Hilbert's inequality

Remarks

Proof idea: the function $\varphi(t) = i(\pi - t)$ on $(0, 2\pi)$ has Fourier coefficients $1/k$ for $k \ne 0$ and $0$ for $k = 0$, and $\abs{\varphi} \le \pi$. So the double sum equals $\frac{1}{2\pi} \int_0^{2\pi} \varphi(t) \, \abs{P(t)}^2 \, dt$ for the trigonometric polynomial $P(t) = \sum_r u_r e^{-i r t}$. Bounding $\abs{\varphi}$ by $\pi$ and using Parseval's identity $\frac{1}{2\pi} \int_0^{2\pi} \abs{P}^2 = \sum_r \abs{u_r}^2$ gives the result.

The constant $\pi$ cannot be replaced by a smaller one that works for all $n$.

A closely related inequality, also called Hilbert's inequality, says that for nonnegative $a_m, b_n$, $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_m b_n}{m + n} \le \pi \Big( \sum_{m=1}^{\infty} a_m^2 \Big)^{1/2} \Big( \sum_{n=1}^{\infty} b_n^2 \Big)^{1/2}.$$

Going deeper

For a student

The matrix with entries $1/(r - s)$ for $r \ne s$ is a discrete version of the Hilbert transform. The inequality says that this matrix has operator norm at most $\pi$ on $\C^n$ for every $n$, uniformly in $n$. The proof is a model of a common technique: write a matrix as the Fourier coefficients of a bounded function, then use Parseval's identity.