Let sum from p=0 to infinity x_p²=1 and sum from q=0
Let $\sum_{p=0}^{\infty}x_p^2=1$ and $\sum_{q=0}^{\infty}y_q^2=1$. Prove that: $$ \sum_{p=0}^{\infty}\sum_{q=0}^{\infty} \frac{x_py_q}{p+q+1} \le \pi. $$
Explanation of Method
This is Hilbert's classical double series theorem. The sharp constant $\pi$ comes from a weighted Cauchy–Schwarz (Schur test): inserting the weight $\left(\frac{p+1}{q+1}\right)^{1/4}$ and its reciprocal splits the bilinear sum into two square roots, each bounded using the key integral $\int_0^{\infty}\frac{dv}{(1+v)\sqrt v}=\pi$, evaluated by comparing the discrete row/column sums to this integral.
- Insert the weight $(p+1)^{1/4}/(q+1)^{1/4}$ and its reciprocal, then apply Cauchy–Schwarz.
- Bound each resulting weighted row sum by the integral $\int_0^{\infty}\frac{dv}{(1+v)\sqrt v}=\pi$.
- Combine both bounds to get the sharp constant $\pi$.
Step 1 · Weighted Cauchy–Schwarz (Schur test)
Write $$ \frac{x_py_q}{p+q+1} = \left[\frac{x_p}{\sqrt{p+q+1}}\left(\frac{p+1}{q+1}\right)^{1/4}\right]\left[\frac{y_q}{\sqrt{p+q+1}}\left(\frac{q+1}{p+1}\right)^{1/4}\right]. $$ By Cauchy–Schwarz applied to the double sum: $$ \left|\sum_{p,q}\frac{x_py_q}{p+q+1}\right| \le \left(\sum_px_p^2\sum_q\frac1{p+q+1}\sqrt{\frac{p+1}{q+1}}\right)^{1/2}\left(\sum_qy_q^2\sum_p\frac1{p+q+1}\sqrt{\frac{q+1}{p+1}}\right)^{1/2}. $$
Step 2 · Bound the weighted row sum via an integral
For fixed $p$, compare the (decreasing) summand to the integral with $u=q+1$: $$ \sum_{q=0}^{\infty}\frac1{p+q+1}\sqrt{\frac{p+1}{q+1}} \le \int_0^{\infty}\frac1{p+1+u}\sqrt{\frac{p+1}u}\,du. $$ Substituting $u=(p+1)v$: $$ = \int_0^{\infty}\frac{dv}{(1+v)\sqrt v} = \pi $$ (via $v=\tan^2\theta$: $\int_0^{\pi/2}\frac{2\tan\theta\sec^2\theta}{\sec^2\theta\tan\theta}d\theta=2\int_0^{\pi/2}d\theta=\pi$). By symmetry, the same bound $\pi$ holds for the mirrored sum over $p$.
Step 3 · Combine
$$ \left|\sum_{p,q}\frac{x_py_q}{p+q+1}\right| \le \left(\pi\sum_px_p^2\right)^{1/2}\left(\pi\sum_qy_q^2\right)^{1/2} = \pi\cdot1\cdot1 = \pi. $$
Summary Table
| Quantity | Value |
|---|---|
| $\int_0^{\infty}dv/[(1+v)\sqrt v]$ | $\pi$ |
| Weighted row/column sum bound | $\le\pi$ (each) |
| Final inequality | $\left|\sum\frac{x_py_q}{p+q+1}\right|\le\pi$ |
Additional Notes
- This is exactly Hilbert's double series theorem (Hilbert's inequality), one of the most celebrated results in classical analysis; the constant $\pi$ is sharp — no smaller constant works for all $\ell^2$-normalized $x,y$.
- The Schur-test technique (weighting by a power of $(p+1)/(q+1)$ before applying Cauchy–Schwarz) is the standard method for proving sharp-constant bilinear inequalities of Hilbert type, and is the same idea used to bound $|S|\le\pi$ and $|T|\le2\pi$ in Problem 162.
- Problem 158's positive-definiteness result for the same Hilbert matrix (with $x=y$) is a special case consistent with this bound: the quadratic form $\sum x_px_q/(p+q+1)$ is squeezed between $0$ (exclusive) and $\pi\|x\|_2^2$.
Result
Hilbert's inequality holds with sharp constant $\pi$: