# Functional calculus of self-adjoint operator which has no eigenvalues

I've learn the functional calculus of self-adjoint (compact) operator on finite (infinite) dimensional space using spectral decomposition in which they are has eigenvalues (spectrum).

My question is, Can we define the functional calculus on (self-adjoint) operator which has no eigenvalues and/or no compactness assumption?

Any help would be appreciated.

• Yes. But the correct thing to work with is elements of the spectrum of the operator, i.e., if $T$ is a linear operator, then the spectrum of $T$ is the set of $\lambda\in k$ (here $k$ is the base field) such that $(T-\lambda I)$ is not invertible. – freeRmodule Nov 23 '16 at 4:35

## 1 Answer

Yes, a canonical type of example is the multiplication operator $$(Mf)(x)=xf(x),\;\;\; f\in L^2[0,1].$$ This operator has no eigenvalues because $Mf=\lambda f$ implies $$0=\int_{0}^{1}|xf(x)-\lambda f(x)|^2dx=\int_{0}^{1}|x-\lambda|^2|f(x)|^2dx,$$ which forces $(x-\lambda)f(x)=0$ a.e.. and $f(x)=0$ a.e..

For this operator, there is a projection-valued function $$(P(t)f)(x)=\chi_{[0,t]}(x)f(x),$$ with $$P(0)=0,\;\; P(1)=I, \\ P(t)^*=P(t)^2=P(t), \\ P(t)P(t')=P(\min(t,t')) \\ 0 \le P(t') \le P(t) \le I,\;\;\; 0 \le t' \le t.$$ The operator $M$ can be written as $$Mf= \int_{0}^{1}\lambda dP(\lambda)f.$$ If $F$ is a continuous function on $[0,1]$, then the functional calculus operators with discrete eigenvalues has corresponding integral forms: $$F(M)f = \int_{0}^{1}F(\lambda)dP(\lambda)f, \\ \|F(M)f\|^2 = \int_{0}^{1}|F(\lambda)|^2d\|P(\lambda)f\|^2.$$ This type of integral with respect to a spectral measure $P$ is the classical Spectral Theorem.

• Thank you so much for the answer, I enlightened. However I have some questions, $M$ is represented by $$Mf=\int_{0}^{1}\lambda dP(\lambda)f,$$ this means, in general we have to represent such operator by "some" linear functional with respect to "some projection" P to work in functional calculus? and how to guarantee that the "projection-valued function" above does always exist for any operator which has no eigenvalues in general setting? – Locomotive Bangla Nov 23 '16 at 7:35
• The integral is a vector Riemann-Stieljtes integral $\lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{j}\lambda_j\Delta P_jf$. Sums are operators with discrete spectrum of $\lambda_1,\lambda_2,\cdots,\lambda_N$, and these converge to the general operator in the strong operator topology, but not in the uniform operator topology. The integration interval contains the spectrum, which is $[0,1]$ in this case. $P(\lambda)$ is the projection onto the $(-\infty,\lambda]$ part of the spectrum, so that $P(\lambda_n)-P(\lambda_{n-1})$ is the projection onto the $(\lambda_{n-1},\lambda_n]$ part of the spectrum. – DisintegratingByParts Nov 23 '16 at 13:00