Hardy's integral inequality ($\Vert Hf{\Vert }_p\le \frac{p}{p-1}\Vert f{\Vert }_p$)

Let $1<p<\infty$ and let $f$ be a real or complex measurable function in Lp space $L^p((0,\infty ),dx)$, with Lebesgue measure $dx$. The Hardy operator, denoted here by $H$, takes the running averageHölder's inequality ensures ${\int }_0^x|f(t)|dt\le x^{1-1/p}\Vert f{\Vert }_p<\infty$, so every such average exists. Hardy's integral inequality statesThe real signed and complex extensions of the nonnegative statement follow from the triangle inequality $|Hf|\le H|f|$. Changing $f$ on a null set changes neither average nor Lp norm.^[ref]

Averaging over an expanding interval spreads a localized input across the whole positive half-line. Hardy's integral inequality bounds the total Lp norm of that spread. The Hardy operator is consequently a bounded linear operator on this Lp space. The factor $p/(p-1)$ bounds its operator norm; its $p$th power is the constant in the displayed integral bound. These are different constants: at $p=2$, the L2 norm bound is two and the squared L2 norm bound is four.