# Refinement of mean value conjecture for complex polynomials?

I was playing around with Smale's Mean Value Conjecture and found a curious formulation of it which would be stronger (and which may simply be false). It seems to hold for generic' random polynomials but of course this may mean very little. I'd be interested in whether there is a chance it might be true or whether there are counterexamples (it isn't really clear to me how to try to construct one).

Smale's Mean Value Conjecture. Smale proved in 1981 that if $$f:\mathbb{C} \rightarrow \mathbb{C}$$ is a polynomial with $$f(0) = 0$$ and $$|f'(0)| = 1$$, then there exists a critical point of the polynomial ($$z \in \mathbb{C}$$ such that $$f'(z) = 0$$) satisfying $$|f(z)| \leq 4 |z|.$$ One question is whether 4 can be replaced by 1 (which would be best possible if one is interested in universal constants that do not depend on the degree). The best bound depending on the degree is conjectured to be $$(d-1)/d$$.

I was looking at some examples and noticed a curious pattern which seems to hold at least for polynomials with randomly generated roots with very high probability (meaning I did not find a counterexample but maybe random polynomials are the wrong place to look).

A Stronger Conjecture? Let $$g:\mathbb{C} \rightarrow \mathbb{C}$$ be a polynomial with $$|g(0)|=1$$. Consider the subset $$A = \left\{z \in \mathbb{C}: |g(z)| < 1 \right\} \subset \mathbb{C}$$ and let $$B$$ denote the connected component of $$A$$ whose closure contains the point 0. Is it true that the number of critical points of $$z g(z)$$ in $$B$$ is exactly the same as the number of roots of $$z g(z)$$ in $$B$$? Or, a weaker formulation, that $$B$$ contains at least one critical point?

Note that $$\log |g(z)|$$ is analytic and constant on the boundary, therefore $$B$$ is simply connected and $$B$$ contains at least one root of $$g(z)$$. If this statement were true, it would therefore imply Smale's conjecture. It would also be a bit finer since Smale's conjecture is a priori a statement about the set $$A$$ containing a critical point (and $$B \subset A$$).

The subsequent picture gives an example: the red dots are the roots of $$z g(z)$$, the black dots represent critical points of $$z g(z)$$. The set $$A$$ is shown in blue, the set $$B$$ is the big blue set in the middle. There is one root of $$z g(z)$$ on the boundary of $$B$$ (that is the origin). The set then contains 4 roots and 4 critical points. This seems to be true at least for polynomials with randomly chosen roots. A more complicated example is shown below. What is nice about the formulation is that it seems somewhat more tractable: the function $$g$$ maps the set $$B$$ to the unit disk and the boundary to the boundary. So the question is really whether $$g(z)$$ and $$g(z) + z g'(z)$$ have the same number of roots. I like this formulation because it is much closer to the usual Rouche-type questions and seems a little bit easier to approach (if it were true). • Nice approach! The following is not a counterexample but just an example that shows a potential problem if you whish to use Rouche theorem directly. Take $g(z)=1-\frac{1}{2}z$. The set $A=B$ is the disc of radius $2$ centered at $z=2$. Now $g'(z)=-1/2$ and therefore $|zg'(z)|\leq |g(z)|$ fails on the boudnary $\partial B$, for example at $z=4$ . May 19 at 14:32
• Right, it's not just Rouche but it looks somewhat friendlier' and similar to other problems I have seen solved via a combination of the Schwarz Lemma, Riemann mapping, Rouche, the Argument Principle, Koebe, .... May 19 at 14:36
• The following stronger version on Rouche thm, given by Estermann, could be more useful for this problem: THM: Suppose that $f$ and $g$ are holomorphic on a domain $D$, that $C$ is a simple closed contour in $D$ and that the following "strict" inequality holds $|f(z)-g(z)|< |f(z)|+|g(z)|$ for all $z\in C$. Then $f$ and $g$ have the same number of zeros inside $C$. Now in your case take $f(z)=g(z)+zg'(z)$. May 19 at 14:50
• Interesting! So I guess the desired inequality becomes $|z g'(z)| < | g(z) + z g'(z)| + |g(z)|$ on the boundary. Dividing by $|g(z)|$ (which never vanishes), we get $|z g'(z)/g(z)| < |1 + z g'(z)/g(z)| + 1$. This is true unless $z g'(z)/g(z)$ is on a real number $\leq -1$. If $z g'(z)/g(z) = -1$, we would have found a critical point of $z g(z)$ on the boundary, so it suffices to exclude $z g'(z)/g(z)$ being real and $<-1$... May 19 at 16:04
• Somehow "proving that there exists" seems a lot harder than "there is exactly the same number of roots and critical points". The second statement seems to match so well with the classical complex analysis toolbox... May 24 at 14:36

It is indeed true that among random polynomials only a very tiny fraction of them are counterexamples. Even after a few days my program found only very few examples among thousands of tested cases. In all findings it turned out that the critical point which is supposed to lie outside of $$B$$ actually is very close to the border of $$B$$.

