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Is there some meaningful extension of the following result into complex domain? [1]

Let $n\in \mathbb N$, and let $f:\mathbb R \to \mathbb R$ be such that its $n$-th derivative $f^{(n)}(x)>0, \ \forall x\in \mathbb R$, then $f$ has at most $n$ roots.

Context: The fundamental theorem of algebra can be interpreted that:

If a function $f:\mathbb C \to \mathbb C$ (that is not identical to $0$) has a constant $n$-th derivative, then it has at most $n$ roots in $\mathbb C$.

Restricting attention to the real line only, it implies that $f$ has at most $n$ real roots, which, as shown in [1], holds under a somewhat weaker assumption that the $n$-th derivative of $f$ is strictly positive (and not necessarily constant).

Note: Assume that $f(\mathbb R) \subset \mathbb R$. Then the assumption that $f^{(n)}(x)>0$ is indeed weaker than the assumption that $f^{(n)}$ is constant (and not 0). It is so because if $f^{(n)}(x)<0$, then we could apply the result to $-f$ instead.

Question: What restrictions need to be imposed on $f:\mathbb C \to \mathbb C$ besides that $f^{(n)}(x)>0$ on the real line in order to guarantee that $f$ has at most $n$ roots in $\mathbb C$?

Note: Since considering multiplicity of a root of a function that is not a polynomial is problematic, I don't expect to be able to be able to formulate the result as that there is exactly $n$ roots in $\mathbb C$ considering their multiplicity.

Pavel Kocourek
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    as opposed to the real case when for example $1/(1+x^2)$ is a real analytic and hence infinitely differentiable with no zeroes but tricky to classify, entire functions with at most $n$ zeroes are of the form $f(z)=P(z)e^{g(z)}, \deg P \le n$ and in particular if $f$ is of finite order it then is of integer order $k$ so of the form $P(z)e^{Q(z)}$ where $\deg Q = k$ but other than that not sure what else can be said – Conrad Jan 13 '23 at 01:21
  • @MartinR This would be indeed a very nice characterization. – Pavel Kocourek Jan 13 '23 at 11:38
  • @MartinR I meant the later case: “nonzero $n$-th derivative” (constant is the assumption in FTA which I want to relax). – Pavel Kocourek Jan 13 '23 at 12:19
  • @PavelKocourek: Now you write: “If a function ... has a constant $n$-th derivative” – do you mean “nonzero $n$-th derivative”? – Martin R Jan 13 '23 at 12:52
  • @MartinR Let me clarify that the statement in the ">" block at the beginning of the "Context" paragraph is nothing but restatement of the FTA (since constant $n$-th derivative is equivalent to $f$ being a polynomial of degree at most $n$). As late as in the paragraph below that I discuss the way this statement could be generalizes by replacing the assumption that $f^{(n)(x)}$ is constant by the assumption that it is nonzero (which by https://math.stackexchange.com/q/4617420/1134951 is equivalent to always positive / negative). Does it make sense like this or did I miss something? – Pavel Kocourek Jan 13 '23 at 13:35
  • I will appreciate any suggestions on how to prevent the question being confusing. – Pavel Kocourek Jan 13 '23 at 13:37
  • @PavelKocourek: I was referring to your sentence “If a function $f:\mathbb C \to \mathbb C$ (that is not identical to $0$) has a constant* $n$-th derivative, then it has at most $n$ roots in $\mathbb C$”*  – If the n-th derivative is constant then $f$ is a polynomial, that is not the interesting case. – Martin R Jan 13 '23 at 13:37
  • @MartinR Exactly, that is why I refer to it as an interpretation of FTA. I wanted to emphasize the fact that polynomials can be seen as functions with constant $n$-th derivative, to make the parallel between the claim (assuming that $F^{(n)}(x)>0$) and FTA more apparent. – Pavel Kocourek Jan 13 '23 at 13:45
  • @PavelKocourek: OK, I See now what you mean. But note that “$f^{(n)}(x)>0$ on the real line” makes no sense in general for complex-valued functions. – Martin R Jan 13 '23 at 13:56
  • @MartinR Good point. I just edited the question and clarified that the assumption $f^{(n)}(x)>0 \ \forall x \in \mathbb R$ is weaker than the assumption that $f^{(n)}(x)$ provided by that $f(\mathbb R) \subset \mathbb R$. – Pavel Kocourek Jan 13 '23 at 14:20
  • Let us continue this discussion in chat. – Martin R Jan 13 '23 at 14:28

2 Answers2

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I do not think that there is an analogue statement for holomorphic functions in the complex plane (i.e. for “entire” functions).

The function $f(z) = z^{n-1} + e^z$ has the $n$-th derivative $f^{(n)}(z) = e^z$ which is nonzero everywhere. But $f$ has infinitely many zeros, see for example $e^z-P(z)$ has infinitely many zeros.

Martin R
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  • Thank you for this observation! Yet, this does not satisfy the assumption $Re(f^{(n)}(z))>0$ that you mentioned in the comments. So there is still hope that the assumption $Re(f^{(n)}(z))>0$ might be sufficient. – Pavel Kocourek Jan 13 '23 at 12:32
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    @PavelKocourek: I deleted that comment because it is useless for entire functions. If $Re(f^{(n)}(z))>0$ for all $z$ then $f^{(n)}$ is necessarily constant. – Martin R Jan 13 '23 at 12:37
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If a holomorphic function, say $f(z)$, has at most $n$ roots, say $z_1,...,z_n$ then $g(z) = \frac{f(z)}{(z-z_1)..(z-z_n)}$ has not roots. So it can be lifted, i.e. $g(z) = e^{h(z)}$ for some holomorphic $h(z)$. In words, a holomorphic function with at most $n$ roots has to have form $f(z) = p(z)e^{h(z)}$ where $p$ is a polynomial of degree not bigger than $n$.

Salcio
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  • Thank you for sharing this observation! I'm still wondering however, if there is a way to characterize the fact that $f$ can be represented in the form $f(z)=p(z) e^{h(z)}$ by referring the the $n$-th derivative of $f$. – Pavel Kocourek Jan 13 '23 at 07:52