Hilbert's Inequality

The double-sum bound with the sharp constant pi.

The idea

Theorem (Hilbert's inequality).

For nonnegative sequences $a = (a_m)$ and $b = (b_n)$ with $\|a\|_2 = \bigl(\sum_m a_m^{2}\bigr)^{1/2}$ and $\|b\|_2 = \bigl(\sum_n b_n^{2}\bigr)^{1/2}$ finite, $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{a_m b_n}{m+n} \;\le\; \pi\,\|a\|_2\,\|b\|_2,$ and the constant $\pi$ is sharp: no smaller constant works for every pair of sequences.

No nonzero pair attains equality: sequences exist that bring the ratio of the two sides arbitrarily close to $\pi$, but none reaches it.

That the double sum is finite at all is part of the assertion. For each fixed $m$ the row sum $\sum_n \frac{1}{m+n}$ diverges, so the weights $\frac{1}{m+n}$ alone do not force convergence; the square-summability of $a$ and $b$ does.

The constant deserves attention too. The right-hand side measures the sequences only through their $\ell^{2}$ norms, with no reference to how their entries are distributed, and the sharp constant for a bound of that form is exactly $\pi$.

Ways to work on it

• Walkthrough. The kernel 1/(m+n), the asymmetric Cauchy-Schwarz split, and the constant π.
• Practice. State the sharp constant for a Hilbert-type double sum.
• Hardest. Prove that the constant π is sharp.

Not sure where to start? Take the ten-question placement test.