Picard's little theorem says that
If there exist two complex numbers $a,b$ such that $f: \Bbb{C} \to \Bbb{C}\setminus \{a,b\}$ is holomorphic then $f$ is constant.
I am interested in proofs for this theorem. Until now I found at least two, one in W. Rudin's Real and Complex Analysis and another one in S. Krantz, Geometric Function Theory. Both of them need some preparation before someone not very advanced in Complex Analysis could understand them, especially the one in S. Krantz's book.
My questions are
How many proofs are there for Picard's little theorem? (references if possible)
Is there a "simple" proof for Picard's little theorem? Simple means that it could be presented to an audience which had a one semester course in complex analysis.
Thank you.
Answer
There is an essentially elementary proof that can be presented to an audience having only little background in complex analysis. Apart from miraculous trickery and some simple estimates, the only ingredients are Cauchy's integral formula and the existence of holomorphic logarithms on simply connected domains.
A much more complete exposition can be found in Remmert, Classical Topics in Complex Function Theory, Springer GTM 172, chapter 10. Let me emphasize: the following is only a distillate of the parts from Remmert's chapter 10 needed for a proof of Picard's little theorem. Said chapter contains a lot more: extensive historical remarks and references, variants of the proofs and further developments, improvements of the results, some nice applications of Picard's theorem and it culminates in a proof of Picard's great theorem.
I once presented this argument to a group of talented students in a two hours “Christmas special lecture” and I think it worked quite well, but admittedly it is ambitious and the argument is flabbergasting at various points.
The main ingredient in the proof is the amazing:
Theorem (Bloch). If $f$ is holomorphic in a neighborhood of the closed unit disk $\overline{\mathbb D}$ and $f'(0) = 1$ then $f(\mathbb{D})$ contains a disk of radius $\frac{3}{2} - \sqrt 2 \gt 0$.
Remmert prefaces the section containing this result by a statement of J.E. Littlewood:
One of the queerest things in mathematics, ... the proof itself is crazy.
I'll give a proof at the end of this answer.
The way this is applied is:
Exercise. If $f: \mathbb{C} \to \mathbb{C}$ is holomorphic and non-constant then $f(\mathbb{C})$ contains disks of arbitrary radius.
Hint: If $f'(0) \neq 0$ then $g(z) = \frac{f(rz)}{r |f'(0)|}$ satisfies the hypothesis of Bloch's theorem.
There's a second tricky ingredient, due to Landau and refined by König:
Let $G \subset \mathbb{C}$ be a simply connected domain and let $f: G \to \mathbb{C}$ be holomorphic. If $f(G)$ does not contain $0$ and $1$ then there is a holomorphic $g: G \to \mathbb{C}$ such that $$f = \frac{1}{2}\big(1+ \cos{(\pi\cos{(\pi g)})}\big).$$
Moreover, if $g$ is any such function then $g(G)$ does not contain a disk of radius one.
Simple connectedness is used in guise of existence of roots and logarithms of holomorphic functions omitting the value $0$. Let us show first that for a function $h$ on a simply connected domain $G$ such that $\pm 1 \notin h(G)$ there is a holomorphic $H:G \to \mathbb{C}$ such that $h = \cos{H}$: The trick is that $1-h^2$ has no zero, hence there exists $k$ such that $k^2 = 1-h^2$, so $1 = h^2 + k^2 = (h+ik)(h-ik)$. But this means that $h+ik$ doesn't have a zero either, hence it has a logarithm: $h+ik = e^{iH}$ and thus $h = \frac{1}{2}(e^{iH}+e^{-iH})$. Applying this to $h = 2f-1$ (which leaves out the values $\pm 1$ by hypothesis) we get an $F$ such that $h=\cos{(\pi F)}$, but $F$ must leave out all integer values in its range, hence $F = \cos{(\pi g)}$ and unwinding the construction gives us the desired $f=\frac{1}{2}\big(1+ \cos{(\pi\cos{(\pi g)})}\big)$.
