# On the definition of spectral integrals in Conways "A course in functional analysis"

I am trying to make sense of the spectral integral defined in Conways "A course in functional analysis" but I cant really settle on how to think about it. He does the following,

He just proved that $<E(\Delta)g,h>=E(\Delta)_{g,h}$ is a measure of bounded variation on the spectra or any $X$ with a sigma algebra if one prefers.

Now, the left hand side of the inequlities at the end seems to contain some kind of Riemann sum(which I cant define a integral for w.r.t a arbitrary measure) rather then simple functions as one would have in the case of a Lebgue integral.

So what integral concept is the sum< $\sum \phi(x_{k})E(\Delta_{k})g,h>$ associated to? Is he inventing his own concept here?

Edit!

It might be possible to consider $\phi(x_{k})$ a step function and do Lebague all the way, I am not sure tho.

• It is possible to consider $\phi(x_k)$ a step function, this is done in Rudin, Functional Analysis, ch 13.2x I believe.
– Neal
Commented Jun 2, 2017 at 18:11
• @Neal so you are saying it a lebegue integral? Commented Jun 2, 2017 at 19:47

The expression $\langle \sum_k \phi(\lambda_k) E(\Delta_k) g,h \rangle$ can be connected to the Lebesgue integral with the measure $E_{g,h}$. Note that you have

$$\langle \sum_k \phi(\lambda_k) E(\Delta_k) g,h \rangle = \int \sum_k \phi(\lambda_k) \chi_{\Delta_k} dE_{g,h}.$$

The function $\sum_k \phi(\lambda_k) \chi_{\Delta_k}$ is a simple function, that approaches $\phi$ as you decrease the $\varepsilon > 0$ mentioned in the statement of the proposition. It can also be proven of course that the above sum converges to the Lebesgue integral $\int \phi dE_{g,h}$ when $\varepsilon \to 0$.

• I talked to a proferssor today, she agreed that it was Lebegue. But I think you have a misstake in the first sentance. should there be a sum? Commented Jun 13, 2017 at 16:02
• @user84647 Ah yes of course, that equality shouldn't have been there at all. Commented Jun 13, 2017 at 16:04

The sum you wrote approximates an actual integral: \begin{align} \langle \sum_k \phi(\lambda_k)E(\Delta\lambda_k)x,y\rangle & = \sum_k\phi(\lambda_k)\langle E(\Delta\lambda_k)x,y\rangle \\ & \approx \int \phi(\lambda)d_{\lambda}\langle E(\lambda)x,y\rangle. \end{align} And the complex integral on the far right is sesquilinear form that can be used to define a unique operator $\int \phi(\lambda)dE(\lambda)''$ as the unique operator satisfying $$\int \phi(\lambda)d_{\lambda}\langle E(\lambda)x,y\rangle = \left\langle \int \phi(\lambda)dE(\lambda)x,y\right\rangle,\;\;\; x,y\in H.$$ Such an operator exists because the left side is a bounded sesquilinear form on the Hilbert space. If $E$ is a spectral measure associated with a selfadjoint operator $A$, then $A=\int \lambda dE(\lambda)$, and $A^n = \int \lambda^n dE(\lambda)$ for $n=1,2,3,\cdots$. So the spectral integral represents a function of the operator $A$. For example $e^{A} = \int e^{\lambda}dE(\lambda)$. The rough idea is that $dE(\lambda)$ is the projection onto the part of the space associated with spectral component $\lambda$. That is, $AdE(\lambda)=\lambda dE(\lambda)$. For a selfadjoint with discrete spectrum, such as a matrix on a finite-dimensional space, the measure $E$ has discrete support that is equal to the set of eigenvalues of $A$, and $E(\{\lambda\})$ is the projection onto the eigenspace associated with eigenvalue $\lambda$ of $A$, i.e., $AE\{\lambda\}=\lambda E\{\lambda\}$ is exact. The general case of a selfadjoint operator is well-approximated by an operator with discrete spectrum, even if it has no actual eigenvalues, and this is how the Riemann-Stieltjes sum comes into play to approximate the general by the discrete case.

• I cant find any defintion of the Riemann integral w.r.t this kind of general measure, you have any reference? Also $\Delta \lambda_{k}$ is confusing me. Commented Jun 2, 2017 at 20:09
• @user84647 The sum $\sum_k \phi(\lambda_k)E(\lambda_{k-1},\lambda_k]f$ converges to a vector as the norm of the partition tends to $0$. This looks just like a Riemann-Stieltjes integral because $E(\alpha,\beta]= F(\beta)-F(\alpha)$ where $F(\beta)=E(-\infty,\beta]$. Commented Jun 2, 2017 at 20:51
• For selfadjont yes, but what about normal? Then we leave the real line. I am not sure of the exact nature of the funtion $F$ either in general. Commented Jun 3, 2017 at 4:34
• @user84647 : For the general case, you follow their prescription stated in the theorem. The partitions are choosen so that $\phi$ does not vary more than $\epsilon$ over an element of partition, and you'll have convergence of the operator integral. Commented Jun 3, 2017 at 14:27
• I cannot find a defintion of Riemann type integral w.r.t a complex valued measure defined on the complex plane of bounded variation. This I only know for lebegue type integrals. If you know a reference let me know! Commented Jun 3, 2017 at 15:21