Negative Lipschitz (Hölder) exponent: Intuition

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Negative Lipschitz (Hölder) exponent: Intuition!!!

Hi everybody!

Sorry for double posting... :(

I have some problem with Hölder (Lipschitz) exponent!
From what I know Lipschitz refer to integer values whereas Hölder to non integer ones.
The usual definition roughly states that:

A function [tex]f[/tex] is pointwise Lipschitz (local Hölder) [tex] \alpha \geq 0 [/tex] at [tex] \nu [/tex] if
there exist K>0 and a polynomial [tex]P_{\nu}[/tex] of degree [tex]m = \lfloor \alpha \rfloor [/tex], where [tex]\lfloor \alpha
\rfloor [/tex] is defined as the largest integer less than or equal to [tex]\alpha [/tex], such that
[tex] \forall x \in \mathbb{R} , |f(x)-P_{\nu}(x)|\leq
K|x-\nu|^{\alpha} . [/tex]

The above definition refers only to functions which have a positive alpha. I read somewhere , but I can't find no more references, that "if we want to consider negative values
too, as in this case the concept of function is not longer valid, because of the divergence at the singular
point, we have to shift to the class of tempered distribution"

What does it mean? Why if such exponent is negative the concept of function is no longer valid? Can you give me some intuition for that? :(

Many many thanks in advance!!
:)
 

Answers and Replies

  • #2
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For ##\alpha < 0## we would have a condition ##|f(x)-P_\nu (x)|\leq K\cdot \dfrac{1}{|x-\nu|^{|\alpha|}}## which hasn't anything to do with continuity no longer. All of a sudden we are dealing with poles, the closer we get to ##x=\nu##. The boundary would become trivially true in case of defined functions, and poles otherwise.
 

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