}(D)$, then
$$f(0)=\frac{1}{\pi} \iint_{D} f(\zeta) d A(\zeta) .$$
This is the standard, two-dimensional area form of the mean value property for holomorphic or harmonic functions.

Of course the constant function $u(z) \equiv 1$ is in $A^{2}(D)$, so it is reproduced by integration against the Bergman kernel. Hence, for any $w \in D$,
$$1=u(w)=\iint_{D} K(w, \zeta) u(\zeta) d A(\zeta)=\iint_{D} K(w, \zeta) d A(\zeta)$$
or
$$\frac{1}{\pi}=\frac{1}{\pi} \iint_{D} K(w, \zeta) d A(\zeta) .$$
By (1.3.1), we may conclude that
$$\frac{1}{\pi}=K(w, 0)$$
for any $w \in D$.
Now, for $a \in D$ fixed, consider the Möbius transformation
$$h(z)=\frac{z-a}{1-\bar{a} z} .$$
We know that
$$h^{\prime}(z)=\frac{1-|a|^{2}}{(1-\bar{a} z)^{2}} .$$
We may thus apply Proposition $1.2 .11$ with $\phi=h$ to find that
\begin{aligned} K(w, a) &=h^{\prime}(w) \cdot K(h(w), h(a)) \cdot \overline{h^{\prime}(a)} \ &=\frac{1-|a|^{2}}{(1-\bar{a} w)^{2}} \cdot K(h(w), 0) \cdot \frac{1}{1-|a|^{2}} \ &=\frac{1}{(1-\bar{a} w)^{2}} \cdot \frac{1}{\pi} \ &=\frac{1}{\pi} \cdot \frac{1}{(1-w \bar{a})^{2}} . \end{aligned}
This is our formula for the Bergman kernel. The formula for the Bergman metric follows immediately by differentiation.

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Calculation of the Bergman Kernel for the Disk

$$K(z, \zeta)=\frac{1}{\pi} \cdot \frac{1}{(1-z \bar{\zeta})^{2}} .$$

$$g(z)=\frac{2}{\left(1-|z|^{2}\right)^{2}}$$

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Construction of the Bergman Kernel for the Disk by ConformalInvariance

$$f(0)=\frac{1}{\pi} \iint_{D} f(\zeta) d A(\zeta) .$$

$$1=u(w)=\iint_{D} K(w, \zeta) u(\zeta) d A(\zeta)=\iint_{D} K(w, \zeta) d A(\zeta)$$

$$\frac{1}{\pi}=\frac{1}{\pi} \iint_{D} K(w, \zeta) d A(\zeta)$$

$$\frac{1}{\pi}=K(w, 0)$$

$$h(z)=\frac{z-a}{1-\bar{a} z} .$$

$$h^{\prime}(z)=\frac{1-|a|^{2}}{(1-\bar{a} z)^{2}} .$$

$$K(w, a)=h^{\prime}(w) \cdot K(h(w), h(a)) \cdot \overline{h^{\prime}(a)} \quad=\frac{1-|a|^{2}}{(1-\bar{a} w)^{2}} \cdot K(h(w), 0) \cdot \frac{1}{1-|a|^{2}}=\frac{1}{(1-\bar{a} w)^{2}} \cdot \frac{1}{\pi} \quad=\frac{1}{\pi} \cdot \frac{1}{(1-w \bar{a})^{2}}$$