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The best bound in the $ L\,{\rm log}\,L$ inequality of Hardy and Littlewood and its martingale counterpart

Author: David Gilat
Journal: Proc. Amer. Math. Soc. 97 (1986), 429-436
MSC: Primary 26D15; Secondary 60G46
MathSciNet review: 840624
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Abstract: The smallest positive constant $ c$ for which the Hardy and Littlewood [8] $ L\log L$ inequality (1)

$\displaystyle M(f) \equiv \int \frac{{dx}}{x}\int_0^x {\vert f\vert \leq c(1 + \int \vert f\vert{{\log }^ + }\vert f\vert)} \quad \equiv c(1 + L(f))$

is valid is proved to be the unique positive solution $ {c_0}$ of the equation $ {e^{ - c}} = {(c - 1)^2}$. This settles a question raised by Dubins and Gilat (1978) [6] and, again, more recently, by D. Cox (1984) [3].

Numerically, $ {c_0} \doteq 1.478$. This should be compared with $ c = e{(e - 1)^{ - 1}} \doteq 1.582$, obtained by Doob (1953) [5] in the context of martingale theory and, since then, widely used in the probability literature. Curiously enough, Doob's coefficient is the best upper bound, but for a slightly different inequality. If only the plus sign is removed from $ \log ^ + \left\vert f \right\vert$ in (1), then $ c$ must be at least $ e{(e - 1)^{ - 1}}$ for (1), so modified, to be valid. The original inequality (1) is a normalized form of the two-parameter inequality (2)

$\displaystyle M(f) \leq cL(f) + d,\quad {\text{for all integrable}}\quad f.$

The set of all ordered pairs $ (c,d)$ for which (2) holds is identified as

$\displaystyle \{ (c,d):c > 1,d \geq 1 + {e^{ - c}}/(c - 1)\} .$

Furthermore, for each point on the lower boundary of this set, there is a unique $ f$ (up to null sets) which attains equality in (2).

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Keywords: Hardy-Littlewood maximal function, $ L\log L$, $ {L_p}$, martingale inequalities, Doob
Article copyright: © Copyright 1986 American Mathematical Society

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