4

Is the following statement true?

Suppose, $ f:D\to \mathbb C $ is an analytic function where $ D $ is the unit disc of radius $ 1 $ around $0 $. Suppose, $ f $ is analytic on the boundary of $ D $ as well. Then prove that, if $ f $ is one-one on the boundary of $ D $, then $ f $ is one-one on $ D $.

P.S. I think I got a solution. But I was wondering if this problem has a trivial solution.

Here is my solution, since nobody has posted a solution yet:-

Suppose, $ \gamma $ is the boundary of $ D $(assuming $ \gamma(t)=e^{2\pi it} $). Then $ f\circ \gamma $ is a simple continuously differentiable curve. So by Jordan curve theorem, $ f\circ \gamma $ divides $ \mathbb C $ into 2 path connected regions $ A_1, A_2 $ such that if $ a\in A_1 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=2\pi i$ and if $ a\in A_2 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=0$. So, $ f(z)=a $ for some $ a\in A_1,z\in D $ implies $ \int_{\gamma} \frac {f'(z)}{f(z)-a} dz=2\pi i $. So $ f(z)=a $ has only one solution in $ D $. Similarly if $ a\in A_2 $, then $\int_{\gamma} \frac {f'(z)}{f(z)-a}=0 $, so $ f(z)=a $ has no solution in $ D $. If $ f(z)=a $ for some $ z\in D, a\in f\circ \gamma $, then by open mapping theorem, $ f(z)=a' $ for some $ z\in D, a'\in A_2 $, which is not possible. So $ f $ is one-one in $ D $.

Chandan
  • 229

1 Answers1

8

Yes, the statement is true, and your solution is the standard argument. I'm not aware of a more elementary proof.

Let's write down the proof a little more systematically.

Since $f$ is by assumption injective (and analytic) on the boundary of the unit disk, $f\circ \gamma$ - where $\gamma$ is a positively oriented parametrisation of the unit circle - is a simple closed (continuously differentiable) curve, hence, by the Jordan curve theorem, it divides the plane into two disjoint domains whose common boundary is the trace of $f\circ\gamma$. Let $A_1$ denote the bounded domain (the interior of $f\circ\gamma$), and $A_2$ the unbounded domain (the exterior). Then the winding number

$$n(f\circ\gamma,z)$$

is constant on both domains, and $n(f\circ\gamma,z) = 0$ on $A_2$, while $n(f\circ\gamma,z) = \pm 1$ on $A_1$. But

$$n(f\circ\gamma,z) = \frac{1}{2\pi i}\int_{f\circ\gamma} \frac{dw}{w-z} = \frac{1}{2\pi i}\int_\gamma \frac{f'(\zeta)}{f(\zeta)-z}\,d\zeta$$

is, by the argument principle, the number of times $f$ attains the value $z$ in the open unit disk, hence non-negative, and it follows that $n(f\circ\gamma,z) \equiv 1$ on $A_1$. From $n(f\circ\gamma,z) \equiv 0$ on $A_2$, it follows that $f(D) \subset \overline{A}_1$, and by the open mapping theorem, it follows that $f(D)\subset A_1 = (\overline{A}_1)^{\Large\circ}$. Thus $f$ maps $D$ biholomorphically onto $A_1$.

Daniel Fischer
  • 206,697