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a all all and and and and and and antiholomorphic at

because By by by by

conclude condi- continuous

definition derivatives

everywhere everywhere.

find find for for for for for for for formula $F R E$ ) $E E C K$

GARDINER

Hence H61der holomorphic holomorphic

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$\frac{\partial P h (η^{)}}{\partial η} = 0 \frac{\partial}{\partial \bar{η}} P h (η) = 0$ Ph(q) $=$ $P h (\bar{η})$ Ph(yy) $=$ $P h (\bar{η})$ $\frac{\partial^{k}}{\partial z^{k}} φ (z_{0}) = 0$

$⌞ k$ , L. U. y7 $ε L$ y7 $ε L φ \equiv 0 η = z 0$ y7 $ε$ L. y7 $ε$ L. y7 $ε$ R. y7 $ε U$ , y7 $ε U,$ $k \geq 0$ . $\bar{η} \in U$

$φ = 0$ . $φ \equiv 0$ . Ph(η) Ph(yy) Ph(yy) $h (η) = 0$ $P h (\bar{η})$ $φ (z_{O}) = 0$ ,

$\int_{U} K (η, ζ) K (ζ, z_{0}) λ^{- 2} (ζ) φ (\bar{ζ)} d ζ d ζ = 0 h (η) = \frac{\partial}{\partial \bar{η}} P h (η) = \frac{\partial}{\partial \bar{η}} P h (\bar{η}) = \bar{\frac{\partial}{\partial η} P h (η)} = T h (η) = 0$

of of order over,

172 [2] $)$ . (see

prove proved.

reproducing

satisfies show So so

Lhat that that the the then theorem Therefore Therefore, Thus tion, To

use

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