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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 $\Bbb{D}=\{z\in \Bbb{C} : |z|<1\}$

Laura
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1 Answers1

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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.

Ehsan M. Kermani
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    CIT requires $z f(z)$ to be holomorphic on a neighborhood of the closed disc. That requires some more explanation. – WimC May 02 '13 at 10:38
  • The contradiction should be $0=\int_{S^1}f(z)dz=\int_{S^1}\frac{1}{z}dz=2\pi i$. – 23rd May 02 '13 at 10:39
  • @WimC: CIT(a stronger version) only assumes the function is holomorphic in the interior and continuous on the closure. – 23rd May 02 '13 at 10:44
  • @WimC: See the middle of the page http://en.wikipedia.org/wiki/Cauchy's_integral_theorem – Ehsan M. Kermani May 02 '13 at 10:50
  • @Landscape: Yes, I wanted to first write it, but this way has some hidden aspect :) – Ehsan M. Kermani May 02 '13 at 10:52
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    For solution 2 you could use $z f(z) - 1$ instead, which vanishes on the unit circle. – WimC May 02 '13 at 10:54
  • The present version of solution 1 looks good to me, but "by Cauchy" seems not in an explicit way. Maybe "by mean value theorem" sounds better. For solution 2, I agree with @WimC, but I don't understand why you can assume $f(0)=0$. – 23rd May 02 '13 at 11:15
  • @Landscape: I've edited the first one to "mean value property". For the second one, doesn't $h(z)=f(z)-f(0)$ works equally the same? – Ehsan M. Kermani May 02 '13 at 11:46
  • Then the boundary condition $f(z)=\frac{1}{z}$ will change. – 23rd May 02 '13 at 12:07
  • @Landscape: For $h$ we'd have $h(z)=az$ so $f(z)=az+f(0)=\frac{1}{z}$ again different values for $a,$ right? – Ehsan M. Kermani May 02 '13 at 12:49
  • @EhsanM.Kermani: Yes, and you can also get a contradiction from that. – 23rd May 02 '13 at 15:18
  • Schwarz's lemma does not imply $f(z)=az$: for example $f(z)=z^n$ still satisfies your hypotesis, i.e. $|f(z)|=1$ on $S^1$, $f(\mathbb{D})\subseteq\mathbb{D}$ and $f(0)=0$. –  Jun 11 '20 at 09:01