On the Gradient Inequality by S. L Ã ojasiewicz and M.-A. Zurro
Summary. Some proof of the Gradient Inequality is given.
K. Kurdyka and A. Parusi´ nski in [1] have done a very elegant proof of the Gradient Inequeliy. The aim of this note is to give another some other version of their proof. It is rather conceptual. Let us recall this inequality : Let f be an analytic function in a neighborhood of 0 ∈ Rn such that f (0) = 0, then we have the inequality: |f (x)|ϑ 6 |grad f (x)|
,
with
0 < ϑ < 1,
in a neighborhood of 0. As a consequence of this proof, we obtain a qualitative statement which is easily equivalent to the above inequality (see the remark 1). Proof. We can assume that grad f (0) = 0 (the remaining case being trivial). Let us consider the mapping g : B 37−→ ( |f (x)| , |grad f (x)| ) ∈ R2uv , with B a compact ball sufficientely small and centered at 0. Consider the set E = g(B) ⊂ [0, ∞) × [0, ∞) ; it is compact and subanalytic. It is enough to prove that for every L > 0 the point 0 does not belong to the closure of the set F = E ∩ {0 6 v < Lu} . In fact, the function ϕ(u) = inf Eu for small u > 0 is bouded semianalytic, hence then we must have ϕ(u) = 0 0 a + cuϑ + o (uϑ ) with 0 < ϑ0 < 1 . It follows that some neighborhood of 0 in E is contained in {v > uϑ } with some 0 < ϑ < 1 and this implies the Gradient Inequality. Let us suppose the contrary i.e. that with some L > 0 we have 0 ∈ F . Then the set g −1 (F ) ∩ g −1 (0) is non empty and so it contains a point b. By the curve sellecting lemma, there is a semianalytic arc {x = γ(t) : 0 < t 6 ε} contained in g −1 (F ) with γ analytic in some neighborhood of 0 and such that γ(0) = b . Then we have g ◦ γ ⊂ {0 6 v < u} , and therefore for the function h = f ◦ γ we have the inequalities (1)
|h0 (t)| 6 M |grad f (γ(t))| 6 M L|h(t)| ,
with a constant M . But h(0) = 0 and h 6≡ 0, hence h(t) = ak tk + . . . with ak 6= 0 , k > 1 , and so h0 (t) = kak tk−1 + . . . , which contradicts the inequality (1). 1
The argument of the above proof implies the following :
Remark 1. The gradient inequality (for any f 6≡ 0) is equivalent to each of the following qualitative statements: (A) Let F be an analytic function in a neighborhood of 0 ∈ Rn , such that F 6≡ 0 and F (0) = 0. Then |F (x)| −→ 0 | grad F (x) |
for x → 0 , grad F (x) 6= 0 .
(B) Let F be an analytic function in a neighborhood of 0 ∈ Rn , such that F 6≡ 0 and F (0) = 0. For any analytic function x(t) ∈ Rn in a neighborhood of 0 ∈ R , such that x(0) = 0 and grad F (x(t)) 6≡ 0 , we have |F (x(t))| −→ 0 for t → 0 . | grad F (x(t)) | In fact, the gradient inequality implies of course the statement (A) which implies the statement (B). Assume now the statement (B) with grad f (x) = 0 (for any function F). We have the semianalytic function 0 0 ϕ(u) = inf Eu = cuϑ + o(uϑ ) for small u > 0 , with c > 0 and some ϑ0 > 0 . Then the set g −1 (ϕ) ∩ g −1 (0) is non empty and so it contains a point b. By the curve sellecting lemma, there is a semianalytic arc {x = ξ(t) : 0 < t 6 ε} contained in g −1 (ϕ) with ξ(t) analytic in some neighborhood of¢ 0 and such that ξ(0) = b . As g(ξ(t)) ∈ ϕ for small t > 0 , ¡ (i.e. ϕ |f (ξ(t))| = |grad f (ξ(t))|) the statement for F (x) = f (x + b) and x(t) = z(t) − b implies that we must have ϑ0 < 1 . This implies the gradient inequality with some ϑ < ϑ0 . Obsere that (A) can be easily shown by proving by courve sellecting lemma that for every ε > 0 zero can not be an adherent point of the set {x : ε|grad f (x)| < |f (x)|} . Observe that the equivalence of both statements is a particular case of the following fact : 2
Proposition. Let g : M → N be a subanalytic mapping of real analytic manifolds such the image g(M ) is relatively compact. Let a ∈ M b ∈ N . Then g(x) → b for x → a+ iff for any analytic function x(t) in a neighborhood of 0 ∈ R and such that x(0) = a we have g(x(t)) −→ b for t → 0 + . One shows the sufficiency of the condition in the following way. It is enaugh to consider the case where M is a neighborhood of 0 in Rn , N = R , a = 0 , b = 0 and g is bounded and 6≡ 0 . One considers the function γ(r) = sup |x|=r |g(x)| and the set {x : |g(x)| > γ(|x|) − |x|} . It is subanalytic, and so by curve sellecting lemma it contains a semianalytic arc {x(t) : t > 0} where x(t) is analytic ¡ in¢a neighborhood of 0 in R and such that x(0) = 0 . It implies that γ x(t) → 0 for t → 0+ . Observe that if we do not assume that the image is relatively compact, the proposition does not hold. One checks it on the following exemple : one takes the caracteristic function ϕn of the set {0
0 .
StanisÃlaw L Ã ojasiewicz, Instytut Matematyki Uniwersytetu Jagiello´ nskiego, Reymonta 4, 30059 Krak´ ow, Poland. Mar´ıa Angeles Zurro Moro, Departamento de Matem´ aticas, Universidad Aut´ onoma de Madrid, Canto Blanco, 28106 Madrid, Spain.
References [1]
´ ski A. , wf -stratification of subanalytic functions Kurdyka K., Parusin and the L Ã ojasiewicz inequality , C.R.Acad.Sci.Paris, I, 318 (1994) . 3
[2]
L Ã ojasiewicz S. , Ensembles semi-analytiques , Preprint IHES (1965) .
[3]
L Ã ojasiewicz S., Zurro M. A. , Una introducci´on a la geometr´ıa Semiy Sub- anal´ıtica , Publicaciones de la Universidad de Valladolid (Spain) (1993) .
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