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

Abstract:

The smallest positive constant $c$ for which the Hardy and Littlewood [8] $L\log L$ inequality (1) \[ M(f) \equiv \int \frac {{dx}}{x}\int _0^x {|f| \leq c(1 + \int |f|{{\log }^ + }|f|)} \quad \equiv c(1 + L(f))\] 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 $\log ^ + \left | f \right |$ 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) \[ M(f) \leq cL(f) + d,\quad {\text {for all integrable}}\quad f.\] The set of all ordered pairs $(c,d)$ for which (2) holds is identified as \[ \{ (c,d):c > 1,d \geq 1 + {e^{ - c}}/(c - 1)\} .\] 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).