The “moreover” part follows from the observation that $g(G)$ must not hit the set $$A = \left\{m \pm \frac{i}{\pi} \log{\big(n+\sqrt{n^2 - 1}\big)},\;m\in\mathbb{Z},\;n \in \mathbb{N}\smallsetminus\{0\}\right\}$$
since for $a \in A$ we have $\cos{(\pi a)} = (-1)^m \cdot n$ by a short calculation. Thus $\cos{(\pi\cos{(\pi a)})} = \pm 1$ and if there were $z \in G$ such that $g(z) \in A$ we would have $f(z) \in \{0,1\}$ contradicting the assumptions. It is not hard to convince oneself that every point $w \in \mathbb{C}$ is within distance $\lt 1$ of some point of $A$ (a picture would help!), hence $g(G)$ can't contain a disk of radius $1$.
Armed with these two ingredients the proof of Picard's little theorem is immediate:
Picard's little theorem. If there exist two complex numbers $a,b$ such that $f: \mathbb{C} \to \mathbb{C}\smallsetminus \{a,b\}$ is holomorphic then $f$ is constant.
Proof. We may assume $\{a,b\} = \{0,1\}$. By the Landau–König theorem we have $f(z) = \frac{1}{2}\big(1+ \cos{(\pi\cos{(\pi g)})}\big)$ for some $g$ whose image does not contain a disk of radius $1$ and by the exercise to Bloch's theorem $g$ must be constant.
Now for the proof of Bloch's theorem:
Lemma. Let $f$ be holomorphic in a neighborhood of the closure of the disk $D = B_r(a)$ and assume that $|f'(z)| \lt 2|f'(a)|$ for $z \in D$. Put $\rho = (3-2\sqrt{2})\cdot r \cdot |f'(a)|$ then $B_{\rho}(f(a)) \subset f(D)$.
Proof. Assume $a = f(a) = 0$ for simplicity of notation and write $C = \sup\limits_{z \in D}{\,|f'(z)|}$.
Put $A(z) = f(z) - f'(0)\cdot z$. Then $A(z) = \int_{0}^{z} (f'(w) - f'(0))\,dw$, so
$$|A(z)| \leq \int_{0}^{1} |f'(zt) - f'(0)|\,|z|\,dt.$$
For $d \in D$ we have by Cauchy's integral formula
$$f'(d) - f'(0) = \frac{d}{2\pi i} \int_{|w|= r} \frac{f'(w)}{w(w-d)} \,dw,$$
hence
$$|f'(d) - f(0)| \leq \frac{|d|}{r - |d|} C$$
and thus
$$|A(z)| \leq \int_{0}^{1} \left(\frac{|zt|}{r - |zt|}C\right)|z|\,dt \leq \frac{1}{2} \frac{|z|^2}{r - |z|} C.$$
Let $x = |z| \in (0,r)$ and observe that $|f(z) - f'(0)z| \geq |f'(0)| x - |f(z)|$. The last inequality together with the hypothesis $C \leq 2 |f'(0)|$ gives
$$|f(z)| \geq \underbrace{\left(x - \frac{x^2}{r- x}\right)}_{h(x)} |f'(0)|.$$
Now $h(x)$ assumes its maximum $(3 - 2\sqrt{2})r$ at the point $\tilde{x} = (1-\frac{\sqrt{2}}{2})r$. Thus we have shown that for $|z| = \tilde{x}$ we have
$$|f(z)| \geq (3 - 2\sqrt{2})\cdot r \cdot |f'(0)| = \rho.$$
But this implies that $B_{\rho}(f(0)) \supset f(B_{\tilde{x}}(0))$. Why? This is because $B_{\tilde{x}}(0)$ is a domain whose boundary is mapped outside the ball $B_{\rho}(f(0))$ by $f$, as $f(0) = 0$, see here (1) at the bottom of the page for more details.
Proof of Bloch's theorem. Assume that $f$ is holomorphic in a neighborhood of the closed unit disk and assume that $f'(0) = 1$. Consider the function $z \mapsto |f'(z)|(1-|z|)$. It takes on its maximum at some point $p \in \mathbb{D}$. Putting $t = \frac{1}{2}(1-|p|)$ we have $B_{t}(p) \subset \mathbb{D}$ and $1-|z| \geq t$ for all $z \in B_{t}(p)$. Therefore $|f'(z)|(1-|z|) \leq 2t|f'(p)|$ and hence $|f'(z)| \leq 2|f'(p)|$ for all $z \in B_t(p)$. Hence the lemma gives us $B_{\rho}(f(p)) \subset f(\mathbb{D})$ for $\rho = (3-2\sqrt{2}) \frac{1}{2} t |f'(p)| \geq \frac{3}{2} - \sqrt{2}$.
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