UDC 513.881

MATHEMATICS

M. A. GOLDMAN, S. N. KRACHKOVSKII

ON PRODUCTS, POWERS, AND CONTRACTIONS OF HOMOMORPHISMS

(Presented by Academician V. I. Smirnov on 8 XII 1967)

A linear operator \(A\) with domain of definition \(\mathfrak D(A)\) in a linear topological space (l.t.s.) \(X\) and range \(\mathfrak R(A)\) in an l.t.s. \(Y\) is called a homomorphism if the image of every set open relative to \(\mathfrak D(A)\) is open relative to \(\mathfrak R(A)\). Numerous works have been devoted to the study of homomorphisms, especially in connection with the theory of linear closed operators. Already S. Banach established that if \(X\) and \(Y\) are spaces of type \(F\), then a linear continuous operator \(A\) is a homomorphism if and only if \(\mathfrak R(A)\) is closed in \(Y\). This result was subsequently extended to a broader class of operators and to spaces of a more general type. For example, in the paper of F. E. Browder \((^1)\), among a large number of different results on linear operators, conditions are established under which a closed operator in certain locally convex spaces turns out to be a homomorphism.

One of the results of the present paper is a theorem stating that in arbitrary l.t.s. a homomorphism with closed null set and closed range is a closed operator (Theorem 1). In addition, certain general properties of homomorphisms are studied (Theorems 2 and 3 on the product of homomorphisms and Theorem 4 on the contraction of a homomorphism). Theorems 1–4 (independent of one another) then make it possible to formulate conditions under which the product of closed homomorphisms is a closed homomorphism (Theorem 5), and also to prove that the contractions of a homomorphism \(A\) to the sets \(\mathfrak D(A^n)\) \((n=2,3,\ldots)\), \(\mathfrak L(A)=\bigcap_{n=1}^{\infty}\mathfrak D(A^n)\), and \(\mathfrak D(A)\cap\mathfrak M(A)\), where \(\mathfrak M(A)=\bigcap_{n=1}^{\infty}\mathfrak R(A^n)\), are again homomorphisms (Corollary of Theorem 4 and Theorem 6). In Lemmas 1, 2 and Theorem 7, certain properties of the sets \(\mathfrak L(A)\) and \(\mathfrak M(A)\) are considered (for an operator in an arbitrary set in Lemmas 1, 2, and for a closed linear operator in a Banach space in Theorem 7).

Theorem 1. Let \(X\) and \(Y\) be l.t.s., \(A\) a homomorphism from \(X\) into \(Y\), and \(\mathfrak Z(A)\) the kernel of the homomorphism \(A\). If the sets \(\mathfrak Z(A)\) and \(\mathfrak R(A)\) are closed, then \(A\) is a closed operator.

Proof. It is required to prove that the graph \(G(A)\) of the operator \(A\) is closed in \(X\times Y\). Let \((x,y)\in \overline{G(A)}\); then \(y\in\mathfrak R(A)\) by virtue of the closedness of \(\mathfrak R(A)\). Let \(z\) be some preimage of the point \(y\). Take in \(\mathfrak D(A)\) an arbitrary neighborhood \(u(x)\) of the point \(x\); shifting it by \(z-x\), we obtain a neighborhood \(u(z)\) of the point \(z\). Since \(A\) is a homomorphism, the set \(v_u(y)=A(u(z))\) is a neighborhood of the point \(y\) in \(\mathfrak R(A)\). Since \((x,y)\in \overline{G(A)}\), in \(u(x)\) there is a point \(x_u\) such that \(Ax_u=y_u\in v_u(y)\). Let \(z_u\) be some preimage in \(u(z)\) of the point \(y_u\). Consider the generalized sequences \(\{x_u\}\) and \(\{z_u\}\), where \(u\) runs through the partially ordered by inclusion set of all neighborhoods of the point \(x\). Then \(x_u\to x\), \(z_u\to z\) and, consequently, \(x_u-z_u\to x-z\). But \(A(x_u-z_u)=\theta\), i.e. \(x_u-z_u\in\mathfrak Z(A)\),

whence, by virtue of the closedness of \(\mathfrak Z(A)\), we conclude that \(x-z \in \mathfrak Z(A)\). Since \(z \in \mathfrak D(A)\), it follows that \(x \in \mathfrak D(A)\) and \(Ax=Az=y\). Hence \((x,y)\in G(A)\) and, thus, \(G(A)=\overline{G(A)}\).

