Hi
Let HH be a Hilbert space. Let S∈B(H) and let T be a densely defined closed operator such that TS⊂ST. Assume further that T is boundedly invertible and that both S and T are self-adjoint and positive.
Is it true that ST is closed?
Many thanks.
Hichem
(STx,y)=(Tx,Sy)=(x,TSy)=(x,STy) for all x,y in D(ST)
Therefore ST is symmetric and, consequently, ST⊂(ST)*. Further, using that T is boundedly invertible, I think it's possible to prove that ST is closed.
But why we need positivity of S and T?
Yes, if you understand by "T is boundedly invertible", that T is invertible as operator from D(T) to H and, T-1 is bounded. T is densely defined and it results that, if u is in H, then we can converge to u with sequences from D(T), so with sequences of the form T-1(un), where (un)⊂H.
Mohammed Hichem Mortad
Dear Mohammed,
Sorry, but my above assertions are wrong, my ideea to prove the proposed result was wrong! Can you add, please, some connections with this result! Why you think to obtain "ST is closed", under above assumptions?
Yes, ST is closed.
In general if X is a Banach space, S a bounded operator in X, T a closed
operator in X with bounded inverse and TS⊂ST then ST is closed.
The proof follows from these facts:
1) TS is closed.
2) TS=ST.
See the attached file for the details.
I have now read the proof thoroughly...Can we pass from $TS-\lambda T\subseteq ST-\lambda T$ to $T(S-\lambda I)\subseteq (S-\lambda I)T$ without any further condition?
Using your idea by assuming the resolvent set of $T$ is non-empty (which is true when $T$ is e.g. self-adjoint) with an extra condition on $S$ should fix this issue I think...
Mohammed Hichem Mortad
Unfortunately you are right, the passage is not justified. It seems that some other condition is necessary.
- Badges
- Science topic
- Similar topics
- Mathematics