The Hilbert-transform characterisation of $\pi$

For a real-valued $f$ in $L^2(\mathbb{R})$, the Hilbert transform is the singular integral taken as a Cauchy principal value,

$Hf(t) = \frac{1}{\pi} \, \mathrm{p.v.} \int_{-\infty}^{\infty} \frac{f(x)}{x - t} \, dx.$

Characterisation of $\pi$. The constant $\pi$ is the unique positive normalising factor $c$ for which the operator $f \mapsto \frac{1}{c}\,\mathrm{p.v.}\int f(x)/(x - t)\,dx$ satisfies $H^2 = -I$, that is, for which $H$ defines a linear complex structure on the real Hilbert space of square-integrable real-valued functions on the line. Equivalently, up to a normalising factor $H$ is the unique bounded linear operator on $L^2(\mathbb{R})$ that commutes with positive dilations and anticommutes with reflection of the line, and $\pi$ is the factor that makes it unitary.

Status: settled, but outside this library's stack. This is classical, not open. It is out of reach here because it needs the deferred measure and integration track (Lebesgue measure, the space $L^2$, principal values of singular integrals) and the deferred functional analysis track (bounded operators on a Hilbert space, unitarity, complex structures) at the same time.

Not proved in this library. The characterisation is recorded and cited, and no page here may use it.

What is known, and what would prove it here. Everything is known; the obstacle is again purely prerequisite. Discharging this item requires both deferred analysis tracks: Lebesgue measure and $L^p$ spaces on one side, and bounded operators, the Plancherel theorem and the Fourier-multiplier description of $H$ on the other. On the Fourier side $H$ is multiplication by $\pm i \,\mathrm{sgn}(\xi)$, the sign depending on which of the two standard sign conventions is taken for the kernel (the one displayed above, with $x - t$ in the denominator, gives $+i \,\mathrm{sgn}(\xi)$; writing the kernel as $t - x$ instead gives $-i \,\mathrm{sgn}(\xi)$, and the two operators differ by a sign). Either way the multiplier has modulus one and squares to $-1$, so $H$ is unitary and $H^2 = -I$ is immediate, which is why the choice of convention does not affect the characterisation. Once those tracks exist the proof is short, which is exactly why the item is a prerequisite problem and not a difficulty problem.

Why it matters here. This library treats the many equivalent characterisations of $\pi$ as a subject in its own right: the least positive zero of the sine, half the period of the complex exponential, the area of the unit disc, the value of Wallis's product, and so on. That collection is supposed to be a complete answer to "what is $\pi$", and it cannot be, because at least one of the standard characterisations lives in a subject the library does not build. Recording this one keeps the claim of completeness honest, and marks precisely which two tracks would have to exist before it could be added.