Theorem 2. If \(X,Y,Z\) are l.t.s., \(A\) is a homomorphism from \(X\) into \(Y\), and \(B\) is a homomorphism from \(Y\) into \(Z\), then \(BA\) is a homomorphism from \(X\) into \(Z\).

Proof. Let \(M\) be an open set in \(\mathfrak D(BA)\), i.e. \(M=u\cap \mathfrak D(BA)\), where \(u\) is open in \(X\). It is required to prove that \(BA(M)\) is open in \(\mathfrak R(BA)\). For this we note that
\[ A(u\cap \mathfrak D(BA))=A(u\cap \mathfrak D(A)\cap \mathfrak D(B)); \]
indeed, the inclusion
\[ A(u\cap \mathfrak D(BA))\subseteq A(u\cap \mathfrak D(A))\cap \mathfrak D(B) \]
is obvious; conversely, let
\[ y\in A(u\cap \mathfrak D(A))\cap \mathfrak D(B), \]
i.e. \(y=Ax\), where \(x\in u\cap \mathfrak D(A)\), and \(y\in \mathfrak D(B)\); then \(x\in u\cap \mathfrak D(BA)\) and \(y\in A(u\cap \mathfrak D(BA))\). Hence
\[ BA(M)=B(A(M))=B\bigl(A(u\cap \mathfrak D(BA))\bigr) =B\bigl(A(u\cap \mathfrak D(A))\cap \mathfrak D(B)\bigr). \]
But by hypothesis
\[ A(u\cap \mathfrak D(A))=v\cap \mathfrak R(A), \]
where \(v\) is open in \(Y\); consequently
\[ BA(M)=B(v\cap \mathfrak R(A)\cap \mathfrak D(B)). \]
Next we observe that
\[ B(v\cap \mathfrak D(B))\cap \mathfrak R(BA) \]
(the inclusion \(\subseteq\) is obvious); conversely, let
\[ z\in B(v\cap \mathfrak D(B))\cap \mathfrak R(BA), \]
i.e. \(z=By\), \(y\in v\cap \mathfrak D(B)\), and \(z\in \mathfrak R(BA)\); then
\[ y\in v\cap \mathfrak R(A)\cap \mathfrak D(B) \]
and
\[ z\in B(v\cap \mathfrak R(A)\cap \mathfrak D(B)), \]
and since, by hypothesis,
\[ B(v\cap \mathfrak D(B))=w\cap \mathfrak R(B), \]
where \(w\) is open in \(Z\), we have
\[ BA(M)=w\cap \mathfrak R(B)\cap \mathfrak R(BA)=w\cap \mathfrak R(BA). \]

Corollary. If \(A\) is a homomorphism in the l.t.s. \(X\), then \(A^n\) \((n=2,3,\ldots)\) are also homomorphisms in \(X\).

Theorem 3. Let \(A\) be a homomorphism from the l.t.s. \(X\) into the l.t.s. \(Y\), and \(B\) a homomorphism from \(Y\) into the l.t.s. \(Z\). If \(\mathfrak R(A)\) and \(\mathfrak R(B)\) are closed respectively in \(Y\) and in \(Z\), and
\[ \gamma(B,A)\overset{\mathrm{def}}{=}\dim\bigl(\mathfrak Z(B)\ominus \mathfrak Z(B)\cap \mathfrak R(A)\bigr)<\infty, \]
then \(\mathfrak R(BA)\) is closed in \(Z\) (the symbol \(\ominus\) serves to denote the algebraic complement).

