The complex gradient inequality with parameter

We prove that given a holomorphic family of holomorphic functions with isolated singularities at zero and constant Milnor number, it is possible to obtain the gradient inequality with a uniform exponent.


Introduction
This note was inspired by a recent article of A. P loski [P2] concerning the semi-continuity of the Lojasiewicz exponent in a family of multiplicity-constant deformation of a finite holomorphic germ (actually, our main tool is a method of P loski used already in [P1]). Of course, it would probably be possible to derive the result presented herein from this article, but it seems interesting to give a direct, elementary and self-consistent proof, since it is a starting point for tackling a much more general problem.
We are interested in the Lojasiewicz Gradient Inequality: given an (complex or real) analytic (or subanalytic and of class C 1 ) function f of m variables and such that f (0) = 0 we are able to find an exponent θ ∈ (0, 1) such that in a neighbourhood of zero |f (x)| θ ≤ const.||∇f (x)|| where ∇f (x) denotes the (complex or real) gradient of f . The classical version (for a real analytic function) was obtained by Lojasiewicz from his famous Lojasiewicz inequality that he used to solve L. Schwartz's Division Problem. One of the simplest and most important applications of this inequality is the study of the analytic gradient dynamics, see [ L] (the fact that θ ∈ (0, 1) is of utmost importance; to make the paper self-consistent, we give a simple proof of this in our special case in Section 2).
Our general aim is to obtain a parameter version of this inequality, i.e. to prove that given a well-parametrized family of functions we can find (locally) a uniform exponent for this inequality to be satisfied by each of these functions. Probably the simplest case is presented in this paper and it concerns a µ-constant holomorphic unfolding.
A general subanalytic parameter version of the gradient inequality will be given in a forthcoming paper.

On the gradient exponent
For convenience sake, we will give -using only complex analytic geometry tools in the spirit of P loski -yet another and elementary proof of the gradient inequality for a holomorphic germ f : (C m , 0) → (C, 0) satisfying the condition dim 0 ∇f −1 (0) = 0, where is the complex gradient. What will be of greatest interest for our purposes is the bound on the exponent we obtain. We may assume m > 1, since the case m = 1 is a simple exercise involving power series. Hereafter we will insist on details that for simplicity we omit in the next section.
Our approach is based on [P1]. Take a connected neighbourhood Hence ∇f is a proper map from U onto a connected neighbourhood V of zero. This means that the projection π(z, w) = w is proper on Γ ∇f ∩ U × V and sends this part of the graph onto V . It is thence a branched covering (see [Ch]) with covering number µ. Its singular locus 1 σ V is analytic and µ is in fact the Milnor number of f at zero. For w ∈ V \ σ we have ∇f −1 (w) = {z (1) , . . . , z (µ) } consisting of exactly µ points and so we can define for t ∈ C P (w, t) : = are holomorphic functions on V \ σ admitting continuous extensions onto V . Therefore, by the Riemann Extension Theorem, This polynomial is called the characteristic polynomial of f with respect to ∇f . Clearly, P (∇f, f ) ≡ 0 and a j (0) = 0.
Remark 2.1. Note that in the same way we can define the characteristic polynomial of f with respect to any holomorphic map germ The Viète formulae yield |f (z)| ≤ 2 max j |a j (∇f (z))| 1/j . On the other hand, since in a neighbourhood of zer there is |a j (w)| ≤ c j ||w|| ord 0 a j for some c j > 0 (ord 0 h denotes the order of vanishing at zero 2 of the holomorphic function h), this leads to in a neighbourhood of zero. We have obtained the gradient inequality with the exponent θ = µ max j=1 j ord 0 a j and we shall prove in addition that θ < 1 (i.e. ord 0 a j > j for all j). This is still rather delicate (see the crucial article [T] of B. Teissier, Corollaire 2 p. 270 and compare with [PW] Theorem 2).
2 ord 0 h is the degree of the initial form in the expansion of h ≡ 0 into homogeneous forms near zero. There is ord 0 h = max{η > 0 | |h(z)| ≤ const.||z|| η in a neighbourhood of 0}, see [Ch].
3 From the expressions for a j and the Viète formulae it is easy to see that P (∇f, f ) ≡ 0 implies |f (z)| ≤ const.||∇f (z)||. For the other way round it suffices to observe that a j belongs to m j = z α : |α| = j , where m is the maximal ideal of the ring O m . But, P −1 (0) ∩ (Cw × C) = {(∇f (z), f (z)) | z ∈ ∇f −1 (Cw)} is a curve as the image by (∇f, f ) of the polar curve X := ∇f −1 (Cw) (∇f is surjective and we are in a neighbourhood of the origin). Fix (w, t) and denote by Γ an irreducible component of (∇f, f )(X) for which (w, t) ∈ C 0 (Γ). Then there exists v ∈ C m+1 \{0} such that C 0 (Γ) = Cv. Now we need the following lemma.
Lemma 2.2. Let h : W → C n be a holomorphic function defined in a neighbourhood W ⊂ C of zero, satisfying h(0) = 0 and such that h(W ) defines an irreducible curve germ at zero. Then there is a sequence W ∋ t ν → 0 such that h(t ν )/||h(t ν )|| converge to a vector v for which C 0 (h(W )) = Cv.
Summing up, we have proved the following Proposition (cf. [T] Corollaire 2 p. 270 and [P1], [PW]): 3. If f ∈ O m is a holomorphic germ such that ∇f has an isolated zero at the origin, then the coefficients of the characteristic polynomial P (w, t) = t µ + a 1 (w)t µ−1 + . . . + a µ (w) ∈ O m [t], of f with respect to ∇f , where µ is the Milnor number of f , all satisfy ord 0 a j ≥ j + 1 and we have in a neighbourhood of zero the gradient inequality

