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  "body": "\n## Example\n\nLet $(\\Omega, \\mu)$ be a $\\sigma$-finite measure space and let $\\varphi \\colon \\Omega \\to \\C$ be an essentially bounded measurable function.\nThe *multiplication operator* $M_\\varphi$ on the [[Lebesgue space|SASP14MA1Z]] $L^2(\\Omega, \\mu)$ is\n$$(M_\\varphi f)(x) = \\varphi(x) f(x).$$\n\nCheck of the main properties:\n\n1. $M_\\varphi$ is bounded with $\\norm{M_\\varphi} = \\norm{\\varphi}_\\infty$, since $\\int \\abs{\\varphi f}^2 \\, d\\mu \\le \\norm{\\varphi}_\\infty^2 \\int \\abs{f}^2 \\, d\\mu$, and $\\sigma$-finiteness gives sets of finite positive measure on which $\\abs{\\varphi}$ is close to $\\norm{\\varphi}_\\infty$.\n2. Its adjoint is $M_{\\bar\\varphi}$, because $\\int \\varphi f \\bar g \\, d\\mu = \\int f \\, \\overline{\\bar\\varphi g} \\, d\\mu$.\n3. It is a normal operator, since $M_\\varphi M_{\\bar\\varphi} = M_{\\abs{\\varphi}^2} = M_{\\bar\\varphi} M_\\varphi$.\n4. Its [[spectrum|BMKATJ802A]] is the essential range of $\\varphi$.\n\nA concrete case: on $L^2[0, 1]$ with Lebesgue measure, take $\\varphi(x) = x$.\nThen $M_\\varphi$ is self-adjoint and positive, its spectrum is $[0, 1]$, and it has no eigenvalues: $x f(x) = \\lambda f(x)$ almost everywhere forces $f = 0$ almost everywhere.\n\n## Remarks\n\nThe spectral theorem for bounded normal operators says that every bounded normal operator on a separable Hilbert space is unitarily equivalent to a multiplication operator of this form.\nSo multiplication operators are, up to unitary equivalence, all the normal operators.\n\n## Explanations\n\n### L1 human\n\nA multiplication operator multiplies every function by one fixed function $\\varphi$.\nIt is the infinite-dimensional version of a diagonal matrix, with the values of $\\varphi$ playing the role of the diagonal entries.\nMultiplying by $x$ on $[0, 1]$ has a whole interval as its spectrum, yet no function is an eigenvector, which shows how the spectrum can differ from the set of eigenvalues.\n",
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