The following is a degree $$4$$ example which we check purely algebraically without worrying about numerical issues.

For $$S=\{10, 53-21i, 31-67i, -\frac{41483090}{1864897}-\frac{121204944}{1864897}i\}$$ set $$g(z)=\prod_{u\in S}(1-z/u)$$. Then in the same coloring as in the question the picture looks as follows: We prove that all four roots are inside $$B$$ by picking a point $$\beta=15-50i$$ and checking that the (green) open straight lines connecting $$\beta$$ with the zeros $$\alpha$$ of $$g(z)$$ do not intersect the curve $$|g(z)|=1$$, and neither does the (magenta) line connecting $$\beta$$ and $$0$$.

We do this using Sturm sequences as follows: Pick points $$\alpha$$ and $$\beta$$ with exact coordinates in $$\mathbb Q(i)$$. Let $$F(t)$$ be the real polynomial $$F(t)=g(t\alpha+(1-t)\beta)\bar g(t\bar\alpha+(1-t)\bar\beta)-1$$. So the straight open line from $$\alpha$$ to $$\beta$$ lies completely in a connected component of the set $$A$$ of those $$z$$ with $$|g(z)|<1$$ if $$F(t)$$ hat no real root $$0. As $$F(t)$$ has rational coefficients, the Sturm sequence criterion is exact. Using this we see that the four roots of $$g(z)$$ are in $$B$$.

Next, $$\gamma=-8-54i$$ is a root of the derivative $$(zg(z))'$$. Furthermore, $$g(\gamma)\bar g(\bar\gamma)=1026140636196/1005896485625>1$$, so $$\gamma$$ lies outside $$B$$.

Here is the SageMath code which algebraically verifies the example:

k.<I> = QuadraticField(-1)
ki.<t, z, x, y, s> = k[]

def conj(f):
""" complex conjugate of polynomial f """
return ki(str(f).replace('^', '**').replace('I', '-I'))

def line_in_B(g, a, b):
""" check if the line [a, b] intersects curve |g(z)=1| """
fab = g(z=t*a+(1-t)*b)
f = fab*conj(fab)-1
f_pari = pari(PolynomialRing(QQ, 'T')(f.polynomial(t)))
return f_pari.polsturm([0,1]) == 0

roots = [10, -21*I + 53, -67*I + 31, -121204944/1864897*I - 41483090/1864897]
g = prod(r-z for r in roots)/prod(roots)
b = 15-50*I

# Check if the lines connecting b with the roots a of g(z) are in B:
for a in roots:
print(line_in_B(g, a, b))

# Check that a critical point lies outside B
crit  = -54*I - 8
print((z*g).derivative(z)(z=crit) == 0) # crit is critical point
critval = g(z=crit)
print(critval*conj(critval) > 1)


There are also real counterexamples $$g(z)\in\mathbb R[z]$$. The images for the degree $$6$$ and degree $$7$$ polynomials $$\frac{1}{244925}(z^2 - 8z + 97)(z^2 + 2z + 101)(z^2 + 8z + 25)$$ and $$\frac{1}{12873250}(z^2 + 28z + 221)(z^2 + 16z + 233)(z^2 + 4z + 125)(z + 2)$$ are:

Furthermore, there are counterexamples where the set $$A$$ is disconnected, like $$g(z)=\prod_{u\in S}(1-z/u)$$ for $$S=\{12 i - 20, -21 i + 1, -20 i - 12, -15 i + 20, -10 i - 18, -3 i + 4, 9 i + 7\}$$ or $$g(z)=\prod_{u\in S}(1-z/u)$$ for $$S=\{-7 i - 10, -8 i + 8, -7 i + 5, -5 i - 1, -4 i - 8, -4 i + 5, 6 i + 8, 7 i + 1, 7 i + 6\}$$ • Amazing!! I think there is a sign flip in the last root (-45 + 18i instead of -45 - 18i) but the picture shows it very clearly and I was able to reproduce the example numerically (I get that $g(\mbox{critical point}) = 1.001...$ for the critical point in question in question). May 27 at 16:36
• Thanks, I fixed the typo in the last root. May 27 at 16:48
• I first thought that since all the roots have integer coordinates, the example must be somewhat robust -- after playing with it a little, it seems like it is actually fairly delicate. How was it constructed? May 27 at 16:53
• I did nothing real smart, in fact it was a search for random polynomials (with random roots, and also with random coefficients). This one was the 3403th tested case in degree $5$. I ran various degrees up to $30$ in parallel. Starting in $0$ I constructed a polygon which approximates the border of $B$, and used the argument principle along the polygon to count the roots of $g(z)$ and $g(z)+zg'(z)$, respectively. (There seem to be more examples, but haven't checked them.) May 27 at 17:10
• As to integer coordinates: First I computed in the complex floats, and in the first example found, I chose approximations of the coefficients in $\mathbb Q(i)$ which are sufficiently good so that the example remains an example. May 27 at 17:46