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.
 A: 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.
