Core for an unbounded operator.

2,455

Solution 1

No, it's not true that any dense subset of $D(T)$ is a core for $T$. For $D \subseteq D(T)$ to be a core for $T$, what you want is $\{(x,Tx): x \in D\}$ to be dense in the graph $\{(x,Tx): x \in D(T)\}$ of $T$ (as subsets of ${\mathscr H} \times {\mathscr H}$, with the norm topology).

Equivalently, for any $x \in D(T)$ there exists a sequence $x_n \in D$ such that $\|x_n - x\| + \|T x_n - Tx\| \to 0$ as $n \to \infty$.

Solution 2

Yes, there is a nice check:

A subset $\mathcal{D}\subseteq\mathcal{D}(A)$ is a core for a closed operator $A$ iff the following condition is satisfied: $$(x_0,Ax_0)\perp(x,A\restriction_\mathcal{D}x)\text{ for all }x\in\mathcal{D}\implies x_0=0$$

That condition basically comes from the orthogonal decomposition for closed subspaces: $$\mathcal{G}(A)=\overline{\mathcal{G}(A\restriction_\mathcal{D})}\oplus\mathcal{G}(A\restriction_\mathcal{D})^\perp$$ But one has to be aware that the operator must have been closed so that: $$A\text{ closed}\implies\mathcal{G}(A)\text{ closed}\implies\mathcal{G}(A)\text{ complete}$$

And yes, there is the ready core:

For a closed and densely defined operator a core is given by: $\mathcal{D}(A^*A)$

That one basically comes while using the above from a subtle investigation of: $(1+A^*A)$

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preetinder86
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Updated on December 22, 2020

Comments

  • preetinder86
    preetinder86 almost 2 years

    A symmetric operator $T$ is called essentially self-adjoint if its closure $T$ is self-adjoint. If $T$ is closed, a subset $D \subset D(T)$ is called a core for $T$ if $\overline {T\upharpoonleft D} = T$. So any dense subset of $D(T)$ is a core for $T$.

    I would like to know about if there are any other equivalent conditions for a subset $D \subset D(T)$ to be a core for $T$.

  • preetinder86
    preetinder86 over 8 years
    yeah that's true. I was assuming graph norm.
  • shuhalo
    shuhalo over 7 years
    Do you know a reference for the domain of $A^\ast A$ being a core of $A$?
  • C-star-W-star
    C-star-W-star over 7 years
    @shuhalo: Yes, have a look at Weidmann, Lineare Operatoren, Satz 4.11 a.
  • Tom Collinge
    Tom Collinge over 4 years
    I linked to your answer here math.stackexchange.com/questions/2783500/… (and would appreciate any feedback)