Reports of the Academy of Sciences of the USSR

1. Volume 132, No. 5

MATHEMATICS

V. E. LYANTSE

CONDITIONS FOR THE CLOSEDNESS OF A RESTRICTION OF A SELF-ADJOINT OPERATOR

(Presented by Academician A. N. Kolmogorov on 21 I 1960)

In the present paper a description is given of closed operators with domain dense in a Hilbert space and contained in a given self-adjoint operator. For simplicity of exposition we shall restrict ourselves to the case of a simple spectrum. However, the method of description presented is applicable to closed restrictions of any self-adjoint, or, more generally, normal operator with finite-multiplicity spectrum.

As is known, a self-adjoint operator with simple spectrum is unitarily equivalent to the operator \(\Lambda_\sigma\) of multiplication by the independent variable in a certain space \(L_\sigma^2\), where \(\sigma\) is a nonnegative measure given on the real axis;

\[ L_\sigma^2=\left\{x:\int_{-\infty}^{\infty}|x(\lambda)|^2\,d\sigma(\lambda)<\infty\right\}; \]

\[ \mathfrak{D}_{\Lambda_\sigma} =\left\{x:x\in L_\sigma^2,\ \int_{-\infty}^{\infty}|\lambda x(\lambda)|^2\,d\sigma(\lambda)<\infty\right\}, \qquad (\Lambda_\sigma x)(\lambda)=\lambda x(\lambda). \]

Since closedness and density are invariant under unitary transformations, the problem reduces to the description of operators \(A\), acting from \(L_\sigma^2\) into \(L_\sigma^2\), such that \(\mathfrak{D}_A=L_\sigma^2\), \(A^{**}=A\), and \(A\subset\Lambda_\sigma\).

For stating the results obtained, denote by \(M_\sigma\) the collection of all \(\sigma\)-measurable and \(\sigma\)-almost everywhere finite complex-valued functions defined on the real axis. We have \(L_\sigma^2\subset M_\sigma\). Introduce the operation \({}^{\perp}\), assigning to each subset \(U\subset M_\sigma\) the linear manifold \(U^\perp\subset L_\sigma^2\) by the formula

\[ U^\perp=\left\{v:v\in L_\sigma^2,\ \int u(\lambda)\overline{v(\lambda)}\,d\sigma(\lambda)\ \text{exists and is equal to }0\text{ for all }u\in U\right\}. \]

Theorem 1. For every operator \(A\subset\Lambda_\sigma\), \(\overline{\mathfrak{D}}_A=L_\sigma^2\), \(A^{**}=A\), there exists a linear manifold \(U\subset M_\sigma\) such that

\[ \mathfrak{D}_A=\mathfrak{D}_{\Lambda_\sigma}\cap U^\perp; \tag{1} \]

moreover, \(U\) contains no functions from \(L_\sigma^2\) (except the zero function), i.e.

\[ \int_{-\infty}^{\infty}|u(\lambda)|^2\,d\sigma(\lambda)=\infty,\qquad u\ne0,\quad u\in U, \tag{2} \]

but for each function \(u\in U\)

\[ \int_{-\infty}^{\infty}\frac{|u(\lambda)|^2}{1+\lambda^2}\,d\sigma(\lambda)<\infty. \tag{3} \]

It is unknown to the author whether, for a given \(A\), formula (1) determines the linear manifold \(U\subset M_\sigma\) uniquely. Therefore, it may be that relations (2) and (3) are not necessary. Moreover, it is unknown whether relations (2) and (3) are sufficient in order that the relations \(\overline{\mathfrak D}_A=L_\sigma^2,\ A^{**}=A\) should hold, if \(\mathfrak D_A\) is given by formula (1) and \(A\subset\Lambda_\sigma\). However, under certain sufficiently general assumptions, relations (2) and (3) are necessary and sufficient.

