UDC 517.43

INTEGRAL OPERATORS WITH CARLEMAN KERNELS

V. B. Korotkov

In the present article we give a solution of a problem posed by I. Neumann [1] concerning the determination of conditions under which a densely defined linear operator in \(L_2\) is an integral operator with a measurable Hermitian kernel, square-summable in one of the variables (a Carleman kernel [2, 3]). In an important special case this problem was solved by N. I. Akhiezer [3].

The article consists of four sections. In the first, a correspondence is established between measurable abstract functions taking values in \(L_2\), and measurable, though not necessarily Hermitian, kernels square-summable in one of the variables (kernels of Carleman type). This correspondence is used in § 2 to obtain necessary and sufficient conditions for the representability of a linear operator \(T\) in the form

\[ Tf=\int k(s,t)f(t)\,dt \tag{0.1} \]

with a kernel of Carleman type. The main result of the article is based on this criterion—the criterion for the representability of a linear operator in the form (0.1) with a Carleman kernel. Its proof constitutes the content of § 3. The criteria found receive in § 4 a further concretization for operators from a certain class containing, in particular, the class of all self-adjoint operators with purely point spectrum and the class of completely continuous operators.

§ 1. KERNELS OF CARLEMAN TYPE AND ABSTRACT FUNCTIONS

The main purpose of this section is to establish a one-to-one correspondence between kernels of Carleman type and measurable abstract functions.

Definition. Let \(\Omega\) be some Lebesgue-measurable set of nonzero measure in \(n\)-dimensional Euclidean space. A complex-valued function \(K(s,t)\) defined on the direct product \(\Omega\times\Omega\) will be called a kernel of Carleman type if the following two conditions are satisfied: 1) the function \(K(s,t)\) is measurable; 2) for almost all \(s\in\Omega\)

\[ \int |K(s,t)|^2\,dt<\infty. \tag{1.1} \]

The following lemmas hold.

Lemma 1. Let \(K(s,t)\) be a kernel of Carleman type, and let \(\Omega_0\) be the set of those points \(s\) of \(\Omega\) for which inequality (1.1) is satisfied. Denote

through \(\theta\), the zero element of the space \(L_2(\Omega)\), and define on \(\Omega\) an abstract function \(\varphi(s)\) by the following equality:

\[ \varphi(s)= \begin{cases} \theta, & \text{if } s\in \Omega\setminus \Omega_0,\\ \overline{K(s,t)}, & \text{if } s\in \Omega_0 . \end{cases} \tag{1.2} \]

The function \(\varphi(s)\) is a measurable abstract function taking values in \(L_2(\Omega)\).

Proof. From (1.1) and the definition of the function \(\varphi(s)\) it follows that the function \(\varphi(s)\) takes values in \(L_2(\Omega)\). We shall show that \(\varphi(s)\) is measurable. To this end, note that, whatever the function \(y(t)\) from \(L_2(\Omega)\) may be, the numerical function of the point \(s\), defined by the equality

\[ (y,\varphi(s))=\int_\Omega K(s,t)y(t)\,dt, \]

is measurable. This means that the function \(\varphi(s)\) is weakly measurable. But every weakly measurable function taking values in a separable space is measurable (see, for example, [4], p. 87). Consequently, the function \(\varphi(s)\) is measurable.

Lemma 2. If the kernels of Carleman type \(K_1(s,t)\) and \(K_2(s,t)\) are equivalent functions, then the corresponding abstract functions \(\varphi_1(s)\) and \(\varphi_2(s)\) are also equivalent.

