Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

  1. How many proofs are there for Picard's little theorem? (references if possible)

  2. 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.

share|improve this question
    
I'm sure I saw a proof of this using Brownian motion somewhere. –  George Lowther Sep 10 '11 at 22:24
4  
Picard's Theorem and Brownian Motion by Burgess Davis. Link (free access) dx.doi.org/10.1090/S0002-9947-1975-0397900-8 –  George Lowther Sep 10 '11 at 22:31
    
@George: According to Remmert's book I mention in my answer there is also an exposition in R. Durrett, Brownian Motion and Martingales in Analysis, Wadsworth, Inc. 1984, p. 139-143. –  t.b. Sep 11 '11 at 14:57
    
I heard that one can use differential geometry(something like defining first fundamental forms and getting to a contradiction) to prove both theorems of Picard's. –  lee May 23 '13 at 1:52

6 Answers 6

up vote 29 down vote accepted

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 $|f'(z)| \leq |f'(p)|$ for all $z \in B_t(p)$. Hence the lemma gives us $B_{\rho}(f(p)) \subset f(\mathbb{D})$ for $\rho = (3-2\sqrt{2}) \frac{1}{2} t |f'(p)| \geq \frac{3}{2} - \sqrt{2}$.

share|improve this answer
1  
That is awsome. The intermediary results are themselves very interesting. I'll surely take a look at the book you named in your post. Thank you for your great answer. –  Beni Bogosel Sep 11 '11 at 20:04
1  
Thanks! Glad you liked it. I should probably have said that there's a lot more in chapter 10 of that book: alternative proofs and improvements of Bloch's theorem, a marvelous theorem of Schottky and the chapter culminates in a proof of Picard's Great Theorem. The book (Funktionentheorie 2) in the German original and its first part are wonderful resources. They are full of such gems (e.g. this one I presented here) and they contain a lot of historical remarks and references to the original literature. –  t.b. Sep 11 '11 at 22:57

The basic argument behind Picard's little theorem is the following:

If $D$ denotes the open unit disk in $\mathbb C$, then there is a holomorphic map $\pi: D \to \mathbb C\setminus \{a,b\}$ which is a covering map in the sense of topology. (Another way to say this is that this map gives a concrete realization of $D$ as the universal cover of $\mathbb C \setminus \{a,b\}$.)

Now if $f: \mathbb C \to \mathbb C\setminus \{a,b\},$ then since $\mathbb C$ is simply connected, covering space theory implies that $f$ lifts to a holomorphic map $\tilde{f}: \mathbb C \to D$, which then must be constant, by Liouville.


If one wants to present this proof in a class, then the hardest part is to construct the map $\pi$. (The covering space arguments can be made without appeal to general theory, by instead constructing the map $\tilde{f}$ concretely via a contour integral.)

In constructing $\pi$, without loss of generality one may assume that $\{a,b\} = \{0,1\}$. One may also replace the disk $D$ by the upper half-plane (since they are conformally isomorphic). The corresponding map $\pi$ was then first discovered in the context of the theory of elliptic integrals and modular functions; indeed, it can be taken to be a certain modular function traditionally denoted $\lambda$. My memory is that Rudin constructs $\lambda$ by hand, and proves enough about it to have the above argument go through; however, I found his argument somewhat mysterious when I first read it, because most of the effort is spent in constructing $\lambda$, and it is not made terribly clear why it exists, or why the theorem pivots on such a construction.

On the other hand, the map $f$ also exists for general reasons, namely the uniformization theorem. (The Riemann surface $\mathbb C \setminus \{a,b\}$ has negative Euler characteristic, and so its universal cover is the open unit disk.)