In order to formulate these assumptions, introduce the operation\(^1\) which assigns to a set \(V\subset L_\sigma^2\) the linear manifold \(V^\perp\subset M_\sigma\) by the formula

\[ V^\perp=\left\{u:\ u\in M_\sigma,\ \int u(\lambda)\overline{v(\lambda)}\,d\sigma(\lambda)\ \text{exists and is equal to }0\text{ for all }v\in V\right\}. \]

We shall call a manifold \(U\subset M_\sigma\) \(\perp\)-closed if

\[ (\mathfrak D_{\Lambda_\sigma}\cap U^\perp)^\perp=U, \tag{4} \]

and strongly \(\perp\)-closed if

\[ (\mathfrak D_{\Lambda_\sigma}\cap U_1^\perp)^\perp=U_1 \tag{5} \]

for every linear manifold \(U_1\subset U\).

Lemma 1. Every finite-dimensional linear manifold is strongly \(\perp\)-closed and, a fortiori, \(\perp\)-closed.

Lemma 2. Let a linear manifold \(U\subset M_\sigma\) have the following properties:

\(\alpha)\) if, for some function \(y\), there exists an increasing sequence of \(\sigma\)-measurable sets \(\mathcal E_1,\mathcal E_2,\ldots\) with union \(\bigcup \mathcal E_n\) of full \(\sigma\)-measure, on each of which the function \(y\) is equal to some function from \(U\), then \(y\in U\);

\(\beta)\) there exists an increasing sequence of \(\sigma\)-measurable sets \(\mathcal E'_1,\mathcal E'_2,\ldots\), whose union has full \(\sigma\)-measure and on each of which every function \(u\in U\) is square-integrable with respect to the measure \(\sigma\).

Then the manifold \(U\) is \(\perp\)-closed.*

Example of an infinite-dimensional manifold \(U\), strongly \(\perp\)-closed: the set of (finite) linear combinations of the functions \(u_\alpha(\lambda)=\lambda^\alpha\), \(-1/2\leq \alpha<1/2\); \(\sigma\) is Lebesgue measure.

Theorem 2. Let a linear manifold \(U\subset M_\sigma\) be \(\perp\)-closed. Let \(A\subset\Lambda_\sigma\), \(\mathfrak D_A=\mathfrak D_{\Lambda_\sigma}\cap U^\perp\). If every function \(u\in U,\ u\ne0\), satisfies relation (2), then \(\overline{\mathfrak D}_A=L_\sigma^2\). If, in addition, every function \(u\in U\) satisfies condition (3), then \(A^{**}=A\).

Theorem 3. Let a linear manifold \(U\subset M_\sigma\) be strongly \(\perp\)-closed. Let \(A\subset\Lambda_\sigma\) and \(\mathfrak D_A=\mathfrak D_{\Lambda_\sigma}\cap U^\perp\). If \(\overline{\mathfrak D}_A=L_\sigma^2,\ A^{**}=A\), then every function \(u\in U\) satisfies relations (2) and (3).

* Under the conditions of Lemmas 1 and 2, \((\mathfrak D\cap U^\perp)^\perp=U\), where \(\mathfrak D\) is the domain of definition of any function of the operator \(\Lambda_\sigma\).

Proof of Theorem 1. Let \(\mathfrak D_A=L_\sigma^2\) and \(A\subset \Lambda_\sigma\). Then \(A\) is a symmetric operator: \(A\subset A^*\), \(\mathfrak D_{A^*}=L_\sigma^2\), and therefore \(A^{**}\) exists. It is not hard to verify that

\[ \mathfrak D_{A^{**}}=\mathfrak D_{\Lambda_\sigma}\cap\left[(A^*-\widetilde\Lambda)\mathfrak D_{A^*}\right]^\perp, \tag{1'} \]

where \((\widetilde\Lambda x)(\lambda)=\lambda x(\lambda)\), by definition, for every function \(x\in M_\sigma\). Therefore, if \(A^{**}=A\), then, putting

\[ U=(A^*-\widetilde\Lambda)\mathfrak D_{A^*}, \tag{6} \]

we obtain formula (1): \(\mathfrak D_A=\mathfrak D_{\Lambda_\sigma}\cap U^\perp\).

