It is easy to construct surjective locally univalent holomorphic functions $f: {\mathbb D}\to {\mathbb C}$, where ${\mathbb D}$ is the open unit disk.

I am pretty sure that the answer to the following is positive and well-known (to experts) and is somewhere in the literature:

Question. Are there (non-affine) surjective entire functions $f: {\mathbb C}\to {\mathbb C}$ without critical points?

Such functions (up to an additive constant) would have the form $f(z)=\int_0^z e^{h(w)} dw$, where $h(w)$ is another nonconstant entire function. I do not see, however, how to verify surjectivity of such $f$. Of course, it can miss at most one point in ${\mathbb C}$ and, hence, it would be astounding if indeed this was the case for arbitrary $h$. However, I do not see how to rule this out. Applying Cauchy's argument principle in this setting looks extremely messy.


Picards theorem will do the trick. Take $$h(z)=z\cdot e^{\int_0^zf(t)dt}$$ where $f(z)=\frac{e^z-1}{z}.$ Then, clearly $$h'(z)=e^{\int_0^zf(t)dt}(1+zf(z))=e^{\int_0^zf(t)dt+z}\ne 0$$ and $h$ is locally univalent. If $h(z)$ is not surjective, then by Picard it omits just one complex value $A$ and we can write $h(z)=e^{g(z)}+A.$ But then equality $$e^{g(z)}+A=z\cdot e^{\int_0^zf(t)dt}$$ is impossible since the LHS takes value $0$ infinitely often while the RHS does not.

In fact, the same proof works for any function of the form $h(z)=g(z)\cdot e^{\int_0^zf(t)dt}$ with $g'(z)+g(z)f(z)\ne 0$ and $g(z)$ is a polynomial.

  • $\begingroup$ Instead of writing $h(z) - A = e^{g(z)}$, you can conclude that on $\mathbb{C} - \{0\}$, $h$ misses both $0$ and $A$ (as $z = 0$ is the only value for which $f(z) = 0$), and use big Picard theorem. $\endgroup$
    – xyzzyz
    Nov 7 '13 at 23:56

The simplest surjective entire function without critical points is the "probability integral", $\int_0^z e^{-\zeta^2}d\zeta$.

Proof. Suppose it omits $a$. As it is of order 2, it must be of the form $a+e^P$, where $P$ is a polynomial of degree $2$. But this function has critical points:-)

The argument extends to $\int e^Q$ for any polynomial of degree at least $2$.

  • $\begingroup$ But how do you see that this function is surjective? $\endgroup$ Nov 6 '13 at 21:30
  • $\begingroup$ I see, very nice! $\endgroup$ Nov 6 '13 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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