= Hardy's integral inequality for an interval indicator
{c}
Take the <indicator function> $f=\mathbf1_{(0,1)}$. Direct <integration> gives
$$
Hf(x)=\begin{cases}1,&0<x\leq1,\\1/x,&x>1.\end{cases}
$$
Although the <function> vanishes beyond one, its average remembers the unit area accumulated before one. Squaring and integrating both portions gives
$$
\|f\|_2^2=1,\qquad \|Hf\|_2^2=\int_0^1 1\,dx+\int_1^\infty x^{-2}\,dx=1+1=2\leq4\|f\|_2^2=4.
$$
Thus the tail accounts for half of the squared <L2 norm>. Averaging does not always decrease the <L2 norm> with this <Lebesgue measure> on the half-line. The same <indicator function> also explains the restriction $p>1$: $\int_1^\infty x^{-1}\,dx=\infty$, so its average is not in <L1 space>.
\Image[/hardy-interval-indicator-average-tail.png]
{title=An <indicator function>, its <Hardy operator> average, and the tail's squared <L2 norm>}
{description=The upper panel shows the input and its running average. The lower panel splits the squared average at one: each portion has <integral> one, including the tail beyond the plotted range.}
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