Unfortunately, the uniformization theorem is probably outside the scope of a one semseter complex analysis course, and I don't know a proof of the uniformization theorem in this particular case that doesn't hinge on the explicit construction of $\lambda$. (Perhaps it's also worth noting that theory of modular functions such as $\lambda$ played an important historical role in the development of that branch of the theory of Riemann surfaces that culminated in the uniformization theorem.)

share|improve this answer
    
Thank you vrey much for your answer. –  Beni Bogosel Sep 10 '11 at 20:29
    
Very clear. Thank you. –  George Sep 10 '11 at 20:42
1  
@Akhil: Dear Akhil, A remark, more than an answer to your question, is that negative Euler char. implies hyperbolicity, and uniformization gives $3$ simply connected Riemann surfaces: the Riemann sphere, which is positively curved, the complex plane, which is flat, and the disk, which is negatively curved. (I realize that this requires elaboration, but hopefully it is at least suggestive.) Best wishes, –  Matt E Sep 11 '11 at 2:46
1  
@Akhil: Dear Akhil, I just reread your comment/question, and saw that you mentioned Gauss--Bonnet, so clearly you already see the relationship with curvature. The point then is that the holomorphic structure is the same as a conformal structure, which in turn determines a constant curvature metric, whose curvature relates to the Euler characteristic (even in the non-compact case --- at least for a mildly non-compact surface like the thrice-punctured sphere). Best wishes, –  Matt E Sep 11 '11 at 2:52
1  
Dear Matt, You may be interested in having a look at the recent expository paper of P. Arés-Gastesi and T.N. Venkataramana, The big Picard theorem and other results on Riemann surfaces, L'Enseignement Mathématique 55 (2009), 127-137 (link should work if your university has a subscription). It contains two proofs of the big Picard theorem using the ideas you describe in your answer. Best wishes, –  t.b. Sep 11 '11 at 9:23

There is a proof in Ahlfors's book "Complex Analysis", in the chapter on global analytic functions. The proof uses analytic continuation, monodromy theorem and a modular function "$\lambda$".

The monodromy theorem can be stated without proof if the audience is unfamiliar with it. The modular function is not so hard to define. Besides, this proof is instructive in the kind of methods it uses.

I will type in a succinct version of this proof when I have more time. For now, the reference is chapter 8, thm 5. If somebody knows how to link to google books or some such place, it will be even better.

share|improve this answer
    
Thank you. This looks similar to the one in Rudin's book. I will look into it. –  Beni Bogosel Sep 10 '11 at 20:30

There is a proof that makes use of covering spaces, which is very simple. There is a way to prove that $j: \mathbb{D} \to \mathbb{C}$ (the so-called Weierstrass $j$ function [it is a modular form]) is a covering space of $\mathbb{C} \setminus \{ a,b\}$. Then, if you show that any holomorphic function $f: \mathbb{C} \to \mathbb{C}\setminus \{ a,b\}$ admits a lifting to a holomorphic function $\tilde{f}: \mathbb{C} \to \mathbb{D}$, then by the Riemann Mapping Theorem (which states that the disc is conformal to the covering space of $\mathbb{C}\setminus \{a,b\}$) and by Liouville's Theorem (which states that every entire function that is not bounded is constant), you conclude that $j^{-1}\circ f = \tilde{f}$. Once $j$ is a projection (hence holomorphic) and $j^{-1}\circ f : \mathbb{C} \to \mathbb{D}$ is bounded and holomorphic, follows, by Liouville, that $j^{-1}\circ f$ is a constant function , say $j^{-1}\circ f(z)= \tilde{f}(z) = c, \forall z \in \mathbb{C}$. Then $f(z) = j(c), \forall z \in \mathbb{C}$. Hence, $f: \mathbb{C} \to \mathbb{C}\setminus \{a,b\}$ must be the constant function.

share|improve this answer
    
Dear Jean, It is not $j$, but $\lambda$, that one uses, and this argument involving covering spaces and $\lambda$ is discussed in several other answers. Regards, –  Matt E May 23 '13 at 2:51

There is a clever proof using harmonic functions in Ransford's book Potential Theory in the Complex Plane.

share|improve this answer

This records an answer by George Lowther in the comments, with a comment by user @t.b.

There is a proof using Brownian motion here: Picard's Theorem and Brownian Motion by Burgess Davis. Link (free access) dx.doi.org/10.1090/S0002-9947-1975-0397900-8

User @t.b. notes:

According to Remmert's book I mention in my answer there is also an exposition in R. Durrett, Brownian Motion and Martingales in Analysis, Wadsworth, Inc. 1984, p. 139-143.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.