# How exactly does one define the “spectral measure” of an operator?

• I am seeing kind of different definitions of "spectral measure" at different places and its not clear to me as to what is the universal idea. It would be great to get some "standard" definition.

• In my zone of studies now, I am seeing this occur frequently : that given a self-adjoint (possibly unbounded) operator $A$ on a separable Hilbert space and a $\phi \in \cal{H}_{-1}(A)$ then the "spectral measure" associated to possibly this pair $(A,\phi)$ is that unique $\mu$ such that for all $\lambda$ (may be one wants to restrict to $\lambda \in \mathbb{C} \backslash \mathbb{R}$ for some reason) the following holds,

$<(A -\lambda I )^{-1}\phi,\phi > = \int_{\mathbb{R}} \frac{ d\mu(x)}{x - \lambda }$

(0) Does this have some standard name which I can look up in some textbook?

(1) I am not sure what is the innerproduct used in the LHS.

(2) Can someone give an example of how this might actually be calculated?

(3) Can one use $(A-\lambda I)^{-2}$ here to define something analogous?

• For references see: Spectral Measures – C-Star-W-Star Jul 17 '15 at 22:22
• The definition in terms of its resolvent is technically important. (Estimates) – C-Star-W-Star Jul 17 '15 at 22:31

Eigenvalue problems were first seriously considered in separation of variables problems arising out of Partial Differential Equations. The separation parameter was the eignevalue, and general eigenfunction analysis came out of Fourier's method. The matrix methods came out of this method, which is the opposite direction of abstraction that one would naturally expect.

Over time, the eigenvalue problem was generalized to deal with the $\lambda$ for which $(L-\lambda I)$ is not invertible. Techniques of proving the validity of eigenfunction expansions turned out to be related to representing the resolvent operator $\lambda \mapsto (L-\lambda I)^{-1}$ through its singular information. Holomorphic functions are basically determined by their singularities, and $R(\lambda)=(L-\lambda I)^{-1}$ is a holomorphic function of $\lambda$ with values in the operators, and its singularities determine the resolvent completely for "nice" operators such as normal, selfadjoint, and for fairly general matrices.

For selfadjoint and normal operators, the singularities occur where there are eigenvectors and "approximate" eigenvectors. The set of singularities of the resolvent is the spectrum of the operator. For example, consider the first serious spectral problem through the resolvent of the differentiation operator $L = \frac{1}{i}\frac{d}{dt}$ on the domain consisting of periodic absolutely continuous functions $f\in L^{2}[-\pi,\pi]$ for which $f'\in L^{2}[-\pi,\pi]$. For a given $g$ the resolvent $f=R(\lambda)g$, where $R(\lambda)g=(L-\lambda I)^{-1}g$ is the unique solution of $$\frac{1}{i}f'-\lambda f = g\\ f(-\pi)=f(\pi).$$ This is a first order ODE that is easily solved, at least for $\lambda \ne 0,\pm 1,\pm 2,\pm 3,\cdots$, which is the spectrum of $L$. Solving this equation requires only an integrating factor $e^{-i\lambda t}$. The resolent operator is \begin{align} f=R(\lambda)g % & = e^{i\lambda(x+\pi)}f(-\pi)+e^{i\lambda x}\int_{-\pi}^{x}ie^{-i\lambda t}g(t)dt \\ & = e^{i\lambda x}\frac{e^{2\pi i\lambda}}{1-e^{2\pi i\lambda}} \int_{-\pi}^{\pi}ie^{-i\lambda t}g(t)dt+e^{i\lambda x}\int_{-\pi}^{x}ie^{-i\lambda t}g(t)dt \end{align} It's amazing that a first order ODE with a simple periodic condition would be so complex looking. But this is where it gets interesting. Notice that $R(\lambda)g$ has singularities at $\lambda=0,\pm 1,\pm 2,\cdots$, and all of the poles are first order poles (or possibly removable singularities.) And the residue at an integer $n$ is $$\lim_{\lambda\rightarrow n}(\lambda-n)R(\lambda)g = -e^{inx}\frac{1}{2\pi}\int_{-\pi}^{\pi}g(t)e^{-int}dt$$ Notice that the sum of all the residues of $R(\lambda)g$ is the negative of the Fourier series for $g$: $$-\sum_{n=-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}g(t)e^{-int}dt\right] e^{inx}.$$ Apparently Cauchy discovered this formula, and it was the first clue that the resolvent operator played an important role in Fourier analysis. This formula intrigued Cauchy, but he never found a way to leverage it to prove something about the convergence of the Fourier series. Mathematicians a century later did make this work for very general Fourier expansions coming from ODEs of Math-Physics.

