Can we prove that every entire one-to-one function is linear?


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.

  • $\begingroup$ Nice application of Baire. Thanks! $\endgroup$ – Digital Gal Mar 29 '11 at 19:24

By shifting $z$, without loss of generality you can assume $f(0) = 0$. 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 $f(z) = {1 \over c_2} z$.

  • $\begingroup$ Why f(z) has simple zero at z=0? $\endgroup$ – nicksohn Dec 30 '15 at 10:33
  • $\begingroup$ Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$? $\endgroup$ – perlman Oct 12 '17 at 5:55
  • $\begingroup$ @nicksohn That follows from $f(z)$ being one to one $\endgroup$ – Zarrax Oct 12 '17 at 17:40
  • $\begingroup$ @MathewJames yes, thanks, corrected it. $\endgroup$ – Zarrax Oct 12 '17 at 17:40

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... :( )

  • $\begingroup$ 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. $\endgroup$ – User001 Dec 18 '14 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 $c=0$.

  • $\begingroup$ 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 $f(z)=ke^{i\theta}z+b$. $\endgroup$ – user641 Mar 29 '11 at 20:36
  • $\begingroup$ Why is it that $c=0$ if $f$ is entire? $\endgroup$ – user225477 Feb 16 '17 at 18:33
  • $\begingroup$ @Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$. $\endgroup$ – user135520 Apr 27 '18 at 16:43

This proof uses only the Open Mapping Theorem and Cauchy's Estimates.

Suppose $f$ is entire and injective. Consider the injective entire function $$\widetilde{f} \colon z \mapsto f(z) - f(0).$$ We see that $\widetilde{f}(0) = 0$. Let $r > 0$, and consider the ball $B(0,r)$ of radius $r$ centered at the origin. By the Open Mapping Theorem, $\widetilde{f}(B(0,r))$ is an open set containing the origin. By openness, there is some $\epsilon > 0$ such that $B(0,\epsilon) \subset \widetilde{f}(B(0,r))$. Since $\widetilde{f}$ is injective, no point of $B(0,r)^c$ may be sent to any point of $B(0,\epsilon)$. In other words, $$|z| \geq r \Rightarrow |\widetilde{f}(z)| \geq \epsilon.$$ Now $\widetilde{f}$ has a zero at the origin, so let's say this zero has order $k \geq 1$. This means there is some entire function $g \colon \Bbb C \rightarrow \Bbb C$ such that $g(0) \neq 0$ and $$\widetilde{f}(z) = z^k g(z) \ \forall \ z \in \Bbb C.$$ Since $\widetilde{f}$ is injective, $\widetilde{f}$ has no other zeroes apart from its zero at the origin. So $g$ has no zeroes. This means the function $F \colon \Bbb C \rightarrow \Bbb C$ defined by $$F(z) = \frac{1}{g(z)} \ \forall \ z \in \Bbb C$$ is entire and also has no zeroes. Write the Taylor expansion of $F$ at the origin as $$F(z) = \sum_{n=0}^\infty \frac{F^{(n)}(0)}{n!}z^n.$$ Given any $R > r$, we can apply Cauchy's Estimates on the circle of radius $R$ to obtain bounds on the derivatives of $F$ at the origin. Indeed, $$F^{(n)}(0) \leq \max_{|z| = R}|F(z)|\frac{n! }{R^n}.$$ But we can apply the fact that $|\widetilde{f}(z)| \geq \epsilon \ \forall \ |z| \geq r$ to notice that $$\max_{|z| = R} |F(z)| = \max_{|z| = R} \frac{1}{|g(z)|} = \max_{|z| = R} \frac{|z|^k}{|\widetilde{f}(z)|} \leq \frac{R^k}{\epsilon}.$$ So if $n > k$, we have $$F^{(n)}(0) \leq \max_{|z| = R}|F(z)|\frac{n!}{R^n} \leq \frac{R^k n!}{\epsilon R^n} = \frac{n!}{\epsilon R^{n-k}} \xrightarrow{R \to \infty} 0.$$ So we conclude from the Taylor expansion of $F$ that $$F(z) = \sum_{n=0}^k \frac{F^{(n)}(0)}{n!} z^n$$ is a polynomial of degree at most $k$. But as previously noted, $F$ has no zeroes. So $F \equiv c$ for some $c \in\Bbb C \backslash\{ 0\}$! This means $g \equiv c^{-1}$, and furthermore, $$\widetilde{f}(z) = c^{-1}z^k \ \forall \ z \in \Bbb C.$$ Supposing $k \geq 2$, the polynomial $z^k - 1 \in \Bbb C[z]$ has at least two roots $\xi_1, \xi_2 \in \Bbb C$ which are certainly non-zero on account of the fact that $0^k \neq 1$. The multiplicity of the root $\xi_1$ is precisely $1$ because $$\frac{d}{dz}\Big\vert_{z=\xi_1} z^k - 1 = k \xi_1^{k-1} \neq 0.$$ This allows us to conclude that $\xi_1 \neq \xi_2$, and that $$\widetilde{f}(\xi_1) = c^{-1} \xi_1^k = c^{-1} \cdot 1 = c^{-1} \xi_2^k = \widetilde{f}(\xi_2).$$ This result contradicts the fact that $\widetilde{f}$ is injective. So $k=1$, and our final conclusion is that $$f(z) = \widetilde{f}(z) + f(0) = c^{-1} z + f(0) \ \forall \ z \in \Bbb C,$$ where for completeness's sake I recall that $c$ was necessarily non-zero.

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    $\begingroup$ Very nice! And you can even add that $c=1/(f(1)-f(0))$. $\endgroup$ – Unit Mar 3 '16 at 21:15

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 $\infty$).


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