Densely-Defined Unbounded Operators on Hilbert Spaces

Abstract

This chapter will extend the theory seen so far to unbounded operators that are not necessarily defined on the entire space. In section one we will define, in particular, the standard domain of an operator built by composing operators with non-maximal domains. We will introduce closed and closable operators. Then we shall study adjoint operators to unbounded and densely-defined operators, thus generalising the similar notion for bounded operators defined on the whole Hilbert space.

Von Neumann had just about ended his lecture when a student stood up and in a vaguely abashed tone said he hadn’t understood the final argument. Von Neumann answered:

“Young man, in mathematics you don’t understand things. You just get used to them.”

David Wells

Exercises

5.1

Let B be a closable operator on the Hilbert space ${\mathsf H}$ with dense domain D(B). Prove that $\overline{Ran(B)} = \overline{Ran(\overline{B})}$ and therefore $Ker(B^*)= Ker(\overline{B}^*)$.

Solution. If $y \in Ran(\overline{B})$, there is a sequence of elements $x_n \in D(B)$ such that $x_n \rightarrow x \in {\mathsf H}$ and $Bx_n \rightarrow y$, so $y \in \overline{Ran(B)}$. Hence $Ran(\overline{B}) \subset \overline{Ran(B)}$. Since $B\subset \overline{B}$ we finally conclude that: $Ran(B)\subset Ran(\overline{B})\subset \overline{Ran(B)}$. Taking again the closure: $\overline{Ran(B)}\subset \overline{Ran(\overline{B})}\subset \overline{Ran(B)}$, so that $\overline{Ran(B)} = \overline{Ran(\overline{B})}$. Eventually: $Ker(B^*) = [Ran(B)]^\perp = [\overline{Ran(B)}]^\perp = [\overline{Ran(\overline{B})}]^\perp = [Ran(\overline{B})]^\perp = Ker(\overline{B}^*)$.

5.2

Let A be an operator on the Hilbert space ${\mathsf H}$ with dense domain D(A). Take $\alpha ,\beta \in {\mathbb C}$ and consider the standard domain $D(\alpha A + \beta I):= D(A)$. Prove that

(i) $\alpha A+\beta I: D(\alpha A + \beta I) \rightarrow {\mathsf H}$ admits an adjoint and

$$(\alpha A + \beta I)^*= \overline{\alpha } A^*+ \overline{\beta }I\,;$$

(ii) assuming $\alpha ,\beta \in {\mathbb R}$, $\alpha A+ \beta I$ is (respectively) Hermitian, symmetric, self-adjoint, or essentially self-adjoint $\Leftrightarrow$ A is Hermitian, symmetric, self-adjoint, or essentially self-adjoint;

(iii) $\alpha A+ \beta I$ is closable $\Leftrightarrow$ A is closable; in that case

$$\overline{\alpha A+ \beta I} = \alpha \overline{A} + \beta I\,.$$

Hint. Apply directly the definitions.

5.3

Let A and B be densely-defined operators on the Hilbert space ${\mathsf H}$. If $A+B: D(A)\cap D(B) \rightarrow {\mathsf H}$ is densely defined, prove

$$A^*+B^*\subset (A+B)^*\,.$$

5.4

Let A and B be densely-defined operators on the Hilbert space ${\mathsf H}$. If the standard domain D(AB) is densely defined, show $AB : D(AB) \rightarrow {\mathsf H}$ admits an adjoint and

$$B^*A^*\subset (AB)^*\,.$$

5.5

Let A be a densely-defined operator on the Hilbert space ${\mathsf H}$ and $L: {\mathsf H}\rightarrow {\mathsf H}$ a bounded operator. Using the definition of adjoint prove that

$$(LA)^*= A^*L^*\,.$$

Then show

$$(L + A)^*= L^*+ A^*\,.$$

5.6

Let $A: D(A) \rightarrow {\mathsf H}$ be a symmetric operator on the Hilbert space ${\mathsf H}$. Prove that A bijective $\Rightarrow$ A self-adjoint. (Bear in mind that the inverse to a self-adjoint operator, if it exists, is self-adjoint. This falls out of the spectral theorem for unbounded self-adjoint operators, that we shall see later.)

Solution. If A is symmetric so is $A^{-1}: {\mathsf H}\rightarrow D(A)$. The latter is defined on the whole Hilbert space, so it is self-adjoint. Its inverse will, in turn, be self-adjoint.

5.7

Let $A: D(A) \rightarrow {\mathsf H}$ be a symmetric operator on the Hilbert space ${\mathsf H}$. Prove that the closure $\overline{A}$ is symmetric, using the properties of $^*$ in Theorem 5.18.

Solution. $A \subset A^*$ by hypothesis, then $A^{*}\supset A^{**}= \overline{A}$, and finally $A^{**}\subset \overline{A}^*$, that is $\overline{A}\subset \overline{A}^*$.

5.8

Let $A: D(A) \rightarrow {\mathsf H}$ be a symmetric operator on the Hilbert space ${\mathsf H}$. Prove that the closure $\overline{A}$ is symmetric, using the continuity of the inner product and the definition of closure of an operator in terms of sequences.