Gradient inequality with parameter
Let f (x, t) ∈ O m+k be a holomorphic germ. Write f t (x) := f (x, t) and put g t (x) := ∇f t (x) for the complex gradient. The resulting germ g(x, t) := g t (x) is, of course, holomorphic.
We assume that for all t small enough, g −1 t (0) = {0} is isolated at zero (note that this implies that f (0, t) ≡ 0 e.g. by the gradient inequality) and that the Milnor numbers m 0 (g t ) ≡ µ are constant.
The theorem we are aiming at is the following: Theorem 3.1. If f (x, t) ∈ O m+k is such that for all t small enough, f t (x) has an isolated singularity at zero with constant Milnor number µ, then for all (x, t) in a neighbourhood of zero where C > 0. Note that the exponent belongs to [1/2, 1) and f itself satisfies the gradient inequality (with respect to (x, t)) with this exponent.
The proof will be given in several steps.
Lemma 3.2. In the setting introduced, the multiplicity at zero of the branched covering G is µ.
Step 2 Let now P (y, t, s) ∈ O m+k [s] be the characteristic polynomial of f with respect to G. Therefore, we have P (t, g(x, t), f (x, t)) ≡ 0 and so by the Viète formulae: where we write P (y, t, s) = s µ + a 1 (y, t)s µ−1 + . . . + a µ (y, t).
Lemma 3.3. P (·, t, ·) is the characteristic polynomial of f t with respect to g t .
Proof. For a generic point y near zero there is #g −1 t (y) = µ. Let x (1) , . . . , x (µ) be the points in this fibre. Then since the fibre G −1 (y, t) consists precisely of the points (x (j) , t), it is maximal and we get which gives the assertion.
In particular we have a j (0, t) = 0.
for all j and (x, t) in a neighbourhood U of zero such that G(U) ⊂ V . This gives the gradient inequality with parameter and µ max j=1 j j + 1 = µ µ + 1 as exponent. The proof is complete.