Finite Hilbert transform

Remarks

If $\tilde f$ is $f$ on $(-1, 1)$ and $0$ elsewhere, then $T f = -H\tilde f$ on $(-1, 1)$, where $H$ is the Hilbert transform with kernel $\frac{1}{\pi (x - t)}$. By the theorem of M. Riesz on $H$, the operator $T$ is bounded on $L^p(-1, 1)$ for $1 < p < \infty$. Some authors use the kernel $\frac{1}{x - t}$ instead, which changes the sign of $T$.

Example: $f(t) = \sqrt{1 - t^2}$ gives $T f(x) = -x$. Check: for $\abs{z} > 1$, $\int_{-1}^{1} \frac{\sqrt{1 - t^2}}{t - z}\, dt = \pi\bigl(\sqrt{z^2 - 1} - z\bigr)$ with the branch of the root that behaves like $z$ at infinity. Its boundary values from above and below on $(-1, 1)$ are $\pi\bigl(\pm i\sqrt{1 - x^2} - x\bigr)$, and the principal value is their average $-\pi x$.

Non-example of injectivity: $g(t) = (1 - t^2)^{-1/2}$ lies in $L^p(-1, 1)$ for $1 < p < 2$ and $T g = 0$ on $(-1, 1)$. So $T$ is not injective on these spaces. For $p \ge 2$ the function $g$ is not in $L^p(-1, 1)$.