The proof is based on Lemma 2 of \((^2)\), according to which every linear set containing all zeros of a homomorphism and closed relative to its domain is mapped by this homomorphism onto a set closed relative to its range. Since
\[ \mathfrak R(BA)=B(\mathfrak D(B)\cap \mathfrak R(A)) =B\bigl(\mathfrak D(B)\cap(\mathfrak Z(B)+\mathfrak R(A))\bigr), \]
the closedness of \(\mathfrak R(BA)\) follows, according to the lemma just mentioned, from the closedness of the set \(\mathfrak Z(B)+\mathfrak R(A)\), which follows from the closedness of \(\mathfrak R(A)\) and the condition \(\gamma(B,A)<\infty\) \(\bigl((^3),\) p. 50, Corollary 4\(\bigr)\).

Corollary. If \(A\) is a homomorphism in the l.t.s. \(X\),
\[ \gamma(A)\overset{\mathrm{def}}{=}\dim\bigl(\mathfrak Z(A)\ominus \mathfrak Z(A)\cap \mathfrak R(A)\bigr)<\infty \]
and \(\mathfrak R(A)\) is closed, then \(\mathfrak R(A^n)\) \((n=2,3)\) is also closed.

Indeed, since
\[ \gamma(A,A^n)=\dim\bigl(\mathfrak Z(A)\ominus \mathfrak Z(A)\cap \mathfrak R(A^n)\bigr) \le \dim\bigl(\mathfrak Z(A)\ominus \mathfrak Z(A)\cap \mathfrak R(A)\bigr) =\gamma(A)<\infty, \]
\(\mathfrak R(A^{n+1})\) is closed, provided \(\mathfrak R(A^n)\) is closed.

Theorem 4. Let \(X\) and \(Y\) be l.t.s., and let \(A\) be a homomorphism from \(X\) into \(Y\). If \(E\) is a linear set in \(X\) and \(\mathfrak Z(A)\subseteq E\subseteq \mathfrak D(A)\), then the restriction of \(A\) to \(E\) is a homomorphism from \(X\) into \(Y\).

Proof. Let \(u\) be an open set in \(X\). The equality
\[ A(u\cap E)=A(u\cap \mathfrak D(A))\cap A(E) \]
holds. Indeed, the inclusion
\[ A(u\cap E)\subseteq A(u\cap \mathfrak D(A))\cap A(E) \]
is obvious. Conversely, let
\[ y\in A(u\cap \mathfrak D(A))\cap A(E), \]
i.e. \(y=Ax\), where \(x\in u\cap \mathfrak D(A)\), and \(y=Az\), where \(z\in E\); then \(z-x\in \mathfrak Z(A)\), and, since \(z\in E\) and \(\mathfrak Z(A)\subseteq E\), we have \(x\in E\); consequently,
\[ y\in A(u\cap E), \]
and hence
\[ A(u\cap E)\supseteq A(u\cap \mathfrak D(A))\cap A(E). \]
According to the hypothesis,
\[ A(u\cap \mathfrak D(A))=v\cap \mathfrak R(A), \]
where \(v\) is open in \(Y\). But \(A(E)\subseteq \mathfrak R(A)\), whence
\[ A(u\cap E)=v\cap A(E). \]

Corollary. If \(A\) is a homomorphism in the l.t.s. \(X\), then the restrictions of \(A\) to \(\mathfrak D(A^n)\) \((n=2,3,\ldots)\) and to \(\mathfrak L(A)\), and also the restrictions of \(A^m\) to \(\mathfrak D(A^n)\) \((m<n)\) and to \(\mathfrak L(A)\), are homomorphisms in \(X\).

