Smale's mean value constant (1981) exceeds 1.1559

AI Disclosure. This work was done overwhelmingly by AI models and supervised by the author. The models were Anthropic’s Claude and OpenAI’s ChatGPT. The author takes full responsibility for the paper.

Introduction

We prove that Smale’s mean value constant $K^*$, defined below, exceeds 1.1559837. The best lower bound proved before was 1.1018, due to Jatar and Ng [10].

Smale's mean value problem

For $d \ge2$ let $\mathcal{P}_d$ be the set of complex polynomials $p$ of degree $d$ with $p(0) = 0$ and $p'(0) = 1$. Every critical point $\beta$ of such a $p$ is nonzero, as $p'(0) = 1$, and we put

If $P$ is a polynomial of degree $d$ and $P'(z) \ne0$, then $p(u) = (P(z+u)-P(z))/P'(z)$ lies in $\mathcal{P}_d$, and its critical points are the points $\beta= \zeta-z$ with $\zeta$ a critical point of $P$. So $K(d)$ is the best constant $K$ in the inequality

over all polynomials $P$ of degree $d$ and all $z$ with $P'(z) \ne0$, and $K^*$ is the best constant over all degrees: Smale’s mean value constant. In words, (1) says that the mean rate of change of $P$ between $z$ and some critical point $\zeta$ is at most $K$ times its rate of change at $z$; hence the name.

Smale proved (1) with $K = 4$ [13], so $K^* \le4$. The polynomial $p(z) = z-z^d/d$ gives $K(d) \ge(d-1)/d$ (its critical points $\beta$ are the $(d-1)$-st roots of unity, and $p(\beta)/\beta= 1-\beta^{d-1}/d = (d-1)/d$ at each). Smale conjectured that $K^* = 1$, and even that $K(d) = (d-1)/d$ for every $d$; this is Smale’s mean value conjecture. The mean value problem is the first of the additional problems in Smale’s list of problems for the next century [14]. The equality $K(d) = (d - 1)/d$ holds for $d \le4$ (Tischler [15]; see also [12 Sect. 7.2]), and Crane announced a computer-assisted proof for $d = 5$ [6 Sect. 1]. Every later improvement of the constant 4 is a bound for $K(d)$ that depends on $d$. For example, Beardon, Minda and Ng proved $K(d) \le4^{(d-2)/(d-1)}$ using an estimate for the hyperbolic metric [3]; Conte, Fujikawa and Lakic proved $K(d) \le4(d-1)/(d+1)$ using Bieberbach’s bound for the second coefficient of univalent functions [4]; Fujikawa and Sugawa combined the two methods [8]; and Crane proved $K(d) < 4 - 2.263/\sqrt{d}$ for $d \ge8$ [6]. All these bounds tend to 4 as $d \to\infty$, so 4 is still the best known upper bound for $K^*$. Crane also showed that $K(d)$ is nondecreasing in $d$ [5].

Smale’s conjecture is false: a proof that $K^* > 1$ was made public in September 2026 [1]; it shows that some $p$ has $m(p) > 1$, but does not make the degree of $p$ explicit. Jatar and Ng [10] then recast the problem in terms of pairs of a univalent map of the unit disk and a finite Blaschke product (Section 2) and proved

this bound comes from an explicit pair. (In [10], $m(p)$, $K(d)$ and $K^*$ are written $S(P)$, $\mathcal{M}_d$ and $\mathcal{M}$.) Adamczewski and Colbrook [2] explain the mechanism behind the proof in [1] and give an explicit polynomial $p$ of degree 29 984 with $m(p) > 1.0000004$, verified by interval arithmetic; it gives $K(d) > 1$ for every $d \ge29 984$.

Main result

The table compares the best earlier bounds with ours.

