Borel functional calculus

Remarks

The map $f \mapsto f(N)$ is linear and multiplicative, sends $\overline{f}$ to $f(N)^*$, and satisfies $\norm{f(N)} \le \sup_{\sigma(N)} \abs{f}$. For $f(\lambda) = \lambda$ it gives $N$, and on continuous functions it agrees with the continuous functional calculus. If $f_n \to f$ pointwise with $\sup_n \sup_{\sigma(N)} \abs{f_n} < \infty$, then $f_n(N) x \to f(N) x$ for every $x \in H$. Example: for a Borel set $B$, the indicator function $1_B$ gives $1_B(N) = E(B)$, the spectral projection of $N$ for $B$. For instance, if $N$ is the multiplication by $t$ on $L^2([0,1])$, then $f(N)$ is the multiplication by $f(t)$. Non-example: let $N$ be the diagonal operator $N e_n = e_n / n$ on $\ell^2$, so $\sigma(N) = \set{1/n : n \ge 1} \cup \set{0}$. The 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. Changing $f$ on a set $B$ with $E(B) = 0$ does not change $f(N)$. For 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.