Show that there is no holomorphic function $f$ in the unit disc $\Bbb{D}$ that extends continuously to $\partial \Bbb{D}$ such that $$f(z)=\frac{1}{z}$$ for $z \in \partial \Bbb{D}$.
where $\Bbb{D}=\{z\in \Bbb{C} : |z|<1\}$
Complex Analysis – Why Holomorphic Function $f$ Can’t Extend Continuously to Boundary
complex-analysis
Best Answer
Assume such an extension exists, then
Solution 1:Define $g(z)=zf(z),$ by the mean value property of holomorphic functions
$$0=g(0)=\frac{1}{2\pi}\int_0^{2\pi} g(e^{it}) dt = \frac{1}{2\pi}\int_0^{2\pi} 1 \,dt = 1$$
which is a contradiction.
Solution 2: We have $|f(z)|=1$ on $S^1.$ Since $f$ is continuous on the boundary $\partial \overline{D}=S^1$ then, by maximum modulus principle, it attains its maximum on the boundary $S^1.$ ($f$ is not constant.) Consequently, $f(D) \subseteq D.$ WLOG assume that $f(0)=0$, therefore, by Schwarz lemma $f(z)=az$ where $|a|=1.$ Moreover, we should have $az=1/z$ on $S^1$, that is, $a.1=1/1 \rightarrow a=1$ and $ai=1/i=-i \rightarrow a=-1.$ Hence, a contradiction.