= Hardy's integral inequality
{c}
{title2=$\|Hf\|_p\leq\frac{p}{p-1}\|f\|_p$}
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 average
$$
Hf(x)=\frac1x\int_0^x f(t)\,dt,\qquad x>0.
$$
<Hölder's inequality> ensures $\int_0^x|f(t)|\,dt\leq x^{1-1/p}\|f\|_p<\infty$, so every such average exists. <Hardy's integral inequality> states
$$
\int_0^\infty |Hf(x)|^p\,dx\leq\left(\frac{p}{p-1}\right)^p\int_0^\infty |f(x)|^p\,dx.
$$
The real signed and complex extensions of the nonnegative statement follow from the <triangle inequality> $|Hf|\leq H|f|$. Changing $f$ on a <null set> changes neither average nor <Lp norm>.<Inequalities by Godfrey Harold Hardy and others>[Inequalities, second edition, §9.8, Theorem 327, equation (9.8.2), printed p. 240]{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.
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