Proof. Let \(\Omega_0^i\) \((i=1,2)\) be the set of those points \(s\in \Omega\) for which the inequality

\[ \int_\Omega |K_i(s,t)|^2\,dt<\infty \quad (i=1,2) \]

is satisfied. By hypothesis,

\[ m(\Omega\setminus \Omega_0^i)=0. \tag{1.3} \]

Let \(\omega\) be the set of those points \((s,t)\in \Omega\times\Omega\) for which \(K_1(s,t)=K_2(s,t)\). To each point \(s\in \Omega\) assign the set \(\omega_s\) of those points \(t\in \Omega\) for which \((s,t)\in \omega\), and denote by \(\sigma\) the totality of all points \(s\in \Omega\) for which \(m(\Omega\setminus \omega_s)=0\). Since, by hypothesis, \(m[(\Omega\times\Omega)\setminus \omega]=0\), it follows that

\[ m(\Omega\setminus \sigma)=0. \tag{1.4} \]

Put \(\Omega_0=\Omega_0^1\cap \Omega_0^2\cap \sigma\). Using (1.3) and (1.4), we obtain

\[ m(\Omega\setminus \Omega_0)=0. \tag{1.5} \]

Let \(s\in \Omega_0\). Then, by virtue of (1.2),

\[ \varphi_i(s)=\overline{K_i(s,t)},\quad i=1,2. \]

Since the set of those points \(t\) for which \(K_1(s,t)\ne K_2(s,t)\) is \(\Omega\setminus \omega_s\), and by the choice of \(s\), \(m(\Omega\setminus \omega_s)=0\), it follows that for \(s\in \Omega_0\) the elements \(\varphi_1(s)\) and \(\varphi_2(s)\) coincide. Hence, by (1.5), the equivalence of the abstract functions \(\varphi_1(s)\) and \(\varphi_2(s)\) follows.

Lemma 3. Let \(\varphi(s)\) be a measurable abstract function taking values in \(L_2(\Omega)\). Then there exists a kernel of Carleman type \(K(s,t)\) such that

\[ \overline{K(s,t)}=\varphi(s). \tag{1.6} \]

Proof. Since \(\varphi(s)\) takes values in \(L_2(\Omega)\), to each point \(s\in \Omega\) the function \(\varphi\) assigns the equivalence class \(\varphi(s)\) equi-

… functions from \(L_2(\Omega)\). From this class choose, in an arbitrary way, a function \(\varphi(s)\) and set

\[ K(s,t)=\overline{f_s(t)}. \tag{1.7} \]

Of course, the function \(K(s,t)\) depends on the choice of the functions \(f_s(t)\); however, this choice does not lead to an essential difference between the kernels, since, as is easy to see, any two kernels obtained in the indicated manner are equivalent.

The function \(K(s,t)\), defined by equality (1.7), satisfies equality (1.6). Let us show that the function \(K(s,t)\) is measurable. To this end, denote by \(\Omega_m\) the set of all points of \(\Omega\) whose distance from the origin is not greater than \(m\), \(m=1,2,\ldots\). It is clear that there is a number \(m_0\) such that \(m\Omega_m>0\). Fix \(m>m_0\). Since \(\Omega_m\) is a bounded set, by a known property of abstract functions (see, for example, [4], p. 86), for any measurable set \(G\subseteq\Omega_m\) and any \(\sigma\), \(0<\sigma<mG\), there exists a set \(E_{\sigma,G}\) such that \(mE_{\sigma,G}=\sigma\), and on \(G_\sigma=G\setminus E_{\sigma,G}\) the function \(\varphi(s)\) can be uniformly approximated with arbitrary accuracy by finitely valued abstract functions. Let \(\sigma=\sigma_p=m\Omega_m\cdot 2^{-p}\), \(p=1,2,\ldots\). Construct the following sequence of sets:
\[ E_1=E_{\sigma_1,G_m},\quad E_2=E_{\sigma_2,E_1},\quad E_3=E_{\sigma_3,E_2},\ldots \]
Put
\[ G_1=\Omega_m\setminus E_1,\quad G_p=E_{p-1}\setminus E_p,\quad p=2,3,\ldots . \]
It is clear that
\[ mG_p=m\Omega_m 2^{-p},\quad p=1,2,\ldots . \]
Let
\[ F_j=\bigcup_{p=1}^{j}G_p,\quad j=1,2,\ldots . \]
Since the \(G_p\) are pairwise disjoint, we have