Solution. If $f, g \in D(\overline{A})$, we have $D(A)\ni f_n \rightarrow f$ for some sequence such that $Af_n \rightarrow \overline{A}f$ and $D(A)\ni g_n \rightarrow g$ for some sequence such that $Ag_n \rightarrow \overline{A}g$. Then

$$(f|\overline{A}g) = \lim _{m\rightarrow +\infty }(f| Ag_m) = \lim _{m\rightarrow +\infty }\lim _{n\rightarrow +\infty }(f_n| Ag_m) = \lim _{n\rightarrow +\infty }\lim _{m\rightarrow +\infty }(f_n| Ag_m)$$

$$\lim _{n\rightarrow +\infty } \lim _{m\rightarrow +\infty }(Af_n| g_m)= \lim _{n\rightarrow +\infty } (Af_n| g) = (\overline{A}f|g)\,.$$

(The two limits can be swapped, because the inner product is jointly continuous.)

5.9

In the sequel the commutant $\{A\}'$ of an operator A on ${\mathsf H}$ indicates the set of operators B in ${\mathfrak B}({\mathsf H})$ such that $BA \subset AB$. Let $A: D(A) \rightarrow {\mathsf H}$ be an operator on the Hilbert space ${\mathsf H}$. If D(A) is dense and A closed, prove that $\{A\}'\cap \{A^*\}'$ is a strongly closed $^*$-subalgebra in ${\mathfrak B}({\mathsf H})$ with unit.

5.10

Prove Proposition 5.15.

5.11

Discuss whether and where the operator $-d^2/dx^2$ is Hermitian, symmetric, or essentially self-adjoint on the Hilbert space ${\mathsf H}=L^2([0,1], dx)$. Take as domain: (i) periodic maps in $C^\infty ([0,1])$, and then (ii) maps in $C^\infty ([0,1])$ that vanish at the endpoints.

5.12

Prove that

$$H: =-\frac{d^2}{dx^2} + x^2\,$$

is essentially self-adjoint on $L^2({\mathbb R}, dx)$ if $D(H):={\mathscr {S}}({\mathbb R})$.

Hint. Seek a basis of $L^2({\mathbb R}, dx)$ made of eigenvectors of H.

5.13

Consider the Laplace operator on ${\mathbb R}^n$ seen in Example 5.48(1):

$$\varDelta := \sum _{i=1}^n \frac{\partial ^2}{\partial x^2_i}\,.$$

Prove explicitly $\varDelta$ is essentially self-adjoint on the Schwartz space ${\mathscr {S}}({\mathbb R}^n)$ inside $L^2({\mathbb R}^n, dx)$, and as such it admits one self-adjoint extension $\overline{\varDelta }$.

Then show that if $\widehat{{\mathscr {F}}}: L^2({\mathbb R}^n, dx) \rightarrow L^2({\mathbb R}^n, dk)$ is the Fourier–Plancherel transform (Sect. 3.7),

$$\left( \widehat{{\mathscr {F}}} \overline{\varDelta } \widehat{{\mathscr {F}}}^{-1} f\right) (k):= -k^2 f(k)\,,$$

where $k^2 = k_1^2 +k_2^2+\ldots + k_n^2$, on the standard domain:

$$\left\{ f \in L^2({\mathbb R}^n, dk)\,\left| \, \int _{{\mathbb R}^n} k^4|f(k)|^2 dk <+\infty \right. \right\} \,.$$

Hint. The operator $\varDelta$ is symmetric on ${\mathscr {S}}({\mathbb R}^n)$, so we can use Theorem 5.19, verifying condition (b). Since the Schwartz space is invariant under the action of the unitary operator $\widehat{{\mathscr {F}}}$ given by the Fourier–Plancherel transform, as seen in Sect. 3.7, we may consider Theorem 5.19(b) for $\widehat{\varDelta }:= \widehat{{\mathscr {F}}} \varDelta \widehat{{\mathscr {F}}}^{-1}$. This operator is essentially self-adjoint on ${\mathscr {S}}({\mathbb R}^3)$ iff $\varDelta$ is defined on ${\mathscr {S}}({\mathbb R}^3)$. Now, $\widehat{\varDelta }$ acts on ${\mathscr {S}}({\mathbb R}^n)$ by multiplication by $-k^2 = -(k_1^2 +k_2^2+\ldots + k_n^2)$, giving a self-adjoint operator on the aforementioned standard domain. Condition (b) can then be verified easily for $\widehat{\varDelta }^*$, by using the definition of adjoint plus the fact that ${\mathscr {S}}({\mathbb R}^n)\supset {\mathscr {D}}({\mathbb R}^n)$. The uniqueness of self-adjoint extensions for essentially self-adjoint operators proves the last part, because $\widehat{{\mathscr {F}}}$ is unitary.

