Holomorphic extension of function

Let $\Omega$ be a simply connected domain in $\mathbb{C}$ and its boundary $\partial\Omega$ be a closed curve. Also let $D$ be a simply connected domain inside $\Omega$ and its boundary $\partial D$ be a closed curve.

If $f$ is a linear function on $\partial\Omega$, i.e. $f(z)=az+b$ and $f$ is a holomorphic function on $\partial D$. Then, I think I can say that $f$ can be holomorphically extend into $\Omega\setminus\overline{D}$. Furthermore $f(z)$ must be also linear function $az+b$ along $\partial D$ Is this correct? or not?

• Can't we define $f(z)=\left\{\begin{matrix} e^z & z\in \overline{D(0,1)} \\ z & |z|=2\end{matrix}\right.$ and have a function that satisfies your conditions, but is not analytically extendable to $D(0,2)$? – Bobby Ocean Jun 5 '13 at 6:10

The function $f(z) = az+b$, initially defined on $\partial \Omega$, admits a holomorphic extension to $\Omega\setminus \overline{D}$, also given by the formula $az+b$. I think the main question here is whether this is the only extension. In other words, if $g$ is holomorphic in $\Omega\setminus \overline{D}$, continuous on $\overline{\Omega}\setminus \overline{D}$, and satisfies $g(z)=az+b$ for $z\in\partial \Omega$, does it follow that $f(z)=az+b$ in $\Omega\setminus \overline{D}$?
The answer is yes. Proof. Let $h(z)=g(z)-az-b$. Our goal is to show that $h$ is identically zero. Let $F$ be a conformal map $F:B\to\Omega$, where $B$ is the unit disk. The composition $\tilde h= h\circ F$ is holomorphic in the doubly-connected domain $B\setminus F^{-1}(\overline{D})$, and $\tilde h(z)\to 0$ as $|z|\to 1$. The domain $B\setminus F^{-1}(\overline{D})$ contains an annulus $\{z:r<|z|<1\}$ for some $r<1$. The function $\tilde h$ is represented by its Laurent series $\sum c_n z^n$ in this annulus. For every $n$ and for every $\rho\in (r,1)$ we have $$c_n=\frac{1}{2\pi i}\int_{|z|=\rho} z^{-n-1}\tilde h(z)\,dz \tag1$$ Letting $\rho\to 1$ in (1) yields $c_n=0$. Thus, $\tilde h$ is identically zero in the annulus $\{z:r<|z|<1\}$. By this identity theorem for holomorphic functions, it is identically zero in $B\setminus F^{-1}(\overline{D})$.
• If $f(z)=0$ on $\partial G$, then the cauchy integral formula gives that $I(z,\partial G)\cdot f(z)=\frac{1}{2\pi i}\int_{\partial G} \frac{f(w)}{w-z}dw$ for every $z\in\mathbb{C}- \partial G$. Thus, $f(z)=0$ everywhere that the winding number is not zero, automatically. This would simplify your proof considerably. – Bobby Ocean Jun 6 '13 at 0:14
• Good point, after staring for awhile, I see now that Laurent series seems to be the only way to complete the problem. However, using the Laurent series coefficient formulas we have $c_n=\frac{1}{2\pi i} \int_{\partial \Omega} z^{-n-1} h(z) dz$ which gives $c_n=0$ for every $n$. I guess I am confused about using a conformal map $F$, is it really needed? – Bobby Ocean Jun 6 '13 at 4:10
• @BobbyOcean Like a Taylor series represents a function only in some disk, a Laurent series can represent it only in some circular annulus, that is, a domain of the form $R_1<|z-z_0|<R_2$. Here we have a general doubly-connected domain, so it's not clear where the function $h$ is represented by a Laurent series (if anywhere). A conformal map helps straighten things out. Here, turning one boundary component into a circle allowed the domain to contain a circular annulus, in which one can work with a Laurent zeries. – ˈjuː.zɚ79365 Jun 6 '13 at 4:15
• Is is true that the series (or Laurent series) will converge in a disc (or annulus) and diverge outside the disc (or annulus). However, the coefficient representations need not have integrals on circles. For example, if $f(z)=\frac{1}{1-z}$ then the $\int_{D(0,0.5)} \frac{f}{z-0} = \int_{D(0,0.7)} \frac{f}{z-0} = \int_R \frac{f}{z-0}=1$ where $R$ is a giant rectangle that goes through the point $0.9$ on the real axis and surrounds the origin. As long as your circle and my $\partial \Omega$ are homotopic in $h$'s regions of analyticity, then we will get the same integral answers. – Bobby Ocean Jun 6 '13 at 4:59