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Let T be a (unbounded) selfadjoint operator in $B(H)$, the bounded operator acting on Hilbert space $H$.

Def: A compression of T is an operator $pTp$, where $p$ is a projection in $B(H)$.

I am thinking the relation between the resolution of identity (spectrum decomposition) of $pTp$ and that of $T$. I guess there should be something like $e_{pTp}(-\infty, t)$ $\le $ $e_T(-\infty, t)$ $\le$ $e_{pTp}(-\infty, t) + (I-p)$.. But I do not know how to prove it. I think someone here might help me out.

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You should define your notations. But let me make a guess: $e_T(-\infty,t)$ is the unitary projector over the subspace $H(T,t)$, invariant under $T$, on which $T\le tI$. And an inequality between projectors means that the range of the first one is included in that of the second. Right ?

If yes to that, then the answer to your question is No. Because there is no reason why $T$ and $pTp$ have an invariant subspace in common. Take the finite dimensional case in dimension $2$, and $$T=\begin{pmatrix} -1 & a \\\\ a & 1 \end{pmatrix},$$ with $a\ne0$. Take $p$ the projection onto the first axis. Finally, take $t=-\frac12$. Then $e_{pTp}(-\infty,t)=p$, while $e_T(-\infty,t)$ is the projection onto a line which is not an axis (the eigenvalues of $T$ are $\pm\sqrt{1+a^2}$). Thus both projectors are not comparable.

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  • $\begingroup$ Denis, Thanks for your answer. You guess is correct. I am sorry I should clarify my notations. If we write $T=\int_R \lambda d(e_{\lambda}) $, then $e_T(-\infty, t)$ means the projections corresponding to unique spectral decomposition of the operator $T$. Similarly, for $e_{pTp}(-\infty, t)$. And $e_{pTp}(-\infty, t)$ $\le$ $e_T(-\infty, t)$ means the range of the first projection is included in the range of the second projection. $\endgroup$
    – Paul Z
    Commented Nov 22, 2010 at 15:22
  • $\begingroup$ But $e_{pTp}(-\infty, t)$ $\le$ $e_T(-\infty, t)$ does not imply that the range of $e_{pTp}(-\infty, t)$ is invariant under $T$. $\endgroup$
    – Paul Z
    Commented Nov 22, 2010 at 15:25
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    $\begingroup$ This is exactly what I mean. In my example, $e_{pTp}(-\infty,t)$ is the eigenspace (a line) of $pTp$ associated with the eigenvalue $-\sqrt{1+a^2}$, whereas $e_{T}(-\infty,t)$ is the eigenspace (a line) of $T$ associated with the eigenvalue $-1$. These lines turn out to be distinct, so there is no inequality between the projectors. $\endgroup$
    – Denis Serre
    Commented Nov 22, 2010 at 15:38

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