Question: Let $E, F,$ and $G$ be three Banach spaces. Let $A:D(A)\subset E\longrightarrow F$ be a densely defined unbounded operator. Let $T\in\mathcal{L}(F,G)$ and consider the operator $B:D(B)\subset E\longrightarrow G$ defined by $D(B)=D(A)$ and $B=T\circ A$.
- Determine $B^*$.
- Prove (by an example) that B need not be closed even if $A$ is closed.
Solution:
- Ok.
- I found this example searching on Google:
If D(A) is not a closed subspace of $A$ and $T=0$, then $B|_{D(B)}\equiv 0$, but there is still a sequence $u_n\longrightarrow u \notin D(A)$ such that $u_n\in D(A)$ and $Bu_n \longrightarrow 0$.
My doubt: I only did not understand why if $D(A)$ is not a closed subspace of $A$ and $T=0$ implies that $B|_{D(B)}\equiv 0$?
Thanks in Advance.
Best Answer
Let's show a different way why this example works, using directly the definition of a closed operator from Brezis. An operator $A : D(A)\subseteq E\to F$ is closed if the graph $G(A)$ of $A$, defined as $$G(A) := \{(u, Au) : u\in D(A)\}\subset E\times F$$ is closed in (the product topology on) $E\times F$. If $D(A)$ is not a closed subspace of $E$ and $T = 0$, then $$G(B) = \{(u, 0) : u\in D(A)\} = D(A)\times \{0\}\subset E\times G$$ which will not be closed in $E\times G$.