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?


For a preexample: Preliminary

  • $\begingroup$ 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

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 $TP = PT$.

  • 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

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