If at least one \(u\ne0\) from \(U\) belongs to \(L_\sigma^2\), then \(U^\perp\) is not dense in \(L_\sigma^2\). Therefore relation (2) holds. In accordance with (6), for every \(u\in U\) there exists such a \(y\in\mathfrak D_{A^*}\) that \(\widetilde\Lambda y+u=A^*y\). Since \(A^*y\in L_\sigma^2\), it follows that

\[ \int |\lambda y(\lambda)+u(\lambda)|^2\,d\sigma(\lambda)<\infty . \tag{7} \]

From \(y\in L_\sigma^2\) it follows that

\[ \int_{|\lambda|<\varepsilon}|\lambda y(\lambda)|\,\sigma(d\lambda)<\infty,\quad \varepsilon>0 . \tag{8} \]

From (7) and (8) we derive

\[ \int_{|\lambda|<\varepsilon}|u(\lambda)|^2\,d\sigma(\lambda)<\infty . \tag{9} \]

Moreover, from (7) we obtain

\[ \int_{|\lambda|\ge\varepsilon}\left|y(\lambda)+\frac{u(\lambda)}{\lambda}\right|^2\,d\sigma(\lambda)<\infty, \]

whence, since \(y\in L_\sigma^2\),

\[ \int_{|\lambda|\ge\varepsilon}\left|\frac{u(\lambda)}{\lambda}\right|^2\,d\sigma(\lambda)<\infty . \tag{10} \]

Relations (9) and (10), taken together, are equivalent to relation (3).

Proof of Theorem 2. Suppose that \(U\) is \(\perp\)-closed. If some element \(x\in L_\sigma^2\) is orthogonal to all elements of \(\mathfrak D_{\Lambda_\sigma}\cap U^\perp\), then \(x\in(\mathfrak D_{\Lambda_\sigma}\cap U^\perp)^\perp=U\) (see (4)), i.e. \(x\in L_\sigma^2\cap U\). If condition (2) is satisfied, then \(x=0\) and \(\mathfrak D_A=\mathfrak D_{\Lambda_\sigma}\cap U^\perp\) is dense in \(L_\sigma^2\). Let us verify that, under our assumptions:

\(\gamma)\) the domain \(\mathfrak D_{A^*}\) of the operator \(A^*\) consists exactly of those \(y\in L_\sigma^2\) for which there exists such a \(u\in U\) that \(\widetilde\Lambda y+u\in L_\sigma^2\), and in this case \(\widetilde\Lambda y+u=A^*y\).

Let \(x\in\mathfrak D_A,\ y\in\mathfrak D_{A^*}\). Put \(u=(A^*-\widetilde\Lambda)y\). The relation \((Ax,y)=(x,A^*y)\) can be rewritten in the form

\[ \int x(\lambda)\overline{u(\lambda)}\,d\sigma(\lambda)=0 . \]

Hence \(u\in\mathfrak D_A^\perp=(\mathfrak D_{\Lambda_\sigma}\cap U^\perp)^\perp=U\) (see (4)). Conversely, suppose that for some \(y\in L_\sigma^2\), \(u\in U\), we have \(\widetilde\Lambda y+u\in L_\sigma^2\). Since

\[ \int x(\lambda)\overline{u(\lambda)}\,d\sigma(\lambda)=0 \]

for \(x\in\mathfrak D_A\), it follows that \((x,\widetilde\Lambda y+u)=(\widetilde\Lambda x,y)=(Ax,y)\) for all \(x\in\mathfrak D_A\), and assertion \(\gamma)\) is proved.

From assertion γ) it follows that

\[ (A^*-\widetilde{\Lambda})\mathfrak{D}_{A^*}\subset U . \tag{11} \]

Let us verify that from relation (3) the reverse inclusion follows:

\[ U\subset (A^*-\widetilde{\Lambda})\mathfrak{D}_{A^*}. \tag{12} \]

