F entire, |f(z)| < exp(|z|), what can you say about f?

[snipez90]

f entire, f =/= 0, |f(z)| < exp(|z|), what can you say about f?

Homework Statement

f entire, f nowhere vanishing, |f(z)| < exp(|z|), what can you say about f?

Homework Equations

Maybe Liouville or something closely related

The Attempt at a Solution

The first thing I considered was applying Liouville to a function of the sort g(z) = f(z)exp(-h(z)) where h is chosen appropriately so that g(z) is bounded, but this basically gave me a lot of trouble. The idea is that |g(z)| < exp[|z| - Re(h(z))], but |z| = sqrt(x^2 + y^2) so it's not clear to me how to choose h. Any ideas? Thanks in advance.

 

[paragrafo]

i think the problem doesn't make much sense: if you have a power series expansion
$$\Sigma$$_n>=0a_nxⁿ
such that a_n<=$$\frac{1}{n!}$$ then |f(z)|<=exp(|z|).
for instance f=sen f=cos, and something more weird like a_n=$$\frac{1}{n^n}$$, verify the initial condition.
so i don't really know what can be said about such functions, maybe de reciprocal?
(|f⁽ⁿ⁾(0)|<=1) i haven't think about it yet but i suspect is false.

 

[snipez90]

Hmm good point, I think I missed a condition which is that f is nowhere vanishing.

 

[paragrafo]

uhh.. ok, then just take g=1/f then g is constant by lioville and so is f, replacing z=0 you get |f|<1 f is a constant with norm less than 1.

 

[snipez90]

Why is 1/f bounded?

 

[paragrafo]

yeah, err sorry big mistake, i'll think the problem.

 

[snipez90]

All right. Could you give me the section that contains the group theory problem you posted earlier. I haven't returned the copy of Herstein I borrowed and might take a stab at it provided it's not completely outside of my range. Thanks.

 

[paragrafo]

the book is "topics in algebra" author I.N Herstein chapter 2 "group theory" the problem is taken from the supplementary problem list at the end of the chapter, problem 18, hope you can find it.

 

[paragrafo]

i thought the problem like promised and now I'm certain its pretty hard, maybe even a typo
all i could do was some technical simplifications, anyways maybe you can use it to solve the problem:
if f(z) is a complex continuous function without zeros, then there exists a continuous function
h(z) such that exp(h(z))=f(z). proof of this is almost a copy of that of the lifting path theorem from algebraic topology, that is for each nonzero complex you can find and open set U containing z such that exp^-1(U)=$$\cup$$_iU_i where the U_i are pairwise disjoint and exp restricted to U_i is a homomorphism; this is (C,exp,C\{0}) is a covering.
now you consider the function f restricted to a square containing the origin, and use lebesgue covering lema to conclude that is posible to subdivide in smaller squares (of size epsilon) such that the f(R_k)$$\subset$$U, R_k a small square and U and open set like before; this way you can define h in each square as the inverse image of exp, in the adequate U_i, determined by the values of h in the border of the small square; what is more if h(0) is given then this definition is unique so that you may extend it (larger and larger rectangles) to the whole plane.
now, if f(z) turns out to be analytic, then h is analytic too, without going in detail the function exp has analytic "local inverse" (something slighty different see above) so that using continuity of h it implies that h is analytic in some given point.
now to use this (rather technical) result your initial condition is equivalent to Re(h(z))<|z| and the problem is reduced to find such h entire.
but this problem seems to be just as hard, I've only founded h linear h(z)=az+b that satisfies the condition, but i haven't been able to proof that linear functions are the only ones that can satisfy the initial condition, and couldn't find a counterexample either. Again is fairly easy to proof that no polynomial of higher degree can verify the condition, so the counterexample, if it exist, is some more complicated function, and the proof if its true doesn't seem to be a straighforward application of lioville theorem at all, in fact it seems to be a nontrivial generalization.
So good luck.

 

[snipez90]

Actually I think this is a job for Weierstrass and Hadamard factorization (c.f. article on wikipedia), though I'm not entirely sure (aha). However I don't find this is very satisfactory since these are not exactly basic theorems in complex analysis. I haven't completely understood your approach, but you start out I think in a similar line of thought as that of the theorems above.

As a side note, I think I saw the first fact you stated on a basic complex analysis final, so there should be some elementary way of proving it that doesn't require notions from algebraic topology. I don't actually know any algebraic topology; in fact I should be sitting in on an elementary course right now but decided to study stochastic calculus instead.

Anyways, I will think about it some more. Unfortunately I haven't gotten anywhere on that algebra problem; sorry.

 

[paragrafo]

uff, i feel owned by weierstrass, it says the exact same thing that me when aplied to a nowhere vanishing function .
guess that's what happens when someone halfway the course tries to solve a problem without the required theory