Theorem 5. Let \(A\) be a homomorphism from a separated l.t.s. \(X\) into the l.t.s. \(Y\), and \(B\) a homomorphism from \(Y\) into the l.t.s. \(Z\). If \(\mathfrak R(A)\) and \(\mathfrak R(B)\) are closed respectively in \(Y\) and \(Z\), \(\alpha(A)<\infty\) and \(\alpha(B)<\infty\) (where \(\alpha(\cdot)=\dim\mathfrak Z(\cdot)\)), then \(BA\) is a closed operator.

The proof reduces to verifying the hypotheses of Theorem 1. By Theorem 2, \(BA\) is a homomorphism, and by Theorem 3, \(\mathfrak R(BA)\) is closed in \(Z\); moreover, \(\mathfrak Z(BA)\) is closed, since \(\alpha(BA)\leq \alpha(A)+\alpha(B)<\infty\) ((3), p. 49, Corollary 1).

Corollary. If \(A\) is a homomorphism in a separated l.t.s. \(X\), \(\mathfrak R(A)\) is closed and \(\alpha(A)<\infty\), then the operators \(A^n\) \((n=1,2,\ldots)\) are closed.

The theorem proved does not extend to the case when \(\alpha(A)=\infty\), as can be seen from the following example. Let \(H\) be an infinite-dimensional Hilbert space, \(T\) a one-to-one linear discontinuous operator defined on \(H\), with nonclosed range \(\mathfrak R(T)\) in \(H\). Then \(T^{-1}\) will be a homomorphism in \(H\). Consider the linear operator \(A\), defined on the nonclosed set \(\mathfrak R(T)\times H\) in the Hilbert space \(H\times H\) by the equalities \(A(x,\theta)=(T^{-1}x,\theta)\) \((x\in \mathfrak R(T))\), \(A(\theta,y)=(\theta,\theta)\) \((y\in H)\). The operator \(A\) is a homomorphism in \(H\times H\); by Theorem 1, \(A\) is a closed operator. However, \(A^2\) is not a closed operator, since it is a bounded operator with nonclosed domain of definition \((A^2=0,\ \mathfrak D(A^2)=\mathfrak R(T)\times H)\).

Theorem 6. Let \(X\) be a locally convex space, \(A\) a homomorphism in \(X\), \(\mathfrak R(A)\) closed, and \(\gamma(A)<\infty\). Then the restriction of \(A\) to \(\mathfrak D(A)\cap \mathfrak M(A)\) is a homomorphism in \(X\).

Proof. By the corollary to Theorem 3, the set \(\mathfrak M(A)\) is closed in \(X\), and according to (4) (Theorem 1)
\[ A(\mathfrak D(A)\cap \mathfrak M(A))=\mathfrak M(A). \]
Let
\[ \mathfrak F(A)=\mathfrak M(A)\oplus(\mathfrak Z(A)\dotplus \mathfrak D(A)\cap \mathfrak M(A)). \]
Since \(\mathfrak Z(A)=\mathfrak D(A)\cap \mathfrak F(A)\), by Theorem 4 the restriction of \(A\) to \(\mathfrak D(A)\cap \mathfrak F(A)\) is a homomorphism in \(X\). Denote by \(P\) the continuous projection operator of \(\mathfrak F(A)\) onto \(\mathfrak M(A)\) (such an operator exists, since \(X\) is locally convex and \(\gamma(A)<\infty\)). Then, for any \(u\) open in \(X\), the set \(P^{-1}(u\cap \mathfrak M(A))\) is open in \(\mathfrak F(A)\), i.e. \(P^{-1}(u\cap \mathfrak M(A))=v\cap \mathfrak F(A)\), where \(v\) is open in \(X\). But
\[ A(u\cap \mathfrak D(A)\cap \mathfrak M(A)) = A(\mathfrak D(A)\cap P^{-1}(u\cap \mathfrak M(A))) = A(v\cap \mathfrak D(A)\cap \mathfrak F(A)), \]
and since the restriction of \(A\) to \(\mathfrak D(A)\cap \mathfrak F(A)\) is a homomorphism, \(A(u\cap \mathfrak D(A)\cap \mathfrak M(A))\) is open in \(A(\mathfrak D(A)\cap \mathfrak M(A))\).