Best result before this paper	New result
$K^* > 1.1018$ [10 Thm. 1.3]; $K^* > 1.1030$, stated without proof in [10 Rem. 6.7]	$K^* > 1.1559837$ (Theorem 5.1)
$K(d) > 1.1018$ for all large $d$ [10 Thm. 1.3]	$K(d) > 1.1559837$ for all large $d$ (Corollary 5.4)

The bound comes from a single explicit admissible pair $(W, B)$ in the sense of Jatar and Ng (Definition 2.1), with $B$ of degree two. We build it from a polynomial $Q$ of degree 18, whose derivative has a zero of order 5 at $z = 1$, by a construction of Jatar and Ng [10 Lemma 6.2] that we extend from one zero of $B$ to several zeros (Lemma 3.1). The coefficients of $Q$ and the zeros of $Q$ used in the construction are Gaussian rationals. That the pair has the required properties reduces, by Dieudonné’s univalence criterion and the Schur–Cohn test, to four finite checks (Lemma 4.3); the main check says that a certain polynomial of degree 578 in one real variable has no zero in $[-1,1]$. All four checks are made in exact arithmetic, on the exact data printed in Section 5. An expanded version of this paper, with further results, is posted on Zenodo [9].

Admissible pairs

Let $\mathbb{D}=\{ |w|<1\}$ be the unit disk and $\mathcal{S}$ the class of univalent (holomorphic and injective) maps $W$ of $\mathbb{D}$ with $W(0)=0$ and $W'(0)=1$. For $a\in\mathbb{D}$ let $B_a(w)=(w-a)/(1-\bar{a}w)$. A finite Blaschke product of degree $k\geq0$ is a function $B=\lambda B_{a_1}\cdots B_{a_k}$ with $|\lambda|=1$ and $a_1,\ldots,a_k\in\mathbb{D}$.

The condition in Definition 2.1 concerns the open disk only: if $WB$ extends holomorphically across the unit circle, $(WB)'$ may vanish there. The pair of Section 5 has such a zero, of order 5. It is still admissible, and [10 Prop. 4.2] is stated for every admissible pair. Its proof first reduces to pairs with no zero of $(WB)'$ on the closed disk [10 Lemma 2.1], as follows. Let $B$ be nonconstant, and let $t<1$ satisfy $t>B(0)$ and $t>|B(z)|$ at every critical point $z$ of $B$ in $\mathbb{D}$. Then the set $\Omega_t=\{z\in\mathbb{D}:|B(z)|<t\}$ is a Jordan domain containing 0, with real-analytic boundary and closure in $\mathbb{D}$. Let $\psi_t$ map $\mathbb{D}$ conformally onto $\Omega_t$ with $\psi_t(0)=0$ and $\psi_t'(0)>0$. Then $\psi_t$ extends conformally across the circle, and $(W^{(t)},B^{(t)})=(W\circ\psi_t/\psi_t'(0),B\circ\psi_t/t)$ is an admissible pair. Moreover $(W^{(t)}B^{(t)})'=(WB)'(\psi_t)\psi_t'/(t\psi_t'(0))$ has no zero even on the closed disk, and $1/B^{(t)}(0)=t/B(0)$ tends to $1/B(0)$ as $t\to1$.

A construction of admissible pairs

For $a,b\in\mathbb{D}$ with $a\neq b$ put

Since $|1-\bar{a}b|^2-|a-b|^2=(1-|a|^2)(1-|b|^2)$, we have $1/|B_a(b)|^2=J(a,b)$. For $b\in\mathbb{D}$ let $\varphi(w)=(w+b)/(1+\bar{b}w)$; it maps $\mathbb{D}$ onto $\mathbb{D}$, with $\varphi(0)=b$ and $\varphi'(0)=1-|b|^2$.

The idea of the next lemma is to start from a polynomial $Q$ without critical points in $\mathbb{D}$ and to divide out the Blaschke factors $B_{a_j}$ of some of its zeros $a_1,\ldots,a_k$ in $\mathbb{D}$. If the quotient $H$ is univalent, then $Q=H\prod_j B_{a_j}$ is a univalent function times a Blaschke product, and the derivative of this product is $Q'$, which has no zero in $\mathbb{D}$ by the choice of $Q$; composition with the automorphism $\varphi$ moves a further zero $b$ of $Q$ to 0. The bound $1/B(0)=\prod_j1/|B_{a_j}(b)|$ is large when the $a_j$ are close to $b$ in the pseudo-hyperbolic distance $|B_{a_j}(b)|$; what limits it is that $Q$ must keep its critical points out of $\mathbb{D}$ while $H$ stays univalent. For $k=1$ the lemma is [10 Lemma 6.2].

