So I am not very well-versed in functional analysis, but while studying a problem from theoretical solid-state physics I came across the following question:
Suppose $\mathcal{H}$ is a complex Hilbert-space and you have an unbounded densely defined self-adjoint operator $H \colon D \subset \mathcal{H} \to \mathcal{H}$ which spectrally decomposes as $$ \begin{align} H = \int_{\lambda \in \sigma(H)} \lambda dP_\lambda, \end{align} $$ where $P_\lambda$ is the projection operator on the eigenvalue $\lambda \in \sigma(H)$ in the spectrum of $H$. For simplicity, assume the spectrum is bounded below, but "includes" $+ \infty$. Now suppose that $P_\lambda \overset{S.O.T}{\to} \tilde{P}$ in the strong operator topology for $\lambda \to + \infty$. Let now $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous (or maybe just Borel-measurable) strictly monotonically increasing function with $\lim_{x \to + \infty} f(x) = 1$. By functional calculus, one could build $$ \begin{align} f(H) = \int_{\lambda \in \sigma(H)} f(\lambda) dP_\lambda. \end{align} $$ Through the S.O.T convergence, the spectrum of this operator should include $1$ and this operator should also be bounded and as such could be extended from $D$ to the whole of $\mathcal{H}$.
Either I am doing something terribly wrong (which could be the case), or it seems that I reduced the operator $H$ to a bounded operator $f(H)$. I am wondering what information gets lost, since I can throw the second operator on any vector of $\mathcal{H}$ and would get a new well-defined vector.
Is there some literature on this? Am I performing certain things which are not allowed?
I would be thankful for any further indication.