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The problem on one of my mock-exams is as follows: Suppose both $f$ and $g$ are entire with both a finite amount of zeros that all lie in a disk $D_r(0)$. The zeros are counted by their multiplicity. Suppose that the amount of zeros of both function is is equal. Prove that there exists an entire function $h$ that has no zeros such that there is a sufficiently large $R > r$ for which it holds that $|f(z) - h(z)g(z)| < |f(z)|$ for all $|z|=R$.

A sub question I have to this is why do we need the condition that the amount of zeros is equal? Why can $h$ and $R>r$ not exist if the number of zeros aren't equal?

I suppose this has to to with some kind of inverse of Rouché's theorem, because we're starting from to fact that the number of zeros is equal and we want to prove an inequality.

$\textbf{Rouché's Theorem:}$ Suppose $f$ and $g$ are holomorphic in an open set containing a circle $C$ and its interior. If $|f(z)| > |g(z)|$ for all $z \in C$ then $f$ and $f+g$ have the same amount of zeros inside the circle $C$.

Could anyone give a hint? Thanks!

This problem is similar to Inverse statement to Rouché's theorem in complex analysis.

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    $\begingroup$ I do not have an answer for your first question right now. But the “sub question“ is a consequence of Rouché's theorem: If $|f(z) - h(z)g(z)| < |f(z)|$ for $|z| = R$ then $f$ and $hg$ have the same number of zeros in $|z| < R$, which means that $f$ and $g$ have the same number of zeros in $|z| < r $. So that is a necessary condition. $\endgroup$
    – Martin R
    Commented Nov 24, 2021 at 13:49
  • $\begingroup$ @MartinR that makes sense, thanks! $\endgroup$
    – Geigercounter
    Commented Nov 24, 2021 at 13:51

1 Answer 1

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Assume first $f=P, g=Q$ non-zero polynomials of the same degree $n \ge 0$ (by hypothesis); if $a \ne 0$ is the ratio of their leading coefficients, $f-ag$ has degree at most $n-1$ (which for example means $f-ag=0$ if $n=0$), so $|f-ag|<|f|$ on a large enough circle $|z|=R$ and we are done.

But now the general case easily follows as $f,g$ entire with finitely many zeroes, means that $f=Pe^{f_1}, g=Qe^{g_1}$ with $P,Q$ non zero monic polynomials of the same degree and $f_1,g_1$ entire (just factor out the finitely many zeroes from $f,g$ with the corresponding monic polynomials and what remains are never vanishing entire functions, so are exponentials of entire functions), so taking $h=e^{f_1-g_1}$ one has $|f-hg|=|e^{f_1}(P-Q)|<|e^{f_1}P|$ for a large enough circle since $P-Q$ has degree strictly less than $P$

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  • $\begingroup$ I think we should set $h=\frac{b_1}{a_1}e^{f_1-g_1}$ where $b_1$ is the leading coefficient of $Q$ and $a_1$ is the leading coefficient of $P$. Otherwise I do not think that $P-Q$ has a degree strictly less than $P$. Even if their zeroes coincide we can just use $P = a\prod (x-\lambda)$ and $Q = \prod (x-\lambda)$ to ensure that $P-Q$ has the same degree as $P$. $\endgroup$
    – mertunsal
    Commented Nov 24, 2021 at 14:16
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    $\begingroup$ We choose $P,Q$ to be monic as we can here - in other words, if $a_k$are the zeroes of $f$ we take $P=(z-a_1)..(z-a_n)$ and the constant is absorbed in the exponential; this is just a matter of convenience though it is standard in similar considerations as this technique is common in the theory of entire functions $\endgroup$
    – Conrad
    Commented Nov 24, 2021 at 14:19
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    $\begingroup$ So if I understand correctly, the assumption that the number of zeros of both functions is equal translates into the fact that the polynomials $P$ and $Q$ are of the same degree? $\endgroup$
    – Geigercounter
    Commented Nov 24, 2021 at 16:52
  • $\begingroup$ yes of course since for polynomials that is true from the fundamental theorem of algebra $\endgroup$
    – Conrad
    Commented Nov 24, 2021 at 17:05

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