Corollary. Under the conditions of Theorem 6, on the basis of Theorem 4 we conclude that the restriction of \(A\) to \(\mathfrak L(A)\cap \mathfrak M(A)\) is a homomorphism in \(X\).

In the following two theorems the \(A\)-invariance of the sets \(\mathfrak L(A)\) and \(\mathfrak L(A)\cap \mathfrak M(A)\) is proved, which makes it possible to consider powers of the restrictions of the operator \(A\) to these sets.

Lemma 1. Let \(X\) be an arbitrary set, \(A\) an operator acting in \(X\), and
\[ \mathfrak L(A)=\bigcap_{n=1}^{\infty}\mathfrak D(A^n). \]
Then the inclusions \(A(\mathfrak L(A))\subseteq \mathfrak L(A)\) and \(A^{-1}(\mathfrak L(A))\subseteq \mathfrak L(A)\) hold, where \(A^{-1}(\mathfrak L(A))\) is the full preimage of the set \(\mathfrak L(A)\).

Proof. Let \(x\in \mathfrak L(A)\); then \(x\in \mathfrak D(A^{n+1})=A^{-1}(\mathfrak D(A^n))\) and \(Ax\in \mathfrak D(A^n)\), whence, in view of the arbitrariness of \(n\), \(Ax\in \mathfrak L(A)\), so that \(A(\mathfrak L(A))\subseteq \mathfrak L(A)\). Let \(x\in A^{-1}(\mathfrak L(A))\); then \(x\in A^{-1}(\mathfrak D(A^{n-1}))=\mathfrak D(A^n)\); since \(n\) is arbitrary, \(x\in \mathfrak L(A)\), whence
\[ A^{-1}(\mathfrak L(A))\subseteq \mathfrak L(A). \]

It is obvious that the inclusions proved in this lemma imply the equality
\[ A^{-1}(\mathfrak L(A))=\mathfrak L(A). \]

Lemma 2. Let \(X\), \(A\), \(\mathfrak L(A)\) have the same meaning as in Lemma 1, and
\[ \mathfrak M(A)=\bigcap_{n=1}^{\infty}\mathfrak R(A^n), \]
where \(\mathfrak R(A^n)\) is the range of the operator \(A^n\). If
\[ A(\mathfrak D(A)\cap \mathfrak M(A))=\mathfrak M(A), \]
then
\[ A(\mathfrak L(A)\cap \mathfrak M(A))=\mathfrak L(A)\cap \mathfrak M(A). \]

Proof. Since \(A(\mathfrak L(A))\subseteq \mathfrak L(A)\), we have \(A(\mathfrak L(A)\cap \mathfrak M(A))\subseteq \mathfrak L(A)\); moreover,
\[ A(\mathfrak L(A)\cap \mathfrak M(A))\subseteq A(\mathfrak D(A)\cap \mathfrak M(A))=\mathfrak M(A). \]
Consequently,
\[ A(\mathfrak L(A)\cap \mathfrak M(A))\subseteq \mathfrak L(A)\cap \mathfrak M(A). \]
On the other hand, if \(x\in \mathfrak L(A)\cap \mathfrak M(A)\), then, by the condition of the theorem, \(x=Ay\), where \(y\in \mathfrak D(A)\cap \mathfrak M(A)\); at the same time, by Lemma 1, \(y\in \mathfrak L(A)\). Thus,
\[ y\in \mathfrak D(A)\cap \mathfrak M(A)\cap \mathfrak L(A)= \]

\[ = \mathfrak{L}(A)\cap \mathfrak{M}(A). \]
Hence, by virtue of the arbitrariness of \(x\in \mathfrak{L}(A)\cap \mathfrak{M}(A)\), it follows that
\[ \mathfrak{L}(A)\cap \mathfrak{M}(A)\subseteq(\mathfrak{L}(A)\cap \mathfrak{M}(A)). \]

