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I have the following question: 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? Hope this question is not too trivial. Thanks in advance.

beno

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I know how its done for the position operator, but can one generalize it ? –  beno Jan 12 '12 at 17:58
    
does anyone have a suggestion ? –  beno Jan 12 '12 at 18:53
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3 Answers

up vote 5 down vote accepted

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 $\Delta$.

Using functional calculus, we have, for any bounded Borel function $g$ on $\sigma(T)$, 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 $1_{\varphi^{-1}(\Delta)}$.

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+1: Nice concise answer. To make the last sentence explicit, the operator $E(\Delta)$ is multiplication by the function $1_{\varphi^{-1}(\Delta)}$. –  Nate Eldredge Jan 13 '12 at 2:06
    
Totally right, Nate: thanks. I was trying to avoid the M_\varphi notation because the OP didn't use it, but then I really needed it in the last sentence. –  Martin Argerami Jan 13 '12 at 6:00
    
I've edited the answer to include Nate's comment. –  Martin Argerami Jan 13 '12 at 7:38
    
how can you apply the functional calculus to show that $g(T)f = (g \circ \varphi) f$ ? –  beno Jan 13 '12 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. –  Norbert Jan 13 '12 at 18:58
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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 $A\cap[a,b]$.

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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.

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I do not see why $\int_{f} dE = A_{f}$. –  beno Jan 13 '12 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 Jan 13 '12 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 Jan 13 '12 at 8:12
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