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Consider a densely defined, self-adjoint operator
$$ H: \mathcal{D} \rightarrow \mathscr{H}. $$ Assume for simplicity that $H$ is nonnegative.

We want to effectively restrict this operator $H$ to a subspace $P\mathscr{H}$. Here $P: \mathscr{H} \rightarrow \mathscr{H}$ is an orthogonal projection. We assume that the subspace $P\mathscr{H} $ to which we want to restrict is not contained in the domain of $H$ (i.e. $P\mathscr{H} \nsubseteq \mathcal{D}$).

If $P$ is finite dimensional, we may define a regularized operator $H_f$ acting on $P\mathscr{H}$ as follows:

Consider a continuous injective function $ f: [0,\infty) \rightarrow [0,\infty) $ satisfying

$$ \lim_{\lambda \rightarrow \infty} f(\lambda) = 0. $$

The operator $f(H): \mathscr{H} \rightarrow \mathscr{H}$ is thus bounded. Hence the operator $$ Pf(H)P: P\mathscr{H} \rightarrow P\mathscr{H} $$

is well-defined. It is not hard to see that it is also non-negative. Thus we may define an effective operator $H_f: P\mathscr{H} \rightarrow P\mathscr{H}$ as $$ H_f = f^{-1}(Pf(H)P). $$

  1. Are there references that discuss these ideas?
  2. Is there a neat characterization of $H_f$. (Even if only for the resolvent case $f(\cdot) = (1 + \cdot)^{-1}$? )
  3. Is anything known in the setting where $P$ is not finite?
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  • $\begingroup$ what is $\mathcal{H}$ as opposed to $\mathscr{H}$? $\endgroup$
    – gmvh
    Commented Dec 6 at 19:14
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    $\begingroup$ It seems like the construction of $H_f$ works just as well if $P$ is not finite ... did I miss something? $\endgroup$
    – Nik Weaver
    Commented Dec 6 at 19:36
  • $\begingroup$ @gmvh: That was a typo; thank you for spotting it! $\endgroup$
    – Qualearn
    Commented Dec 6 at 21:10
  • $\begingroup$ @NikWeaver I think that is true. But generically $H_f$ will then be unbounded and I don't know how to characterize its domain $\mathcal{D}_{H_f} \subseteq P \mathscr{H}$ in a nice way. $\endgroup$
    – Qualearn
    Commented Dec 6 at 21:24

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