Let us note that, on the basis of Lemma 1 and the corollary to Theorem 2, the restrictions to \(\mathfrak{L}(A)\) of a homomorphism \(A\) into \(X\), as well as the powers of this restriction, are homomorphisms into \(\mathfrak{L}(A)\). Analogously, under the conditions of Theorem 6, on the basis of Lemma 2 we conclude that the restriction of \(A\) to \(\mathfrak{L}(A)\cap \mathfrak{M}(A)\) and the powers of this restriction are homomorphisms into \(\mathfrak{L}(A)\cap \mathfrak{M}(A)\).

When considering the restriction of \(A\) to \(\mathfrak{L}(A)\), the question arises whether the set \(\mathfrak{L}(A)\) is not too small—for example, whether it reduces to a single zero point. In the case where \(X\) is a Banach space, \(A\) is a closed linear operator in \(X\), \(\mathfrak{R}(A)=X\), and \(\mathfrak{D}(A)\) is dense in \(X\), an answer to this question is given by a theorem of D. A. Raikov, asserting that under these conditions \(\mathfrak{L}(A)\) is also dense in \(X\). Wishing to consider the case \(\mathfrak{R}(A)\ne X\) (under the same remaining conditions), we shall prove the following theorem.

Theorem 7. Let \(X\) be a Banach space, and let \(A\) be a closed linear operator with closed range \(\mathfrak{R}(A)\) and with \(\gamma(A)\) finite. Then, if \(\mathfrak{D}(A)\) is dense in the set \(\mathfrak{M}(A)\), then
\[ A(\mathfrak{L}(A)\cap \mathfrak{M}(A))=\mathfrak{L}(A)\cap \mathfrak{M}(A) \]
and \(\mathfrak{L}(A)\) is dense in \(\mathfrak{M}(A)\).

Proof. By virtue of the assumptions made (without using the fact that \(\overline{\mathfrak{D}(A)}\subseteq\mathfrak{M}(A)\)), the set \(\mathfrak{M}(A)\) is closed in \(X\). Consider the restriction \(\bar A\) of the operator \(A\) to \(\mathfrak{D}(A)\cap \mathfrak{M}(A)\). This will be a closed operator mapping the set \(\mathfrak{D}(A)\cap \mathfrak{M}(A)\), dense in \(\mathfrak{M}(A)\), onto the whole space \(\mathfrak{M}(A)\). Thus the hypothesis of Lemma 2 is fulfilled, and hence
\[ A(\mathfrak{L}(A)\cap \mathfrak{M}(A))=\mathfrak{L}(A)\cap \mathfrak{M}(A). \]
Further, by virtue of the above-mentioned result of D. A. Raikov, we conclude that the set \(\mathfrak{L}(\bar A)\) is dense in \(\mathfrak{M}(A)\). But
\[ \mathfrak{L}(\bar A)\subseteq \mathfrak{L}(A), \]
for
\[ \mathfrak{D}(\bar A^{\,n})\subseteq \mathfrak{D}(A^n)\qquad (n=1,2,\ldots). \]
Consequently, \(\mathfrak{L}(A)\) is dense in \(\mathfrak{M}(A)\).

Institute of Engineers
of Railway Transport

Received
19 XI 1967

REFERENCES

1. E. E. Browder, Math. Ann., 138, 55 (1959); Russian transl. in Matematika, 4, no. 3, 79 (1960).
2. M. A. Gol’dman, S. N. Krachkovskii, Dokl. Akad. Nauk SSSR, 174, no. 4 (1967).
3. N. Bourbaki, Topological Vector Spaces, IL, 1959.
4. M. A. Gol’dman, S. N. Krachkovskii, Dokl. Akad. Nauk SSSR, 158, no. 3 (1964).