0
$\begingroup$

Problem

Given a Hilbert space $\mathcal{H}$.

Consider normal operators: $$N:\mathcal{D}(N)\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$

Denote for readability: $$\mathcal{D}:=\mathcal{D}(N)=\mathcal{D}(N^*)$$

Regard a closed space: $$\mathcal{S}\leq\mathcal{H}:\quad\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^\perp$$

Suppose invariance: $$N(\mathcal{S}\cap\mathcal{D})\subseteq\mathcal{S}\quad N(\mathcal{S}^\perp\cap\mathcal{D})\subseteq\mathcal{S}^\perp$$

Then it may still happen: $$\mathcal{D}\supsetneq\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$

Does someone have a nonexample?

Reference

For a preexample: Preliminary

$\endgroup$
  • $\begingroup$ Are you asking for a proof of the equivalence, "$TZ....TP$"? $\endgroup$ – Squirtle Mar 12 '15 at 15:54
  • $\begingroup$ @Squirtle: Yep, but let me change my question. (There was a flaw!) $\endgroup$ – C-Star-W-Star Mar 12 '15 at 16:11
-2
$\begingroup$

If $PT = TP$, then for any $x\in Z$ $$ Tx = TP(x) = PT(x) \in Z $$ and so $T(Z) \subset Z$. Now since $$ (I-P)T = T-TP = T-PT = T(I-P) $$ it follows that $T(Z^{\perp}) \subset Z^{\perp}$.

Conversely, if $Z$ is reducing, then for any $x\in H$, write $x = Px + (I-P)x$, then $$ Tx = P(Tx) + (I-P)Tx = T(Px) + T(I-P)x $$ and so $$ P(Tx) - T(Px) = T(I-P)x - (I-P)Tx \in Z\cap Z^{\perp} = \{0\} $$ and so $TP = PT$.

$\endgroup$
  • 1
    $\begingroup$ But the domain is dense only and even worse it may fail to split: $\mathcal{D}\supsetneq\mathcal{D}\cap Z+\mathcal{D}\cap Z^\perp$ $\endgroup$ – C-Star-W-Star Mar 12 '15 at 16:00
  • $\begingroup$ Let me specify my question. Do you mind? $\endgroup$ – C-Star-W-Star Mar 16 '15 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.