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.

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

But how does this show the claim?

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    $\begingroup$ What you have to show (by the usual definition) is $\int \lambda d\langle E(\lambda) x, y\rangle = \langle H x, y\rangle$ for all $x,y$ (as well as $\int d\langle E(\lambda)x, y\rangle = \langle x, y\rangle$). $\endgroup$ – PhoemueX Aug 6 '15 at 16:30
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    $\begingroup$ As far as I know only the first identity is needed according to the definition. Where is the second one needed? $\endgroup$ – R. Swan Aug 6 '15 at 19:10
  • $\begingroup$ This expresses the fact that $E$ is indeed a resolution of the identity. I don't know whether this follows automatically. $\endgroup$ – PhoemueX Aug 6 '15 at 22:05
  • $\begingroup$ I have just "manually" shown this fact. So the first identity should suffice. I wonder why Yosida has also shown that $\int \lambda^2 \, \mathrm{d} \| E(\lambda)x \|^2 = \|Hx\|^2$ holds? $\endgroup$ – R. Swan Aug 6 '15 at 22:07

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}

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    $\begingroup$ @R.Swan : $\int_{-\infty}^{\infty}\lambda d_{\lambda}\int_{-\infty}^{\lambda}x(t)\overline{y(t)}dt = \int_{-\infty}^{\infty}\lambda \frac{d}{d\lambda}\int_{-\infty}^{\lambda}x(t)\overline{y(t)}dt d\lambda = \int_{-\infty}^{\infty} \lambda f(\lambda)\overline{g(\lambda)}d\lambda = (Hf,g)$. $\endgroup$ – DisintegratingByParts Aug 7 '15 at 0:16
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    $\begingroup$ Thank you. I have two questions: (1) What kind of notation ist $\mathrm{d}_\lambda $? I have never seen that. (2) Why do we need two identities? Is it soley to prove that E ist a resolution of the identity? $\endgroup$ – R. Swan Aug 7 '15 at 1:13
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    $\begingroup$ (3) I am also having troubles understanding the formal manipulations of the differentials (i.e. kind of treating them as fractions) $\endgroup$ – R. Swan Aug 7 '15 at 1:22
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    $\begingroup$ @R.Swan : This is a classical work, and I'm using a classical interpretation as a Riemann-Stieltjes integral, $\int_{a}^{b}f(t)d\rho(t)$, though you can elevate this to Lebesgue-Stietljes with no serious issues. When I write $d_{\lambda}$, I am just reminding you that the variable of integration is $\lambda$. If $\rho$ is absolutely continuous and $f$ is continuous, then the Riemann Stieltjes integral of $f$ with respect to $\rho$ is $\int_{a}^{b}f(t)d\rho(t) = \int_{a}^{b}f(t)\rho'(t)dt = \int_{a}^{b}f(t)\frac{d\rho(t)}{dt}dt$. It looks like manipulation of symbols, but it's not. $\endgroup$ – DisintegratingByParts Aug 7 '15 at 3:38
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    $\begingroup$ @R.Swan : $x(t)\overline{y(t)}$ is absolutely integrable and measurable, which means that $\int_{-\infty}^{\lambda}x(t)\overline{y(t)}dt$ is absolutely continuous in $\lambda$ with derivative equal a.e. to $x(\lambda)\overline{y(\lambda)}$. (This is Lebesgue's differentiation theorem.) Note: absolutely continuous means the function is the integral of its derivative (which, of course, requires the derivative to be locally integrable.) $\endgroup$ – DisintegratingByParts Aug 7 '15 at 15:34

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