# Continuously extended holomorphic function on the unit disc.

Let $f$ be continuous on $\bar{\mathbb{D}}$ and holomoprhic on $\mathbb{D}$. How can we show that $$\int_{\partial \mathbb{D}}f(z)dz=0$$?

• $f$ is uniformly continuous. That means you can take an appropriate limit for the integral theorem. (Consider $f_\rho(z) = f(\rho z)$ for $0 < \rho < 1$.) – Daniel Fischer Oct 21 '14 at 18:03
• It seems like I have some flaws in my understadning of Cauchy's integral theorem. The theorem applies to star shaped domains. $\mathbb{D}$ is a star shaped domain. Then integral over any closed path in $\mathbb{D}$ is zero. But $\partial D$ is not $\mathbb{D}$, is that why we cant apply the theorem directly? – L.G Oct 21 '14 at 18:09
• Cauchy's integral theorem applies to much more general situations than star-shaped domains. [Arbitrary domains, and null-homologous cycles in that domain. You do not yet need to have even a vague idea what null-homologous might mean, that comes later.] In the usual formulation, you need a function that is holomorphic in a neighbourhood of the path and the entire region the path surrounds. The thing here is that you can extend it to hold for the boundary of the domain by only requiring that the function has a continuous extension to the boundary [and the boundary is a nice curve], – Daniel Fischer Oct 21 '14 at 18:19
• not requiring that it has a holomorphic extension to a neighbourhood of the closure. The fact that allows this is the uniform continuity of the continuous extension and the compactness of the boundary. Then you can take the limit of integrals over curves approaching the boundary, as in saz's answer. Or, in the situation here, with the disk, take functions that are holomorphic in a neighbourhood of the closed disk - $f(\rho z)$ - and converge uniformly to $f$ on the closed disk. Since all integrals are $0$, the uniform convergence tells you the limit is $0$. – Daniel Fischer Oct 21 '14 at 18:20

Hint: Note that $f$ is uniformly continuous on $\bar{D}$. Use this to conclude that

$$\left| \int_{\partial B(0,r)} f(z) \, dz - \int_{\partial D} f(z) \, dz \right| \to 0$$

as $r \to 1$. Then the claim follows from the fact that

$$\int_{\partial B(0,r)} f(z) \, dz=0$$

for any $r<1$.

Hint: use the Cauchy formula. The difficulty from here is to choose the good function to apply the formula.

Hint 2:

$$\int_{\partial \mathbb{D}}f(z)dz = \int_{\partial \mathbb{D}}\frac{zf(z)}{z-0} dz$$

• I know that it follows from Cauchy's integral theorem, but I dont how to apply it – L.G Oct 21 '14 at 18:02
• use it on the function g(z) = z f(z) and with $a=0$. Also, see my edit. – mookid Oct 21 '14 at 18:03