 Sharp constant in Hardy's integral inequality

The indicator function example reaches a squared L2 norm ratio of two. To approach four, concentrate more strongly near zero. For $0\le a<1/2$, use the measurable functionAll integrals use Lebesgue measure $dx$. Keeping both portions of the squared L2 norm givesConsequently,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.^[ref]

The endpoint is excluded: at $a=1/2$, ${\int }_0^1|f_a{|}^{2}dx={\int }_0^1{x}^{-1}dx=\infty$, and the inner portion of $\Vert H{f}_a{\Vert }_2^2$ 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.