\[ mF_j=\sum_{p=1}^{j}mG_p =m\Omega_m\bigl(1-2^{-j-r}\bigr)\to m\Omega_m \]
as \(j\to\infty\). Fix \(j\) and consider the set \(F_j\), \(j=1,2,\ldots\). By construction, on the set \(F_j\), \(j=1,2,\ldots\), the function \(\varphi(s)\) can be uniformly approximated with arbitrary accuracy by finitely valued functions \(\varphi_{k,j}(s)\). This means that for every \(\varepsilon>0\) there exists a number \(N\), independent of \(\varepsilon\), such that for all \(k>N\) and \(s\in F_j\)

\[ \|\varphi(s)-\varphi_{k,j}(s)\|_{L_2(\Omega)}<\varepsilon/2. \tag{1.8} \]

It follows that for all \(k>N\), \(l>N\) and \(s\in F_j\),

\[ \|\varphi_{k,j}(s)-\varphi_{l,j}(s)\|_{L_2(\Omega)}<\varepsilon . \tag{1.9} \]

Consider the finitely valued function \(\varphi_{k,j}(s)\):

\[ \varphi_{k,j}=\sum_{i=1}^{N_{k,j}} g_i^{(k,j)}\chi_{B_i^{(k,j)}}(s), \tag{1.10} \]

where \(\chi_{B_i}^{(k,j)}(s)\) is the characteristic function of the set \(B_i^{(k,j)}\); \(B_i^{(k,j)}\cap B_l^{(k,j)}=\varnothing\) for \(i\ne l\) and
\[ \bigcup_{i=1}^{N_{k,j}}B_i^{(k,j)}=F_j. \]
Since \(g_i^{(k,j)}(t)\in L_2(\Omega)\) are measurable numerical functions, the functions
\[ g_i^{(k,j)}(t)\chi_{B_i^{(k,j)}}(s) \]
are measurable numerical functions on the direct product \(F_j\times\Omega\). Therefore,

\[ K_{k,j}(s,t)=\overline{\varphi_{k,j}(s)} =\sum_{i=1}^{N_{k,j}}\overline{g_i^{(k,j)}(t)}\,\chi_{B_i^{(k,j)}}(s). \tag{1.11} \]

— measurable numerical functions on the direct product \(F_j\times\Omega_m\). Using (1.11), we rewrite (1.9) in the following form:

\[ \int_{\Omega}\left|K_{k,j}(s,t)-K_{l,j}(s,t)\right|^2\,dt<\varepsilon^2,\qquad s\in F_j. \]

Hence

\[ \int_{F_j}\int_{\Omega_m}\left|K_{k,j}(s,t)-K_{l,j}(s,t)\right|^2\,dt\,ds <\varepsilon^2 m\Omega_m\leq \varepsilon^2 m^n. \]

Thus, the sequence of functions \(K_{k,j}(s,t)\) is fundamental in the space \(L_2(F_j\times\Omega_m)\). By the completeness of this space, there is a function \(K_j(s,t)\in L_2(F_j\times\Omega_m)\) such that

\[ \lim_{l\to\infty}\int_{F_j}\int_{\Omega_m} \left|K_j(s,t)-K_{l,j}(s,t)\right|^2\,dt\,ds=0. \tag{1.12} \]

Let us show that the function \(K(s,t)\), defined above by equality (1.7), satisfies, for almost all \((s,t)\in F_j\times\Omega_m\), the equality

\[ K(s,t)\chi_{F_j\times\Omega_m}=K_j(s,t). \tag{1.13} \]

Put

\[ h_l(s)=\int_{\Omega_m}\left|K_j(s,t)-K_{l,j}(s,t)\right|^2\,dt. \]

It follows from (1.12) that the sequence of functions \(h_l(s)\) defined on \(F_j\) converges in measure to zero; then, by F. Riesz’s theorem, there is a subsequence \(h_{l_i}\) converging to zero for almost all \(s\in F_j\). Denote by \(\hat F_j\) the set of all \(s\in F_j\) for which

\[ \lim_{i\to\infty}h_{l_i}(s)=0. \tag{1.14} \]

