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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.

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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

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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 $0 < \alpha < 2$.

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