# Spectral measure of the multiplication operator

Let $$(X,\mathcal B,\mu)$$ be a finite measure space and consider the operator $$T\colon L^2(X,\mu)\to L^2(X,\mu)$$ given by $$Tf(x)=\varphi(x)f(x)$$, where $$\varphi\colon X\to\mathbb R$$ is a bounded measurable function.

Is there any possibility to determine the spectral measure explicitly?

The spectral measure of $T$ is the measure $E(\Delta)=1_{\Delta}(T)$, where $\Delta$ is any Borel set in the spectrum of $T$, and $1_\Delta$ is the characteristic function of $\Delta$.

In your case, $T=M_\varphi$. Using functional calculus, we have, for any bounded Borel function $g$ on $\sigma(T)=\text {ess ran}\,\varphi$, that $g(T)\,f=(g\circ\varphi)\,f$ for any $f\in L^2(X,\mu)$. Then $$E(\Delta)f=1_\Delta(T)\,f=(1_\Delta\circ\varphi)\,f=1_{\varphi^{-1}(\Delta)}\,f.$$

So, for each Borel set $\Delta$, the projection $E(\Delta)$ is the multiplication operator by the function $1_{\varphi^{-1}(\Delta)}$.

• how can you apply the functional calculus to show that $g(T)f = (g \circ \varphi) f$ ?
– beno
Commented Jan 13, 2012 at 8:24
• @beno An easy, but not rigorous explanation is the following. The equality obviously holds for any $g(t)=t^k$. Hence it is holds for any polynomial. Since polynomials are dense in $C([a,b])$ then the equality $g(T)f=(g\circ\varphi)f$ holds for any continuous $g$. Then for any Borel set $A\subset[a,b]$ its characteristic function can be uniformly approximated by continuous, hence this equality holds for any characteristic function $g$. Since linear span of this characteristic functions is dense in the space of bounded Borel functions, then this equality holds for any bounded Borel function. Commented Jan 13, 2012 at 18:58
• What is the exact measure coming from the spectral projection that you have showed? Is it primary measure $\mu$ that we have started with? Commented Jan 30, 2019 at 11:57
• Not sure what you mean by "exact measure". Commented Jan 30, 2019 at 13:39
• @Phibetakappa: on a quick search, you can check here, here, here, here, here. Commented Jun 24, 2021 at 14:21

Put for $M\in\mathcal B$, where $\mathcal B$ is the $\sigma$-algebra on $X$, $E(M)=A_{\mathbf 1_M}$, where, for $g\in L^{\infty}(\mu)$, $A_g\colon L^2\to L^2$ $A_g(f)=fg$. $E$ is a spectral measure, since

• $E(M)$ is a projection for all $M\in\mathcal B$: $$E(M)(E(M)g)(f)=E(M)(\mathbf 1_Mf)=\mathbf 1_M\cdot \mathbf 1_M \cdot f=E(M)f;$$
• $E(\emptyset)=0,$E(X)=Id$; • If$M$and$N$are disjoint then$E(M)$and$E(N)$are orthogonal, since $$E(M)(E(N)f)=\mathbf 1_M\mathbf 1_N f=0=E(N)(E(M)f).$$ • If$\{A_n\}\subset \mathcal B$are disjoint then $$E\left(\bigcup_{n\in\mathbb N}A_n\right)(f)=\mathbf 1_{\bigcup_nA_n}f=\sum_{n=0}^{+\infty}\mathbf 1_{A_n}f=\sum_{n\in\mathbb N}E(A_n)(f).$$ Now check that$\int_f dE=A_f$for all$f\in L^{\infty}$.$E$is called the standard spectral measure. • I do not see why$\int_{f} dE = A_{f}$. – beno Commented Jan 13, 2012 at 7:37 • I tried to show that$<f,\int_{\sigma(A_{g})} dE f> = <f,A_{g}f>$. But somehow it did not work. I came to the expression – beno Commented Jan 13, 2012 at 8:05 • I tried to show that$<f,\int_{\sigma(A_{g})} dE f> = <f,A_{g}f>$. But somehow it did not work. I came to the expression$<f,A_{g}f> = \int_{X} |f(x)|^{2} g(x) d \mu$and$<f,E(M)f> = \int_{g^{-1}(M)} |f(x)|^{2} d \mu$and from this I could deduce that$d<f,\int \lambda dE f> = \int_{\sigma(A_{g})} \lambda d<f, E(\lambda)f>$and from this I could deduce that$d<f,E(\lambda)f> = |f(x)|^{2} \chi_{g^{-1}(\lambda)}(x) d \mu(\lambda)$. How to go further ? – beno Commented Jan 13, 2012 at 8:12 Consider the space$B([a,b])$of bounded Borel functions. We will endow it locally convex topology. Let$M([a,b])$be a Banach space of complex-valued Borel measures on$[a,b]$. For each$\mu\in M([a,b])$we define a semi-norm $$\Vert\cdot\Vert_\mu:B([a,b])\to\mathbb{R}_+: f\mapsto \int\limits_{[a,b]}f(t)d\mu(t).$$ The family$\{\Vert\cdot\Vert_\mu:\mu\in M([a,b])\}$gives rise to some Hausdorff locally convex topology on$B([a,b])$. We will denote this space$(B([a,b]),wm)$.Consider$\mathcal{B}(H)$with weak operator topology and denote this space$(\mathcal{B}(H),wo)$. A continuous$*$-homomorphism$\gamma_{b,T}:(B([a,b]),wm)\to(\mathcal{B}(H),wo)$such that$\gamma_{b,T}(id_{[a,b]})=T$is called Borel functional calculus of operator$T$. Theorem There exist unique Borel functional calculus$\gamma_{b,T}:(B([a,b]),wm)\to(\mathcal{B}(H),wo)$. Moreover this is contractive involutive homomorphism of involutive algebras which extends continuous calculus on$[a,b]$. Denote$H=L^2(X,\mu)$. Since$\varphi\colon X\to\mathbb{R}$is a bounded measurable function then$T$is a bounded selfadjoint operator. Hence$\sigma(T)\subset\mathbb{R}$. Since$T$is selfadjoint then there exist some interval$[a,b]$such that$\sigma(T)\subset[a,b]$. Now let$\gamma_{b,T}$be the Borel functional calculus on$[a,b]$then we can define spectral measure by the following procedure. For each Borel set$A\subset [a,b]$we define its spectral measure$E(A)$by equality $$E(A)=\gamma_{b,T}(1_{A\cap[a,b]}).$$ where$1_{A\cap[a,b]}$is a characteristic function of$A\cap[a,b]\$.