Proof. Since the $a_j$ are distinct zeros of $Q$, $H$ is a polynomial. As $Q(b)=0$ and $b\ne a_j$ for every $j$, $H(b)=0$. As $H$ is univalent, $H'(b)\ne0$, so $N\ne0$. Since $\varphi$ is a conformal self-map of $\mathbb{D}$, $W$ is univalent in $\mathbb{D}$, with $W(0)=H(b)/N=0$ and $W'(0)=H'(b)\varphi'(0)/N=1$; so $W\in\mathcal{S}$. Each $B_{a_j}\circ\varphi$ is a conformal automorphism of $\mathbb{D}$, hence of the form $\lambda B_\gamma$ with $|\lambda|=1$ and $\gamma\in\mathbb{D}$; so $B$ is a finite Blaschke product of degree $k$. Moreover $\eta\ne0$, since $b\ne a_j$, and $B(0)=(|\eta|/\eta)\eta=|\eta|>0$. Since $H\prod_j B_{a_j}=Q$, we have $WB=(|\eta|/(\eta N))Q\circ\varphi$, so

which has no zero in $\mathbb{D}$, because $\varphi(\mathbb{D})=\mathbb{D}$, $Q'$ has no zero in $\mathbb{D}$ and $\varphi'$ has no zero at all. Finally $1/B(0)^2=1/|\eta|^2=\prod_j1/|B_{a_j}(b)|^2=\prod_jJ(a_j,b)$. □

Reduction to finite checks

Lemma 3.1 needs two facts that are not finite checks as they stand: that $H$ is univalent in $\mathbb{D}$, and that $Q'$ has no zero in $\mathbb{D}$. This section reduces both to finitely many exact computations.

Univalence

Let $H(z)=\sum_{\ell=0}^{n}h_\ell z^\ell$ be a polynomial. The Dieudonné polynomials of $H$ are

where $U_{\ell-1}$ is the Chebyshev polynomial of the second kind, $U_{\ell-1}(\cos\theta)=\sin\ell\theta/\sin\theta$. For $c=\cos\theta$ they satisfy $H(ze^{i\theta})-H(ze^{-i\theta})=2i\sin\theta\,zP_c(z)$; moreover $P_1=H'$, since $U_{\ell-1}(1)=\ell$, and $P_c(0)=h_1$. Since $U_{\ell-1}(-c)=(-1)^{\ell-1}U_{\ell-1}(c)$, also $P_{-c}(w)=P_c(-w)$; in particular $P_{-1}(w)=H'(-w)$. So if $-1<c<1$ and $P_c(z)=0$ with $0<|z|<1$, then $H$ takes the same value at the two distinct points $ze^{\pm i\theta}$ of $\mathbb{D}$, and $H$ is not univalent in $\mathbb{D}$. Dieudonné’s criterion [7] (see also Lemma 6.3 of [10]) is a converse: if no $P_c$ with $c\in[-1,1]$ has a zero in $\mathbb{D}$, then $H$ is univalent in $\mathbb{D}$. It is usually stated for $H(0)=0$ and $H'(0)=1$, but it applies to every polynomial $H$: the hypothesis gives $h_1=P_c(0)\ne0$, and neither adding a constant to $H$ (the $P_c$ do not involve $h_0$) nor dividing $H$ by $h_1$ changes univalence or the zeros of the $P_c$.

That no $P_c$ has a zero in $\mathbb{D}$, for all $c$ in an interval, is not a finite check. The next lemma reduces it to the absence of zeros on the unit circle, which Lemma 4.3 below tests with a resultant.

Proof. Since $P_c(0)=h_1=H'(0)\ne0$, no $P_c$ vanishes identically. Since no $P_c$ has a zero on the circle, the number $\frac{1}{2\pi i}\oint_{|w|=1}P'_c(w)/P_c(w)\,dw$ of zeros of $P_c$ in $\mathbb{D}$ depends continuously on $c\in[-1,1]$; being an integer, it is constant. At $c=1$ it is 0, since $P_1=H'$. So no $P_c$ has a zero in $\mathbb{D}$, and Dieudonné’s criterion applies. □

Zeros in the closed disk

For a polynomial $f(w)=\sum_{\ell=0}^{n}e_\ell w^\ell$ of formal degree $n$ (that is, $e_n$ may vanish) put $f^*(w)=w^n\overline{f(1/\overline{w})}=\sum_{\ell=0}^{n}\overline{e_{n-\ell}}w^\ell$; on the circle, $|f^*|=|f|$.

