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Most densely-defined unbounded linear operators on Hilbert spaces have a very large domain. In fact, for a lot of natural operators the intersection of their domains are still dense.

Let us consider some counterexamples:

  • Let $T : \mathbb Q^n \subset \mathbb R^n \rightarrow \mathbb R^m$ be a densely-defined unbounded operator.

    Let $Q$ be orthogonal.

    Then $T_Q : Q \mathbb Q^n \subset \mathbb R^n \rightarrow \mathbb R^m$, which maps $Qx$ to $Tx$ for $x \in \mathbb Q^n$, is a densely-defined unbounded operator, too.

    Then $\mathbb Q^n \cap Q \mathbb Q^n = \{ 0 \}$ for most choices $Q$.

  • Let $\Omega \subseteq \mathbb R^n$ be a smoothly bounded domain.

    Consider the differential operator $D : C_0^\infty(\Omega) \subset L^2(\Omega) \rightarrow L^2(\Omega)$, which maps compactly-supported smooth functions over $\Omega$ to their total differential.

    Pick an arbitrary weakly differentiable $\psi \in H^1(\Omega)$ that satisfies $\psi > 0$ essentially.

    Then $\psi \cdot C_0^\infty(\Omega)$ is a linear subspace of weakly differentiable functions (partial differentiation), and the operator $D : \psi \cdot C_0^\infty(\Omega) \subset L^2(\Omega) \rightarrow L^2(\Omega)$ is densely defined.

    However, for most $\psi$, we have $C_0^\infty(\Omega) \cap \psi \cdot C_0^\infty(\Omega) = \{ 0 \}$.

These examples are artificial. In both cases, the unbounded are not closed. Taking the closure provides a common intersection for the domains of the operators. In the first case, the closure is even bounded. Most Hilbert spaces of functions have a common subspace of their domains that is dense (e.g., compactly supported smooth functions).

I do not expect that "all closed densely-defined unbounded linear operators have a dense common domain". But is there a theory that provides an abstract point of view on situations where all relevant operators' domains have a large (i.e. dense) intersection?

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    $\begingroup$ I don't understand your first example. $T$ cannot be unbounded as it is an operator on a finite dimensional space. Or am I getting this completely wrong? $\endgroup$
    – user51527
    Jul 18, 2014 at 9:25
  • $\begingroup$ All bounded operators are unbounded operators. en.wikipedia.org/wiki/Unbounded_operator $\endgroup$
    – shuhalo
    Jul 18, 2014 at 9:31

1 Answer 1

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Perhaps not what you're looking for, but you may be interested in the following result. Let $\cal H$ be a separable infinite-dimensional Hilbert space, and $H$ any self-adjoint unbounded linear operator on $\cal H$ with purely discrete spectrum. Consider $T = U H$ where $U$ is a unitary operator. Then the set of $U$ for which ${\cal D}(T) \cap {\cal D}(T^*) = \{0\}$ is a dense $G_\delta$.

See R.B. Israel, ``Some Generic Results in Mathematical Physics'', Markov Processes and Related Fields 10 (2004), 517-521.

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  • $\begingroup$ What topology is being put on the set of all unitary operators? $\endgroup$ Jul 18, 2014 at 19:09
  • $\begingroup$ The paper uses the norm topology, but it seems to me you can use the strong operator topology or the weak operator topology as well. $\endgroup$ Jul 18, 2014 at 19:35

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