The singularities of the resolvent not only characterize the resolvent, but for selfadjoint operators, the "sum" of all of the residues adds up to the identity operator, which is one way the above can be used to show that the Fourier series converges to the original function. Generalized residues turn out to work for arbitrary selfadjoint operators. $$E[a,b]g = \lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\left(\int_{a+i\epsilon}^{b+i\epsilon}-\int_{a-i\epsilon}^{b-i\epsilon}\right)R(\lambda)gd\lambda.$$ The sum of all these residues gives you back $g$: $$\lim_{\begin{array}{c}a\rightarrow -\infty \\ b\rightarrow+\infty\end{array}} E[a,b]g = g.$$ So you get completeness of eigenfunction expansions using this technique of residues. And $E[a,b]f$ is essentially the spectral measure, which is constructively obtained from the resolvent through this generalized residue limit.

One proof of the Spectral Theorem uses a representation theorem proved around 1910 by a Mathematician named Herglotz. The resolvent for a selfadjoint linear operator $A$ satisfies $$\|(A-\lambda I)^{-1}\| \le \frac{1}{|\Im\lambda|} \\ \Im ((A-\lambda I)^{-1}f,f) \ge 0, \\ \lim_{\Im\lambda \uparrow\infty } -\lambda (A-\lambda I)^{-1}f = f \mbox{ in vector norm }.$$

Theorem: Let $\phi(z)$ be a holomorphic function in the upper half plane. Then there exists a finite positive Borel meausre $\mu$ such that $$\phi(z) = \int_{-\infty}^{\infty}\frac{d\mu(t)}{t-z},\;\;\; \Im z > 0,$$ iff there exists a constant $M$ such that $\phi$ satisfies the following for all $z$ in the upper half plane: $$\Im\phi(z) \ge 0 \\ |\phi(z)| \le \frac{M}{\Im z}.$$ For any such $\phi$, the following limits exists $$\lim_{v\uparrow\infty}v\Im\phi(u+iv) = \int_{-\infty}^{\infty}d\mu(t),\\ \lim_{\epsilon\downarrow 0} \frac{1}{\pi}\int_{a}^{b}\Im\phi(u+i\epsilon)du = \frac{1}{2}\mu\{a\}+\mu(a,b)+\frac{1}{2}\mu\{b\}.$$

For the selfadjoint operator, the resolvent $R(\lambda)=(A-\lambda I)^{-1}$ gives rises to functions $\phi$ as described above by defining $$\phi_{x}(\lambda)=\langle(A-\lambda I)^{-1}x,x\rangle = \langle \frac{1}{A-\lambda I}x,x\rangle$$ You then obtain $$\langle \frac{1}{A-\lambda I}x,x\rangle =\int_{-\infty}^{\infty}\frac{d\mu_{x}(t)}{t-\lambda},\;\;\; \lambda\notin\sigma(A).$$ Properties of the resolvent at $\infty$ give you $$\|x\|^{2}=\int_{-\infty}^{\infty}d\mu_{x}(t).$$ That identity is what gives completeness of spectral expansions, just as it does for the ordinary Fourier series using the resolvent given above.

• This is a very nice answer! (Underappreciated, so I'm leaving a comment. =) – petrelharp May 7 '18 at 21:10

Let me leave out technicalities..

Closed Operators

Given a Banach space $E$.

Consider a closed operator: $$T:\mathcal{D}(T)\subseteq E\to E:\quad T=\overline{T}$$

Denote its resolvent: $$R(\lambda):=(\lambda-T)^{-1}\in\mathcal{B}(E)$$

Then one can construct: $$\eta\in\mathcal{H}(\mathbb{C}):\quad\eta(T):=\oint\eta(\lambda)R(\lambda)\mathrm{d}\lambda$$

That is holomorphic calculus.*

*A.k.a. Dunford calculus.

Spectral Measures

Given a Hilbert space $\mathcal{H}$.

Consider an assignement: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$

That are projections: $$E(A)^2=E(A)=E(A)^*\quad(\varphi\in\mathcal{H})$$

As well as 'complete': $$E(\varnothing)=0\quad E(\mathbb{C})=1$$

And countably additive: $$A_k\in\mathcal{B}(\mathbb{C}):\quad E\left(\biguplus_{k=1}^\infty A_k\right)\varphi=\sum_{k=1}^\infty E(A_k)\varphi\quad(\varphi\in\mathcal{H})$$

That are spectral measures.

Normal Operators

Given a Hilbert space $\mathcal{H}$.

Consider a normal operator: $$N:\mathcal{D}N\subseteq\to\mathcal{H}:\quad N^*N=NN^*$$

Then one can construct: $$\eta\in\mathcal{B}(\mathbb{C}):\quad\eta(N):=\int\eta(\lambda)\mathrm{d}E(\lambda)$$

That is Borel calculus.

• Thanks! Any comments about my second point? – user6818 Jul 17 '15 at 23:08
• Oh that is funny: No idea. ^^ (Wait for T.A.E.!) – C-Star-W-Star Jul 17 '15 at 23:10