5.14

Recall ${\mathscr {D}}({\mathbb R}^n)$ denotes the space of smooth complex-valued functions with compact support in ${\mathbb R}^n$. Referring to the previous exercise let $\overline{\varDelta }$ be the unique self-adjoint extension of $\varDelta : {\mathscr {S}}({\mathbb R}^n) \rightarrow L^2({\mathbb R}^n, dx)$. Prove ${\mathscr {D}}({\mathbb R}^n)$ is a core for $\overline{\varDelta }$. In other words show $\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)}$ is essentially self-adjoint and $\overline{\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)}} = \overline{\varDelta }$.

Hint. It suffices to show $(\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*= \overline{\varDelta }$ (because that implies, by taking adjoints, $\overline{\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)}} = ((\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*)^*= \overline{\varDelta }^*= \overline{\varDelta }$). For this identity note that if $\psi \in D((\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*)$ then $(\varDelta \varphi | \psi ) = (\varphi |\psi ')$, with $\psi ' = (\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*\psi \in L^2({\mathbb R}^n, dx)$, for any $\varphi \in {\mathscr {D}}({\mathbb R}^n)$. Applying the Fourier–Plancherel transform immediately gives $\widehat{{\mathscr {F}}}\psi ' = -k^2 \widehat{{\mathscr {F}}}\psi$, since $\widehat{{\mathscr {F}}}({\mathscr {D}}({\mathbb R}^n))$ is dense in $L^2({\mathbb R}^n, dk)$. Therefore we obtained $\psi \in D(\overline{\varDelta })$ and $\psi ' = \overline{\varDelta }\psi$, and so $(\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*\subset \overline{\varDelta }$. Now suppose, conversely, $\psi \in D(\overline{\varDelta })$. Using the Fourier–Plancherel transform gives $-k^2 \widehat{{\mathscr {F}}}\psi \in L^2({\mathbb R}^n, dk)$, and for any $\varphi \in {\mathscr {D}}({\mathbb R}^n)$ we may write $(\varDelta \varphi | \psi ) = -\int dk k^2 \overline{(\widehat{{\mathscr {F}}} \varphi )} \widehat{{\mathscr {F}}} \psi = -\int dk \overline{(\widehat{{\mathscr {F}}} \varphi )} k^2 \widehat{{\mathscr {F}}} \psi = (\varphi | \overline{\varDelta } \psi )$. By definition of adjoint we found $\psi \in D((\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*)$ and $(\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*\psi = \overline{\varDelta }\psi$. Hence we have the other inclusion, $(\varDelta \upharpoonright _{{\mathscr {D}}({\mathbb R}^n)})^*\supset \overline{\varDelta }$.

5.15

Let $A: D(A) \rightarrow {\mathsf H}$ be self-adjoint and T its Cayley transform. Prove that the von Neumann algebra $(\{A\}')'$ generated by A coincides with the von Neumann algebra $(\{T\}')'$ generated by $\{T\}$ (cf. Sect. 3.3.2).

5.16

Prove Proposition 5.45.

5.17

Take a symmetric operator $A: D(A) \rightarrow {\mathsf H}$ on the Hilbert space ${\mathsf H}$ and suppose $\psi \in C^\infty (A)$ is such that the finite linear span of $A^n\psi$, $n\in {\mathbb N}$, is dense in ${\mathsf H}$. Prove that for any chosen $N=0,1,2,\ldots$ and $a_n \in {\mathbb C}$,

$$C: \sum _{n=0}^N a_n A^n \psi \mapsto \sum _{n=0}^N \overline{a_n} A^n \psi $$

determines a conjugation operator $C: {\mathsf H}\rightarrow {\mathsf H}$ (Definition 5.39).

Outline. The first thing to prove is C is well defined as a map, i.e. if $\sum _{n=0}^N a_n A^n \psi = \sum _{n=0}^{N_1} a'_n A^n \psi$ then $\sum _{n=0}^N \overline{a_n} A^n \psi = \sum _{n=0}^{N_1} \overline{a'_n} A^n \psi$. For that it is enough to observe that if $\Psi = \sum _{m=0}^M b_m A^m \psi$, then

$$\left( \psi \left| \sum _{n=0}^N a_n A^n \Psi \right. \right) =\left( \psi \left| \sum _{n=0}^{N_1} a'_n A^n \Psi \right. \right) \text{ so } \text{ that } \left( \left. \sum _{n=0}^N \overline{a_n} A^n \psi \right| \Psi \right) = \left( \left. \sum _{n=0}^{N_1} \overline{a'_n} A^n \psi \right| \Psi \right) \,.$$

Since the vectors $\Psi$ are dense, $\sum _{n=0}^N \overline{a_n} A^n \psi = \sum _{n=0}^{N_1} \overline{a'_n} A^n \psi$, as required. By construction one verifies that if $\Psi$ and $\Psi '$ are as above then $(C\Psi |C\Psi ') = \overline{(\Psi |\Psi ')}$. Since the $\Psi$ are dense in ${\mathsf H}$ and $||C\Psi || = ||\Psi ||$, it is straightforward to see C extends to ${\mathsf H}$ by continuity and antilinearity. The antilinear extension C satisfies $(C\Psi |C\Psi ') = \overline{(\Psi |\Psi ')}$ on ${\mathsf H}$ and is onto, as one obtains by extending the relation $CC\Psi =I\Psi$ by continuity.