Proof. On the circle $|f^*|=|f|$. If $f(w)=0$ there, then $f^*(w)=0$ and $Tf(w)=0$; if $Tf(w)=0$ there, then $|e_0||f(w)|=|e_n||f(w)|$, so $f(w)=0$. If $f$ has no zero on the circle, then $|e_nf^*|<|\overline{e_0}f|$ there, and Rouché’s theorem gives the same number of zeros in $\mathbb{D}$ for $\overline{e_0}f$ and $Tf$. For the last claim, let $1\le j\le n$. The constant term $\delta_j$ of $f_j=Tf_{j-1}$ is $|u_0|^2-|u_{n-j+1}|^2$, where $u_0$ and $u_{n-j+1}$ are the constant and the top coefficient of $f_{j-1}$, of formal degree $n-j+1$. So $\delta_j>0$ is the hypothesis of the first claim for $f_{j-1}$. Now $f_n=\delta_n$ is a nonzero constant, so it has no zeros; going back from $j=n$ to $j=1$, each $f_{j-1}$ has no zero on the circle and none in $\mathbb{D}$. □

The four checks

Proof. Each $P_c$ has formal degree $n-1$. Let $c\in[-1,1]$ and suppose that $P_c(w_0)=0$ for some $w_0$ with $|w_0|=1$. Since $1/\overline{w_0}=w_0$, we get $P_c^*(w_0)=w_0^{n-1}\overline{P_c(w_0)}=0$, so $w_0$ is a common zero of $P_c$ and $P_c^*$, and the resultant vanishes at $c$ (a common zero makes the Sylvester matrix singular, at any formal degrees). This contradicts (R). So no $P_c$ with $c\in[-1,1]$ has a zero on the circle, and by (S) and Lemma 4.1, $H$ is univalent in $\mathbb{D}$. By (C), $Q'$ has no zero in $\mathbb{D}$. So Lemma 3.1 applies, and Theorem 2.2 gives $K^*\ge1/B(0)$. □

The labels (D), (C), (R), (S) stand for the data, the critical points of $Q$, the resultant, and the Schur–Cohn test for $H'$. Since the $h_\ell$ are fixed and each $U_{\ell-1}$ is a polynomial, the formulas for $P_c$ and $P_c^*$ make sense for every $c$, and $\operatorname{Res}_w(P_c,P_c^*)$ is a polynomial in $c$, so (R) is a statement about one polynomial in one real variable. In (C) and (S) the absence of zeros in $\mathbb{D}$ is checked with Lemma 4.2, applied to $V$ and to $H'$. Lemma 4.2 can only show that a polynomial has no zero in the closed disk, while the $Q'$ of § 5 has a zero on the circle, at $z=1$; this is why (C) splits off the factor $Z$.

Proof of Theorem 1.1

The polynomial $Q$ of our pair has degree 18, and we describe it through its critical points. Fifteen of its seventeen critical points are prescribed: $z=1$, with multiplicity 5, and ten Gaussian rationals $\rho_j$; the other two, the zeros of a quadratic $G$, are then fixed by requiring that $Q$ vanish at $a_1$ and $a_2$ as well as at $b$. The numbers $b$, $a_1$, $a_2$, $\rho_1,\ldots,\rho_{10}$ below are Gaussian rationals whose denominators are powers of 2; $G$, $Q$ and $H$ are determined by them exactly. These thirteen numbers were found by a numerical search that made $1/B(0)$ large (see Remark 5.2), with the degree 18 and the critical point of order 5 at $z=1$ held fixed, and were then rounded; the proof uses only the printed numbers. Let

and let $\rho_1,\ldots,\rho_{10}$ be the Gaussian rationals

Put $L(t)=(t-1)^5\prod_{j=1}^{10}(t-\rho_j)$. There is a unique pair $(g_0,g_1)$ of Gaussian rationals such that the quadratic $G(t)=t^2+g_1t+g_0$ satisfies