By virtue of (1.7) and (1.8), for the number \(\varepsilon>0\) there exists an index \(N_1\) such that, for all \(l_i>N_1\), the inequality

\[ \left(\int_{\Omega_m}\left|K(s,t)-K_{l,j}(s,t)\right|^2\,dt\right)^{1/2}<\varepsilon/2 \]

will hold for all \(s\in\hat F_j\). Now fix \(s\in\hat F_j\). According to (1.14), there exists an index \(N_2\) such that, for all \(l_i>N_2\),

\[ h_{l_i}^{1/2}(s)<\varepsilon/2. \]

Let \(N=\max(N_1,N_2)\). Then for all \(l_i>N\),

\[ \left(\int_{\Omega_m}\left|K(s,t)-K_j(s,t)\right|^2\,dt\right)^{1/2} \leq \left(\int_{\Omega_m}\left|K(s,t)-K_{l_i,j}(s,t)\right|^2\,dt\right)^{1/2} +h_{l_i}^{1/2}(s)<\varepsilon. \]

From the obtained inequality, in view of the arbitrariness of \(\varepsilon>0\), the validity of equality (1.13) follows for almost all \(t\in\Omega_m\). But since \(s\) ranges over the set \(\hat F_j\), with \(m(F_j\setminus\hat F_j)=0\), it has thereby been proved that, for almost all \(s\in F_j\),

\[ K_j(s,t)=K(s,t)\chi_{F_j\times\Omega_m} \]

for almost all \(t\in\Omega_m\). It is now not difficult to complete the proof of the measurability of the function \(K(s,t)\). We have \(F_1\subset F_2\subset\ldots\subset\Omega_m\) and \(\lim_{j\to\infty} mF_j=m\Omega_m\); hence it follows that the sequence of measurable functions

\[ K_j(s,t)=K(s,t)\chi_{F_j\times\Omega_m} \]

converges almost everywhere to the function

\[ K^m(s,t)=K(s,t)\chi_{\Omega_m\times\Omega_m}. \]

Consequently, the function \(K^m(s,t)\) is measurable. But the sequence of measurable functions \(K^m(s,t)=K(s,t)\chi_{\Omega_m\times\Omega_m}\), in turn, converges almost everywhere to the function \(K(s,t)\). Hence the function \(K(s,t)\) is also measurable. Moreover, for all \(s\in\Omega\),

\[ \int_\Omega |K(s,t)|^2\,dt=\int_\Omega |f_s(t)|^2\,dt=\|\varphi(s)\|^2<\infty . \]

Consequently, the function \(K(s,t)\) constructed is the desired kernel of Carleman type. The lemma is proved.

Lemma 4. If the measurable abstract functions \(\varphi_1(s)\) and \(\varphi_2(s)\), defined on \(\Omega\) and taking values in \(L_2(\Omega)\), are equivalent, then the corresponding kernels of Carleman type are also equivalent.

Proof. Denote by \(\Omega_0\) the set of those points \(s\in\Omega\) for which \(\varphi_1(s)=\varphi_2(s)\). By hypothesis \(m(\Omega\setminus\Omega_0)=0\). Fix a point \(s\in\Omega_0\). Since \(K_i(s,t)\) \((i=1,2)\), as a function of \(t\), is an element of \(L_2(\Omega)\) coinciding with \(\varphi_i(s)\) \((i=1,2)\), and since for \(s\in\Omega_0\) \(\varphi_1(s)=\varphi_2(s)\), it follows that \(K_1(s,t)=K_2(s,t)\) for almost all \(t\in\Omega\). Thus, for almost all \(s\in\Omega\), \(K_1(s,t)=K_2(s,t)\) for almost all \(t\in\Omega\), and this means precisely that the functions \(K_1(s,t)\) and \(K_2(s,t)\) are equivalent. The lemma is proved.

If one agrees not to distinguish equivalent measurable functions (as is usually done), then the content of the lemmas proved can be briefly summarized as follows: between kernels of Carleman type and measurable abstract functions taking values in \(L_2(\Omega)\) there is a one-to-one correspondence determined by equality (1.6).

