Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove that:

A function $f$ is uniformly Lipschitz $\alpha$ over $\mathbb R$ if $$\int_{-\infty}^{+\infty}|\hat f(\omega)|(1+|\omega|^\alpha)d\omega<+\infty$$ A function $f$ is uniformly Lipschitz over $[a, b]$ if it satisfies $|f(t)-p_v(t)|\le K(t-v)^\alpha$, where $p_v(t)$ is a polynomial of degree $\lfloor\alpha\rfloor,$ for all $v∈[a, b]$ with a constant $K$ that is independent of $v$.

I think that this condition implies $f$ is $\lfloor\alpha\rfloor$ times continuous, but I am still unable to prove it, and I prefer a direct proof.

share|cite|improve this question
I found the answer on page 207 of the book A Wavelet Tour of Signal Processing, 3rd ed. If anyone want to see it, I can write it down. – Ziqian Xie Apr 4 '13 at 21:17
up vote 2 down vote accepted

Yes, this is true. Your continuity condition is basically that of Holder continuity with exponent $\alpha$.

A good reference is the paper "Regularity of the obstacle problem for a fractional power of the laplace operator" by Luis Silvestre, Communications on Pure and Applied Mathematics Volume 60, Issue 1, pages 67–112, January 2007. A version of this paper is online at his web page here.

Your question is a simple corollary of one of Silvestre's preliminary estimates: namely, that if a function $g \in L^\infty$ and $$ |\xi|^{\alpha} \hat{u} = \hat{g} $$ where the hats represent Fourier transforms, then $u$ is $C^\alpha$, where $0 < \alpha < 2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.