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.

Can we prove that given an entire function $f$ that is also one to one then $f$ must be linear?

Thanks for any help.

share|improve this question
1  
en.wikipedia.org/wiki/Picard_theorem (which is overkill if this is homework or something) –  yoyo Mar 29 '11 at 17:06
2  
Maybe you can use the Weierstrass Factorization Theorem? –  Adrián Barquero Mar 29 '11 at 17:06
3  
By 1-1, do you mean injective or bijective? –  lhf Mar 29 '11 at 17:52
1  
@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :) –  Mariano Suárez-Alvarez Mar 29 '11 at 19:56
1  
@Mariano: But it bugs me so... –  Arturo Magidin Mar 29 '11 at 20:40

5 Answers 5

You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.

To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f(\{z:|z|>n\})$ is dense in $\mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=\bigcap_n f(\{z:|z|>n\})$ is dense in $\mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.


I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f(\{z:|z|<1\})$ is open, and $f(\{z:|z|>1\})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.

share|improve this answer
    
Nice application of Baire. Thanks! –  Digital Gal Mar 29 '11 at 19:24

By shifting $z$, without loss of generality you can assume $f(z) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 \over f(z)}$ is bounded outside of $U$. Therefore ${z \over f(z)}$ is an entire function that grows no faster than linearly: $|{z \over f(z)}| < A|z| + B$ for some $A$ and $B$.

It's easy from here to show that $g(z) = {z \over f(z)}$ is linear; for any $z_0$ ${g(z) - g(z_0) \over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z \over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {z\over c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 \over c_2} z$.

share|improve this answer

Let $f:\mathbb C\to\mathbb C$ entire and injective. Let $U=f(\mathbb C)$. $U$ is an open subset of the plane.

$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".

Next, if $U\subsetneq\mathbb C$, from Riemann's theorem we know that there is an biholomorphic map $U\to D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $\mathbb C\to D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.

Using this, one can see that the function $1/f(z)$ is bounded at $\infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.

(This avoids Picard but uses Riemann... :( )

share|improve this answer
    
How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez. –  Lebron James Dec 18 at 6:18

I'll give the "usual" proof.

Note that by Little Picard, $f$ misses at most one point; but it is a homeomorphism onto its image, and the plane minus a point is not simply connected. Thus $f$ is onto $\mathbb{C}$, and hence bijective. Then $f$ has a holomorphic inverse, which is enough to imply $f$ is proper, that is, the pre-image of a compact set is compact. This in turn implies $$ \lim_{z\rightarrow\infty} f(z)=\infty,$$ and thus if we define $f(\infty)=\infty$, $f$ becomes a Möbius transformation of the Riemann sphere. So $f$ has the form $f(z) = \frac{az+b}{cz+d},$ and it is easy to see that if $f$ is entire on $\mathbb{C}$, then $c=0$.

share|improve this answer
    
I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $\mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2\times2$ orthogonal. Since $f$ is holomorphic, $A\in SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $\theta$, so $f(z)=ke^{i\theta}z+b$. –  user641 Mar 29 '11 at 20:36

Here is a (longer) proof using very little of complex analysis. Assume that $f:\mathbb{C}\to\mathbb{C}$ is holomorphic and injective. The function $f$ extends to a holomorphic map of the Riemann sphere to itself, $\mathbb{CP}^1\to\mathbb{CP}^1$. Indeed, choose any $z_0$ such that $f'(z_0)\neq 0$ (if it doesn't exist then $f$ is constant (of course from injectivity we know that actually $f'\neq0$ everywhere)). Then a small neighbourhood $U\ni z_0$ is mapped bijectively to a small neighbourhood $V\ni f(z_0)$. The function $1/(f(z)-f(z_0))$ is therefore bounded in $\mathbb{C}-U$, hence by Riemann removable singularity theorem it extends to a holomorphic function on $\mathbb{CP}^1-U$, therefore $f$ extends to a holomorphic map $\mathbb{CP}^1\to\mathbb{CP}^1$.

As an application of Liouville's theorem, any holomorphic map $F:\mathbb{CP}^1\to\mathbb{CP}^1$ such $F(z)\neq\infty$ for $z\neq\infty$ is a polynomial. If we wish, we can also avoid Liuoville theorem and use some topology. If the order of pole of $f$ at $\infty$ is $>1$ then $f$ is not injective in the neighbourhood of $\infty$. Hence there is $a\in\mathbb{C}$ such that $f-az$ is holomorphic $\mathbb{CP}^1\to\mathbb{C}$, hence it's bounded (being a map from a compact space), hence it's a constant: if $f-az$ is not constant then it is a map $\mathbb{CP}^1\to\mathbb{CP}^1$ which is of positive degree but which is not surjective (as it avoids $\infty$).

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.