Spectral resolution of multiplication operator

Question

Kosaku YOSIDA claims in his book "Functional Analysis" that it is easy to see that the multiplication operator

$Hx(t) = tx(t)$ in $L^2(-\infty,+\infty)$

admits the spectral resolution $H = \int_{-\infty}^{+\infty} \lambda \, \mathrm{d}E(\lambda)$, where the resolution of the identity is defined via

$E(\lambda)x(t) = x(t)$ for $t \leq \lambda$,
$=0$ for $t > \lambda$.

At this point in the book (Chapter XI.6 Normed Rings and Spectral Representation: The Spectral Resolution of a Self-adjoint Operator) the Spectral theorem is NOT yet proven.

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I have problemes with the "easy to see"-part. I think he tries to argue that
$\int_{-\infty}^{+\infty} \lambda^2 \, \mathrm{d} \| E(\lambda) x \|^2 = \|Hx\|^2$
and
$\int_{-\infty}^{+\infty} \, \mathrm{d} \langle E(\lambda) x , y \rangle = \langle Hx, y \rangle$.

But how does this show the claim?

Answer 1

If $\rho$ is a bounded measurable function on $[a,b]$, then $E(t)=\int_{a}^{t}\rho(u)du$ is a function of bounded variation and, for any continuous function $g$, $$ \int_{a}^{b}g(t)dE(t) = \int_{a}^{b}g(t)\frac{dE}{dt}dt = \int_{a}^{b}g(t)\rho(t)dt. $$ In your case, $$ \|E(t)x\|^{2}=\int_{-\infty}^{t}|x(u)|^{2}du $$ The above easily extends to \begin{align} \int_{-\infty}^{\infty}t^{2}d\|E(t)x\|^{2} & =\int_{-\infty}^{\infty}t^{2}\frac{d\|E(t)x\|^{2}}{dt}dt \\ &=\int_{-\infty}^{\infty}t^{2}|x(t)|^{2}dt = \|Hx\|^{2}. \end{align}