Negative Lipschitz (Hölder) exponent: Intuition

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SUMMARY

The discussion centers on the concept of the negative Lipschitz (Hölder) exponent and its implications for function validity. It establishes that while Lipschitz exponents are defined for non-negative values, negative exponents lead to divergence at singular points, necessitating a shift to tempered distributions. The participants clarify that for α < 0, the condition |f(x) - Pν(x)| ≤ K·(1/|x - ν|^|α|) indicates the presence of poles, which disrupts continuity and renders the function concept invalid.

PREREQUISITES
  • Understanding of Lipschitz continuity and Hölder exponents
  • Familiarity with polynomial functions and their properties
  • Knowledge of tempered distributions in mathematical analysis
  • Basic concepts of continuity and singular points in functions
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  • Study the properties of tempered distributions in functional analysis
  • Explore the implications of negative Hölder exponents on function behavior
  • Research the relationship between poles and continuity in mathematical functions
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Mathematicians, analysts, and students studying real analysis, particularly those interested in the behavior of functions under different conditions and the implications of Lipschitz and Hölder exponents.

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

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!
:)
 
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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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