# Hilbert transform and Hilbert matrix

The Hilbert matrix is

\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \dots \\[4pt] \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & & \ddots \\[4pt] \frac{1}{3} & \frac{1}{4} & & \ddots & \\[4pt] \frac{1}{4} & & \ddots & & \\[4pt] \vdots & \ddots & & & \end{bmatrix}

I believe the Hilbert transform of a function $u$ on the circle $\mathbb T$ is, when it exists, the radial limit of the harmonic conjugate of u defined in the open disk $\mathbb D$. This conjugate limit exists in $L^p$ (it also exists pointwise a.e. for Lebesgue measure on $\mathbb T$ then) for all $u\in L^p, p>1$. Here we only need $L^2$.

The Hilbert matrix is a bounded Hankel operator on $\ell^2$, whose entries are therefore the positive Fourier coefficients of some $L^\infty(\mathbb T)$ function. The function $\sum_{n\ge 0}\frac{z^n}{n+1}$ has those Fourier coefficients but is not bounded on the circle, we can add negative Fourier coefficients to obtain $ie^{-it}(\pi-t)$ which is bounded -i.e. in $L^\infty(\mathbb T)$. From this last function we get an $L^\infty$ symbol for the Hilbert matrix as a multiplication operator from Hardy space to negative Hardy space.

The Hilbert transform is a multiplication operator (in $\hat{\ell^2}$, "Fourier" space) but from Hardy space ($H^2(\mathbb T)$) to itself because we just multiply the $n$th Fourier coefficients by $(-1)^{n-1}i$ -if I made no mistake.

Hilbert arrived at both at different times in his career: In 1894 for the matrix, investigating a question of approximation by orthogonal polynomials, and 1905 for the transform, investigating the Riemann-Hilbert problem. However I wonder if they may be related, because they connect to many related concepts. I may expand later, but if anyone has guesses or knows anything, I'd be glad to hear them.

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