On Kolmogorov’s inequalities $\tilde {f}_p \leq C_p$, $f_1$, $0<p<1$

Author:
Burgess Davis

Journal:
Trans. Amer. Math. Soc. **222** (1976), 179-192

MSC:
Primary 42A36; Secondary 60J65

DOI:
https://doi.org/10.1090/S0002-9947-1976-0422983-7

MathSciNet review:
0422983

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Abstract | References | Similar Articles | Additional Information

Abstract: Let $\mu$ be a signed measure on the unit circle *A* of the complex plane satisfying $|\mu |(A) < \infty$, where $|\mu |(A)$ is the total variation of $\mu$, and let $\tilde \mu$ be the conjugate function of $\mu$. A theorem of Kolmogorov states that for each real number *p* between 0 and 1 there is an absolute constant ${C_p}$ such that ${({(2\pi )^{ - 1}}\smallint _0^{2\pi }|\tilde \mu ({e^{i\theta }}){|^p}d\theta )^{1/p}} \leqslant {C_p}|\mu |(A)$. Here it is shown that measures putting equal and opposite mass at points directly opposite from each other on the unit circle, and no mass any place else, are extremal for all of these inequalities, that is, if $\nu$ is one of these measures the number ${({(2\pi )^{ - 1}}\smallint _0^{2\pi }|\tilde \nu ({e^{i\theta }}){|^p}d\theta )^{1/p}}/|\nu |(A)$ is the smallest possible value for ${C_p}$. These constants are also the best possible in the analogous Hilbert transform inequalities. The proof is based on probability theory.

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Keywords:
Conjugate function,
Brownian motion,
Kolmogorov’s inequalities

Article copyright:
© Copyright 1976
American Mathematical Society