Let $D$ be an open disk in $\mathbb{C}$ and let $\overline{D}$ be the closure of $D$. Suppose $f:\overline{D}\to\mathbb{C}$ is continuous on $\overline{D}$ and analytic on $D$, prove that

$$f(w)=\frac{1}{2\pi i}\displaystyle\int_{\partial{D}}\frac{f(z)}{z-w}\mathrm{d}z$$

for any $w\in D$.

Here $\partial{D}$ is the boundary of $D$. In this case $f$ is not analytic on $\overline{D}$ so we can not apply Cauchy's integral formula. Any hints? Thanks!


For ease of notation, let's assume the centre of the disk is $0$. Then for $0 < r < 1$, the function

$$f_r(z) = f (r\cdot z)$$

is holomorphic in a neighbourhood of $\overline{D}$. Thus

$$f_r(w) = \frac{1}{2\pi i} \int_{\partial D} \frac{f_r(z)}{z-w}\,dz.$$

Since $f_r \to f$ uniformly on $\overline{D}$ for $r\to 1$, we can interchange limit and integral to obtain

$$f(w) = \lim_{r\to 1} f_r(w) = \lim_{r\to 1} \frac{1}{2\pi i} \int_{\partial D} \frac{f_r(z)}{z-w}\,dz = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z)}{z-w}\,dz.$$

  • 1
    $\begingroup$ Thanks a lot! I did not think of uniform convergence. $\endgroup$ – Frank Lu Oct 5 '13 at 21:00

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