This is a linear system in $g_0,g_1$ with nonzero determinant; the real and imaginary parts of $g_0$ and $g_1$ have denominators of 451 to 458 digits. Define

Since $GL$ is monic of degree 17, $Q$ is a monic polynomial of degree 18; $Q(b)=0$ by definition, and the two conditions on $G$ say exactly that $Q(a_1)=Q(a_2)=0$. We put $V(z)=18G(z)\prod_j(z-\rho_j)$, of degree 12, and $H(z)=Q(z)(1-\overline{a_1}z)(1-\overline{a_2}z)/((z-a_1)(z-a_2))$, a polynomial of degree 18. Numerically the zeros of $G$ are $1.00000688\ldots-0.00152261\ldots i$ and $-0.41009053\ldots+1.04265219\ldots i$. The exact coefficients of $Q$, $Q'$, $V$ and $H$ are in the repository named on the first page. Figure 2 shows the configuration: ten of the seventeen critical points of $Q$, counted with multiplicity, lie within 0.0123 of $z=1$ (the zero of order 5, the first zero of $G$, and $\rho_1$, $\rho_2$, $\rho_9$, $\rho_{10}$), and the other seven have moduli between 1.017 and 1.238. The two factors of $\Lambda=J(a_1,b)J(a_2,b)$ are $J(a_1,b)=1.30038\ldots$ and $J(a_2,b)=1.02761\ldots$.

Vector (PDF)

Proof. Each of the four facts reduces to finitely many exact computations over $\mathbb{Q}(i)$, which we carry out by computer, from the numbers printed above, as follows. We first solve the linear system for $g_0,g_1$ and expand $Q$, $Q'$ and $H$ into exact lists of coefficients; all the checks are made on these lists. The programs and their output are in the repository named on the first page; the whole computation takes about eight minutes on one processor core.

(D) We compute $|b|^2$, $|a_1|^2$, $|a_2|^2$ exactly, compare them with 1, and check that the three points are distinct and that $Q$ vanishes at each of them. We also check that the coefficient list of $Q'$ is that of the derivative of $Q$, and that $H(z)(z-a_1)(z-a_2)=Q(z)(1-\overline{a_1}z)(1-\overline{a_2}z)$.

(C) We divide $Q'$ by $(z-1)^5$ exactly, with remainder 0, and apply Lemma 4.2 to the quotient $V$: the 12 numbers $\delta_j$ are positive.

(R) After multiplying $H$ by a positive integer, so that its coefficients are Gaussian integers (this multiplies the resultant by a positive constant), we compute $F(c)=\operatorname{Res}_w(P_c,P_c^*)$ as an exact polynomial in $c$. The resultant is the determinant of a $34 \times34$ Sylvester matrix whose entries are polynomials in $c$ of degree at most 17, so $F$ has degree at most 578, and its values at 579 points determine it. We compute the determinant at the 581 integer points $c=0,1,\ldots,580$, where the entries are Gaussian integers since each $U_{\ell-1}$ has integer coefficients. We work modulo the 1128 largest primes $r\equiv1\pmod4$ below $2^{62}$, mapping $i$ to each of the two square roots of $-1$ modulo $r$ (the two images give the real and imaginary parts modulo $r$), and recover the values exactly by the Chinese remainder theorem: Hadamard’s inequality bounds the real and imaginary parts at every point by $2^E$ with $E=69921$, and the product of the primes exceeds $2^{E+2}$. The imaginary parts of all 581 values are 0; since $F$ has degree at most 578, $F$ has real coefficients. We interpolate $F$ from its values at 579 of the points, check the result at the other two, and find that $F$ has degree 578. Finally $F(\pm1)\ne0$, and $F$ has no zero in $(-1,1)$: we split $[-1,1]$ by repeated bisection, checking that $F$ does not vanish at any midpoint, until on each piece $[c_0,c_1]$ the coefficients of $(1+x)^{578}F\left(\frac{c_0+c_1x}{1+x}\right)$ have no sign change; since $x\mapsto\frac{c_0+c_1x}{1+x}$ maps $(0,\infty)$ onto $(c_0,c_1)$, Descartes’ rule of signs shows that $F$ has no zero in $(c_0,c_1)$. For our $F$ this takes 47 applications of Descartes’ rule, 24 of them on final subintervals.

