= Sharp constant in Hardy's integral inequality
{c}
{title2=$4$ for $p=2$}
The <indicator function> example reaches a squared <L2 norm> ratio of two. To approach four, concentrate more strongly near zero. For $0\leq a<1/2$, use the <measurable function>
$$
f_a(x)=x^{-a}\mathbf1_{(0,1)}(x),\qquad Hf_a(x)=\begin{cases}x^{-a}/(1-a),&0<x\leq1,\\1/((1-a)x),&x>1.\end{cases}
$$
All <integrals> use <Lebesgue measure> $dx$. Keeping both portions of the squared <L2 norm> gives
$$
\|f_a\|_2^2=\frac1{1-2a},\qquad \int_0^1|Hf_a|^2\,dx=\frac1{(1-a)^2(1-2a)},\qquad \int_1^\infty|Hf_a|^2\,dx=\frac1{(1-a)^2}.
$$
Consequently,
$$
\frac{\|Hf_a\|_2^2}{\|f_a\|_2^2}=\frac1{(1-a)^2}+\frac{1-2a}{(1-a)^2}=\frac2{1-a}\longrightarrow4\quad\text{as }a\uparrow\frac12.
$$
Every allowed member has ratio strictly below four. For any proposed constant $C<4$, a sufficiently large $a<1/2$ has ratio greater than $C$. Combined with <Hardy's integral inequality>, this proves that four is the smallest possible squared <L2 norm> constant.<Inequalities by Godfrey Harold Hardy and others>[Inequalities, second edition, §9.8, Theorem 327, printed p. 240, and sharpness argument on printed p. 243]{ref}
The endpoint is excluded: at $a=1/2$, $\int_0^1|f_a|^2\,dx=\int_0^1x^{-1}\,dx=\infty$, and the inner portion of $\|Hf_a\|_2^2$ also diverges. Its tail still has finite <integral> four. There is no finite <L2 norm> ratio at this endpoint, hence no equality example obtained by substituting it. The <limit> four is unattained by this family. Multiplying each $f_a$ by $\sqrt{1-2a}$ makes its <L2 norm> one without changing the ratio.
\Image[/hardy-power-family-sharp-constant.png]
{title=A power family approaches the <Sharp constant in Hardy's integral inequality>}
{description=Dashed input curves and solid <Hardy operator> averages show three allowed <exponents>. The right panel splits the exact squared <L2 norm> ratio into inner and tail contributions. The hollow endpoint marks a <limit>, not an admissible <function> in <L2 space>.}
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