2005
DOI: 10.2996/kmj/1111588040
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On the Łojasiewicz exponent and Newton polyhedron

Abstract: The object of this paper is to give an estimation of the Łojasiewicz exponent of the gradient of a holomorphic function under Kouchnirenko's nondegeneracy condition, using information from the Newton polyhedron.Let f : ðC n ; 0Þ ! ðC; 0Þ be a germ of holomorphic function. The Łojasiewicz exponent of gradient of f , Lð f Þ is by definitionIt is well-known that Lð f Þ < y if and only if f has an isolated singularity at the origin. Chang and Lu [1] proved that for any integer r greater than Lð f Þ, f is a C 0 -s… Show more

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Cited by 22 publications
(24 citation statements)
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“…where m 0 ð f Þ is a combinatorial number calculated from G þ ð f Þ: Also Abderrahmane [1] gave a similar estimation in terms of nð f Þ: In the case n ¼ 2; 3 there are more exact formula for the Łojasiewicz exponent of non-degenerate holomorphic functions (see [11] and [14]).…”
Section: Resultsmentioning
confidence: 88%
“…where m 0 ð f Þ is a combinatorial number calculated from G þ ð f Þ: Also Abderrahmane [1] gave a similar estimation in terms of nð f Þ: In the case n ¼ 2; 3 there are more exact formula for the Łojasiewicz exponent of non-degenerate holomorphic functions (see [11] and [14]).…”
Section: Resultsmentioning
confidence: 88%
“…There are two type of inequalities which are shown by S. Lojasiewicz [16,17]. ∂f (z) ≥ c|f (z)| θ , c = 0, 0 ≤ ∃θ < 1, ∀z ∈ U, (1) ∂f (z) ≥ c z η , c = 0, ∃η > 0, ∀z ∈ U (2) where U is a sufficiently small neighborhood of the origin. Here ∂f (z) is the gradient vector ( ∂f ∂z 1 , .…”
Section: Holomorphic Functions and Lojasiewicz Exponentsmentioning
confidence: 99%
“…, I r n n ), respectively. By an argument analogous to the previous discussion and considering the definitions of r and r , we have that if p ≥ max{r, r } then σ (I r 1 1 , . .…”
Section: Lemma 26mentioning
confidence: 99%