Convergence in the formulation of the spectral theorem

Question

Let $\mathcal H$ be a complex (separable, if needed) Hilbert space and $B(\mathcal H)$ the ring of bounded operators on $\mathcal H$. I am interested in understanding the formulation of the spectral theorem (for self-adjoint, possibly unbounded operators) in terms of resolutions of the identity.

A function $E:\mathbb R\rightarrow B(\mathcal H)$ is a resolution of the identity if

1. for each $\lambda\in\mathbb R$, $E(\lambda)$ is an orthogonal projection;
2. for each $\lambda_0<\lambda_1$, $E(\lambda_0)\le E(\lambda_1)$;
3. the function $E(\lambda)$ is right-continuous;
4. $\lim_{\lambda\rightarrow-\infty} E(\lambda)=0$ and $\lim_{\lambda\rightarrow\infty}E(\lambda)=I$ where $I$ is the identity operator.

One then defines a self-adjoint (possibly unbounded) operator $A$ through $$ A=\int_{-\infty}^{+\infty}\lambda\, \mathrm dE(\lambda). \qquad(\ast)$$

The spectral theorem then states that corresponding to every densely defined self-adjoint operator on $\mathcal H$ there is a unique resolution of the identity such that $(\ast)$ is true.

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I do have access to a number of references that treat the spectral theorem this way in a reasonably rigorous manner but nonetheless I am a bit confused about the various definitions of convergence that appear in the statements above.

So first, as I understand, if $A_n$ is a sequence of bounded operators on $\mathcal H$, we have the strong convergence $A_{n}\overset{s}{\longrightarrow}A$ if for each $x\in\mathcal H$, $A_nx\longrightarrow Ax$ and the weak convergence $ A_{n}\overset{w}{\longrightarrow}A $ if for each $x,y\in\mathcal H$ we have $ \langle x,A_n y\rangle\rightarrow \langle x,Ay\rangle $ (I am a physicist so for me the inner product is linear in the second variable).

Now I assume that since each operator $A$ is uniquely determined by the sesquilinear form $\langle x,Ax\rangle$ through the polarization identity, it is true that if $\langle x,A_nx\rangle\rightarrow\langle x,Ax\rangle$, then $A_n\overset{w}{\longrightarrow A}$ is that correct?

Then the first question is

• Q1: In property 3. and 4. of the resolution of identity, the right continuity of the family $E(\lambda)$ and the limits $E(\lambda)$ as $\lambda\rightarrow\pm\infty$ are meant in the sense of weak or strong convergence, and why?

I think it is irrelevant, because from what I can tell, for orthogonal projections the weak and strong convergence coincides, but I often find functional analysis so counterintuitive I am not sure I trust my proof. Basically suppose that $P_n$ is a sequence of orthogonal projections converging weakly to $P$, i.e. $\langle x,P_ny\rangle\rightarrow\langle x,Py\rangle$, then we have $$ \Vert P_nx-Px\Vert^2=\Vert P_nx\Vert^2+\Vert Px\Vert^2-2\mathrm{Re}\langle P_nx,Px\rangle, $$ but $\langle x,P_nx\rangle=\langle P_nx,P_nx\rangle=\Vert P_nx\Vert^2$, hence $\Vert P_nx\Vert\rightarrow \Vert Px\Vert$ and thus $$ \lim_{n\rightarrow\infty}\Vert P_nx-Px\Vert^2=2\Vert Px\Vert^2-2\langle Px,Px\rangle=0, $$ thus $P_n\overset{s}{\longrightarrow} P$ as well. Is this correct?

I have more problems with interpreting the integral $(\ast)$ defining $A$. I have often seen it being meant that for any $x\in \mathcal H$, we have $$ \langle x,Ax\rangle=\int_{-\infty}^{+\infty}\lambda\,\mathrm d\langle x,E(\lambda)x\rangle, \qquad(\ast\ast)$$ where the RHS is an ordinary Riemann-Stieltjes integral with respect to the function $P_x(\lambda):=\langle x,E(\lambda)x\rangle$.

Due to the polarization identity, this does determine $A$ uniquely and I guess $x$ belongs to the domain of $A$ if and only if the above Stieltjes integral converges.

• Q2: Is there any direct expression for the action $Ax$? I would intuitively think that the formula for $\langle x,Ax\rangle$ implies that for any $x\in\mathcal H$ (or at least in the domain of $A$) we have $$ Ax=\int_{-\infty}^{+\infty}\lambda\,\mathrm d(E(\lambda)x), \quad(\ast\ast\ast)$$ and as far as I am aware, Stieltjes integrals with values in a Hilbert space do make sense (hence "$\mathrm d(E(\lambda)x)$" can be interpreted), but the fact that weak convergence does not coincide with strong convergence in general makes me think that $(\ast\ast)$ does not imply $(\ast\ast\ast)$. So does $(\ast\ast\ast)$ makes sense and if so when?