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  "body": "\n## Statement\n\nLet $1 \u003c p \u003c \\infty$ and let $f \\in L^p(\\R)$, an element of the [[$L^p$ space|SASP14MA1Z]].\nLet $H$ denote the [[Hilbert transform|NSCEBXM7ZM]].\nThen $Hf \\in L^p(\\R)$ and\n$$H(Hf) = -f$$\nalmost everywhere.\nSo $H$ is a bijection of $L^p(\\R)$ onto itself, with inverse $H^{-1} = -H$.\n\n## Remarks\n\nFor $f \\in L^2(\\R)$ the [[Fourier transform|DPVM34ADZV]] of $Hf$ is $-i \\sgn(\\xi)\\, \\hat f(\\xi)$.\nApplying this twice multiplies $\\hat f$ by $(-i \\sgn \\xi)^2 = -1$ for $\\xi \\ne 0$, which gives $H^2 f = -f$ on $L^2$.\nFor general $p$, [[M. Riesz's theorem|PJFH51Z48M]] says that $H$ is bounded on $L^p$, so $H^2$ is bounded too, and the identity extends from the dense subspace $L^2 \\cap L^p$ by continuity.\nThe hypothesis $p \u003e 1$ is needed: for $f \\in L^1(\\R)$ the function $Hf$ need not lie in $L^1$.\nFor example, the indicator function of $[0, 1]$ has $Hf(x) = \\frac{1}{\\pi} \\log \\abs{\\frac{x}{x - 1}}$, which behaves like $\\frac{1}{\\pi x}$ as $\\abs{x} \\to \\infty$ and is not integrable.\n\nExample: for $f(x) = \\frac{1}{1 + x^2}$ one has $Hf(x) = \\frac{x}{1 + x^2}$ and then $H\\big[\\frac{x}{1 + x^2}\\big] = -\\frac{1}{1 + x^2}$.\n\n## Explanations\n\n### L1 human\n\nDoing the Hilbert transform twice gives back the original function with a minus sign.\nSo the transform loses no information, and its inverse is the same operation with a minus sign.\nOn waves, it is a quarter-period shift, and two quarter-period shifts turn a wave upside down.\n\n### L2 human\n\nThe inversion formula says that $H$ is, up to sign, its own inverse; on $L^2$ it is a unitary operator with $H^* = -H$.\nIt explains the symmetry between the two halves of a Hilbert transform pair.\nAn example is the pair of real and imaginary parts of the boundary values of a function in the [[Hardy space|PNNNFJVF6W]] of the upper half-plane.\nTogether with the boundedness on $L^p$ for $1 \u003c p \u003c \\infty$, it is the starting point of the theory of singular integral operators.\n",
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