Continuous spectral theorem example

Question

The spectral theorem can be explicitly expressed for an hermitian matrix by providing its eigen decomposition. In the more general case of a bounded self-adjoint operator with a continuous spectrum, the expression of the spectral theorem is $Ax=\int \lambda dF(\lambda) x$. This expression is rather abstract. Can someone provide a simple example of the continuous case, taking $x$ from a space of function, a differential operator A, and the detailed expressions of $F(\lambda)$ and $dF(\lambda)$ ?

Answer 1

Yes, let $X=L^{2}[0,1]$, and let $(Tx)(t)=tx(t)$, i.e., multiplication by $t$. This is the prototypical example in many regards. You don't have any eigenfunctions because $Tx=\lambda x$ would require $(t-\lambda)x(t)=0$ for a.e. $t\in[0,1]$ which means that $x=0$ a.e..

However, you have approximate eigenfunctions. For example, if $\lambda\in(0,1)$ and $\epsilon > 0$ is smaller than the distance of $\lambda$ to $\mathbb{R}\setminus[0,1]$, then $$ \varphi_{\epsilon} = \frac{1}{\sqrt{2\epsilon}}\chi_{[\lambda-\epsilon,\lambda+\epsilon]} $$ is a unit vector in $X$ (here $\chi_{S}$ means characteristic function of set $S$). And, $$ \|(T-\lambda I)\varphi_{\epsilon}\|^{2}=\frac{1}{2\epsilon}\int_{0}^{\epsilon}t^{2}\,dt = \frac{\epsilon^{2}}{6},\\ \|(T-\lambda I)\varphi_{\epsilon}\| = \frac{\epsilon}{\sqrt{6}}. $$ As $\epsilon\downarrow 0$, the function $\varphi_{\epsilon}$ looks more and more like an eigenvector with eigenvalue $\lambda$, but it does not converge to an actual eigenvector. Continuous spectrum looks like that.

In this case, the spectral measure $F$ is $$ F(S)x = \chi_{S}(t)x(t). $$ So, $F(\lambda)x=\chi_{[0,\lambda]}(t)x(t)$ and $F(\lambda+\delta)x-F(\lambda)x=\chi_{(\lambda,\lambda+\delta]}(t)x(t)$, which is related to the discussion of the previous paragraph.