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  "body": "\n## Definition\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]], and let $N$ be a bounded normal operator on $H$ with [[spectrum|BMKATJ802A]] $\\sigma(N)$.\nLet $E$ be the [[projection-valued measure|ATGT25E5FF]] of $N$ given by the [[spectral theorem for bounded normal operators|X71XRPJR7D]], and write $E_{x,y}(B) = \\ip{E(B) x}{y}$.\nFor a bounded [[Borel|0WQXY32Q8D]] measurable function $f \\colon \\sigma(N) \\to \\C$, let $f(N)$ be the unique bounded operator on $H$ with\n$$\\ip{f(N) x}{y} = \\int_{\\sigma(N)} f(\\lambda) \\, d E_{x,y}(\\lambda) \\quad \\text{for all } x, y \\in H.$$\nThe map $f \\mapsto f(N)$, from the bounded Borel functions on $\\sigma(N)$ to the bounded operators on $H$, is the *Borel functional calculus* of $N$.\n\n## Remarks\n\nThe map $f \\mapsto f(N)$ is linear and multiplicative, sends $\\overline{f}$ to $f(N)^*$, and satisfies $\\norm{f(N)} \\le \\sup_{\\sigma(N)} \\abs{f}$.\nFor $f(\\lambda) = \\lambda$ it gives $N$, and on continuous functions it agrees with the [[continuous functional calculus|JB7S6WPZ77]].\nIf $f_n \\to f$ pointwise with $\\sup_n \\sup_{\\sigma(N)} \\abs{f_n} \u003c \\infty$, then $f_n(N) x \\to f(N) x$ for every $x \\in H$.\nExample: for a Borel set $B$, the indicator function $1_B$ gives $1_B(N) = E(B)$, the spectral projection of $N$ for $B$.\nFor instance, if $N$ is the multiplication by $t$ on $L^2([0,1])$, then $f(N)$ is the multiplication by $f(t)$.\nNon-example: let $N$ be the diagonal operator $N e_n = e_n / n$ on $\\ell^2$, so $\\sigma(N) = \\set{1/n : n \\ge 1} \\cup \\set{0}$.\nThe function $f(\\lambda) = 1/\\lambda$ for $\\lambda \\ne 0$, $f(0) = 0$, is Borel but not bounded, and the formula would give $f(N) e_n = n e_n$, which is not a bounded operator.\nChanging $f$ on a set $B$ with $E(B) = 0$ does not change $f(N)$.\nFor unbounded self-adjoint operators, the same formula defines $f(A)$ for every bounded Borel function on $\\R$, and for unbounded $f$ on a suitable dense domain.\n",
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