Let \(u\in U\). Put \(y(\lambda)=0\) for \(|\lambda|<\varepsilon\) and \(y(\lambda)=-u(\lambda)/\lambda\) for \(|\lambda|\geqslant\varepsilon\), \(\varepsilon>0\). Relation (3) is equivalent to (9) and (10). From (10) it follows that \(y\in L_\sigma^2\). We have \(\widetilde{\Lambda}y+u=v\), where \(v(\lambda)=0\) for \(|\lambda|\geqslant\varepsilon\) and \(v(\lambda)=u(\lambda)\) for \(|\lambda|<\varepsilon\). By virtue of (9), \(v\in L_\sigma^2\). In accordance with assertion γ), \(y\in\mathfrak{D}_{A^*}\), \(\widetilde{\Lambda}y+\mu=A^*y\), \(u=(A^*-\widetilde{\Lambda})y\in (A^*-\widetilde{\Lambda})\mathfrak{D}_{A^*}\). Inclusion (12) is proved. From (11), (12), and (1) it follows that
\[ \mathfrak{D}_{A^{**}}=\mathfrak{D}_{\Lambda_\sigma}\cap [(A^*-\widetilde{\Lambda})\mathfrak{D}_{A^*}]^\perp =\mathfrak{D}_{\Lambda_\sigma}\cap U^\perp=\mathfrak{D}_A. \]
Consequently, \(A^{**}=A\).

Proof of Theorem 3 is similar to the proof of Theorem 1.

Proof of Lemma 2. Obviously, \(U\subset(\mathfrak{D}_{\Lambda_\sigma}\cap U^\perp)^\top\), and therefore it is sufficient to prove that the inclusion sign can be directed in the opposite direction. Consider an arbitrary element
\[ w\in(\mathfrak{D}_{\Lambda_\sigma}\cap U^\perp)^\top \]
and put
\[ \mathcal{E}_k=\left\{\lambda:\ |\lambda|<k,\ \frac{|w(\lambda)|}{s(\lambda)}<k\right\}\cap \mathcal{E}'_k, \]
where \(\mathcal{E}'_1,\mathcal{E}'_2,\ldots\) is a sequence of sets possessing the property described in condition β), and \(s\) is some almost everywhere positive function from \(L_\sigma^2\): \(s(\lambda)>0\), \(\int [s(\lambda)]^2\,d\sigma(\lambda)<\infty\). Denote by \(\chi_k\) the characteristic function of the set \(\mathcal{E}_k\), and also the operator of multiplication by this function in \(M_\sigma\). Put \(L_\sigma^2(\mathcal{E}_k)=\chi_k L_\sigma^2\): all functions from \(L_\sigma^2(\mathcal{E}_k)\) are equal to zero outside \(\mathcal{E}_k\) and \(L_\sigma^2(\mathcal{E}_k)\subset L_\sigma^2\). Moreover,

\[ \chi_k U\subset L_\sigma^2(\mathcal{E}_k)\subset \mathfrak{D}_{\Lambda_\sigma}, \tag{13} \]

for every function \(u\in U\) is square integrable on each of the sets \(\mathcal{E}_k\) (see condition β)) and \(|\lambda|<k\) on \(\mathcal{E}_k\). Therefore, as is not difficult to verify,

\[ L_\sigma^2(\mathcal{E}_k)\cap U^\perp = L_\sigma^2(\mathcal{E}_k)\ominus \chi_k U . \tag{14} \]

Obviously, \(w\in (L_\sigma^2(\mathcal{E}_k)\cap U^\perp)^\top\), since, by (13), \(L_\sigma^2(\mathcal{E}_k)\cap U^\perp\) is contained in \(\mathfrak{D}_{\Lambda_\sigma}\cap U^\perp\). Consequently, also
\[ \chi_k w\in (L_\sigma^2(\mathcal{E}_k)\cap U^\perp)^\top . \]
But, in accordance with the definition of the set \(\mathcal{E}_k\), \(\chi_k w\in L_\sigma^2(\mathcal{E}_k)\):
\[ \int_{\mathcal{E}_k}|w(\lambda)|^2\,d\sigma(\lambda) \leqslant k^2\int_{\mathcal{E}_k}[s(\lambda)]^2\,d\sigma(\lambda). \]
Therefore, by virtue of (14), \(\chi_k w\in \chi_k U\). This means that on each of the sets \(\mathcal{E}_k\) the function \(w\) is equal to some function from \(U\). Taking into account condition α), we obtain \(w\in U\), since the complement of the set \(\bigcup \mathcal{E}_k\) has \(\sigma\)-measure zero.

Proof of Lemma 1. Every finite-dimensional linear manifold \(U\subset M_\sigma\) satisfies the conditions of Lemma 2.

Lviv Polytechnic Institute

Received
28 XII 1959