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  "body": "\n## Definition\n\nLet $1 \u003c p \u003c \\infty$ and let $f \\in L^p(-1, 1)$, a function in the [[$L^p$ space|SASP14MA1Z]] of the interval.\nThe *finite Hilbert transform* of $f$ is the function $T f$ on $(-1, 1)$ given by\n$$(T f)(x) = \\frac{1}{\\pi}\\, \\mathrm{PV}\\!\\int_{-1}^{1} \\frac{f(t)}{t - x}\\, dt = \\frac{1}{\\pi} \\lim_{\\varepsilon \\to 0^+} \\int_{\\set{t \\in (-1, 1) : \\abs{t - x} \u003e \\varepsilon}} \\frac{f(t)}{t - x}\\, dt,$$\nthe [[Cauchy principal value|0SYR10HD4R]] integral.\nThe limit exists for almost every $x \\in (-1, 1)$.\n\n## Remarks\n\nIf $\\tilde f$ is $f$ on $(-1, 1)$ and $0$ elsewhere, then $T f = -H\\tilde f$ on $(-1, 1)$, where $H$ is the [[Hilbert transform|NSCEBXM7ZM]] with kernel $\\frac{1}{\\pi (x - t)}$.\nBy the theorem of M. Riesz on $H$, the operator $T$ is bounded on $L^p(-1, 1)$ for $1 \u003c p \u003c \\infty$.\nSome authors use the kernel $\\frac{1}{x - t}$ instead, which changes the sign of $T$.\n\nExample: $f(t) = \\sqrt{1 - t^2}$ gives $T f(x) = -x$.\nCheck: for $\\abs{z} \u003e 1$, $\\int_{-1}^{1} \\frac{\\sqrt{1 - t^2}}{t - z}\\, dt = \\pi\\bigl(\\sqrt{z^2 - 1} - z\\bigr)$ with the branch of the root that behaves like $z$ at infinity.\nIts 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 $-\\pi x$.\n\nNon-example of injectivity: $g(t) = (1 - t^2)^{-1/2}$ lies in $L^p(-1, 1)$ for $1 \u003c p \u003c 2$ and $T g = 0$ on $(-1, 1)$.\nSo $T$ is not injective on these spaces.\nFor $p \\ge 2$ the function $g$ is not in $L^p(-1, 1)$.\n",
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