(S) We apply Lemma 4.2 to $H'$, of degree 17: the 17 numbers $\delta_j$ are positive.

We also check that $J(a_1,b)J(a_2,b)$ equals the rational number $\Lambda$ of Theorem 1.1. In exact arithmetic, $\Lambda-1.1559837^2=6.75\cdots\cdot10^{-8}>0$, so $\sqrt{\Lambda}>1.1559837$. Lemma 4.3 gives the rest.

Remark 5.2 (The margins). The zeros of $V$, $H'$ and the $P_c$ lie outside $\mathbb{D}$. We call the distance from the unit circle to the nearest zero of such a polynomial its margin. Some margins are very small, which is why every step is done in exact arithmetic. They are small because the data were chosen to make $1/B(0)$ large, and, as explained before Lemma 3.1, what limits $1/B(0)$ is that these zeros must stay out of $\mathbb{D}$. The zero of $V$ nearest to the circle, a zero of $G$, has modulus $1.00000804\ldots$. The zero of $H'$ nearest to the circle has modulus $1.00001141\ldots$; since $P_{\pm1}(w) = H'(\pm w)$, this gives the margin at $c = \pm1$. Numerically the zeros of the $P_c$ come closest to the circle near $c = \pm0.9231$, within about $7.03 \cdot10^{-6}$; see Figure 3. A small margin has a geometric meaning. For $-1 < c < 1$ and $c = \cos\theta$, a zero $w$ of $P_c$ gives $H(we^{i\theta}) = H(we^{-i\theta})$ with $|w| > 1$; so when $|w|$ is close to $1$, two points of the boundary curve of $H(\mathbb{D})$ nearly coincide. Numerically, at the dips of Figure 3 this happens along the narrow inlet of Figure 2, whose two sides almost touch.

Vector (PDF)

The degree statement of Theorem 1.1 rests on the monotonicity of $K(d)$, which is due to Crane [5] (see also [10 proof of Cor. 4.6]); we include the short proof.

Proof. Let $p \in\mathcal{P}_d$, and for $\varepsilon\ne0$ put $p_\varepsilon(z) = p(z) + \varepsilon z^{d+1}$, which lies in $\mathcal{P}_{d+1}$. As $\varepsilon\to0$, $d - 1$ of the $d$ critical points of $p_\varepsilon$, counted with multiplicity, tend to those of $p$, and the remaining one, $\beta_\varepsilon$, tends to $\infty$. The first claim follows from Hurwitz’s theorem, as $p_\varepsilon' \to p'$ locally uniformly; the second because the product of the $d$ critical points of $p_\varepsilon$ has modulus $1/((d + 1)|\varepsilon|) \to\infty$, as $p_\varepsilon'(0) = 1$, while the other $d - 1$ stay bounded. Since $p_\varepsilon'(\beta_\varepsilon) = 0$ gives $\varepsilon\beta_\varepsilon^d = -p'(\beta_\varepsilon)/(d + 1)$, we have

and if $\alpha$ is the leading coefficient of $p$, the right side is $\alpha\beta_\varepsilon^{d-1}/(d + 1) + O(\beta_\varepsilon^{d-2})$, which tends to $\infty$ in modulus, since $d \ge2$. As the critical points of $p$ are nonzero, it follows that $m(p_\varepsilon) \to m(p)$. Hence $K(d + 1) \ge m(p)$ for every $p \in\mathcal{P}_d$, and $K(d + 1) \ge K(d)$.

Proof. Let $A = 1.1559837$. By Theorem 5.1, $A < K^* = \sup_d K(d)$, so $K(d_0) > A$ for some $d_0$. By Lemma 5.3, $K(d) > A$ for every $d \ge d_0$.

The proof gives no explicit value of $d_0$, since Theorem 2.2 does not bound the degree of the polynomial it gives. Theorem 5.1 and Corollary 5.4 prove Theorem 1.1.

Acknowledgments

The author is immensely grateful to Edward Crane for mentorship, and for feedback and suggestions that greatly improved this work and its exposition.

References