§ 2. OPERATORS OF CARLEMAN TYPE

Let \(L_2(\Omega)\) be a linear (generally speaking, unbounded) operator acting in \(L_2(\Omega)\) and defined on a linear set \(D_T\) everywhere dense in \(L_2(\Omega)\). We shall call the operator \(T\) an integral operator of Carleman type if, for each function \(f\in D_T\), there is a set \(\Omega_f\), \(m(\Omega\setminus\Omega_f)\), such that for all \(s\in\Omega_f\)

\[ (Tf)(s)=\int_\Omega K(s,t)f(t)\,dt, \tag{2.1} \]

where \(K(s,t)\) is a kernel of Carleman type.

In this section we shall determine the conditions under which a linear operator is an integral operator of Carleman type. We shall need the following

Definition. Let \(T\) be a linear operator and let \(D_T\) be its domain of definition. A measurable nonnegative almost everywhere finite function \(\Lambda(s)\) will be called a majorant of the operator \(T\) if, for every function \(f\in D\), \(\|f\|=1\), there is a set \(\omega_f\), \(m(\Omega\setminus\omega_f)=0\), such that for all \(s\in\omega_f\)

\[ |Tf(s)|\leqslant \Lambda(s). \]

Let us note that there exist, and moreover quite simple, operators having no majorants. The identity operator in \(L_2(\Omega)\) may serve as an example. Examples of operators having a majorant are furnished by integral operators of Carleman type. This is not difficult to verify by applying the Bunyakovsky–Schwarz inequality to (2.1). It turns out that the converse assertion is also true; namely, the following is valid:

Theorem 1. A linear operator \(T\), defined on an everywhere dense linear manifold \(D_T\) in \(L_2(\Omega)\), is an integral operator of Carleman type if and only if \(T\) has a majorant.

Necessity. Let \(T\) be an integral operator of Carleman type and let \(K(s,t)\) be the corresponding kernel. Let \(f \in D_T\) and \(\|f\|=1\). Denote by \(\widehat{\Omega}_f\) the intersection of the set \(\Omega_f\) of points \(s\) for which (2.1) holds with the set \(\Omega_0\) of points \(s\) for which inequality (1.1) holds. It is clear that \(m(\Omega\setminus \widehat{\Omega}_f)=0\). Using the Cauchy--Schwarz inequality, we obtain, for all \(s \in \widehat{\Omega}_f\),

\[ |(Tf)(s)|=\left|\int_\Omega K(s,t)f(t)\,dt\right| \leq \left(\int_\Omega |K(s,t)|^2\,dt\right)^{1/2}=K(s). \]

The function \(K(s)\) is nonnegative, measurable, and finite almost everywhere. Consequently, the operator \(T\) has a majorant.

Sufficiency. Let \(\Lambda(s)\) be a majorant of the linear operator \(T\). Denote by \(E_k\) the set of those points \(s\in\Omega\) for which \(\Lambda(s)\leq k\), \(k=1,2,\ldots\). The sets \(E_k\) form an expanding sequence of sets, and, on the basis of the fact that the function \(\Lambda(s)\) is finite almost everywhere,

\[ \lim_{k\to\infty} m(\Omega\setminus E_k)=0. \tag{2.2} \]

Let \(V_k\) be the ball of radius \(k\) with center at the origin. We note that

\[ \lim_{k\to\infty} m(\Omega\setminus V_k)=0. \tag{2.3} \]

Set \(\Omega_k=E_k\cap V_k\), \(k=1,2,\ldots\). Then \(\Omega_1\subseteq\Omega_2\subseteq\cdots\subseteq\Omega\). Taking (2.2) and (2.3) into account, we obtain

\[ m\left(\Omega\setminus\bigcup_{k=1}^{\infty}\Omega_k\right) = \lim_{k\to\infty} m(\Omega\setminus\Omega_k) = \lim_{k\to\infty} m\bigl[(\Omega\setminus E_k)\cup(\Omega\setminus V_k)\bigr] \leq \]

\[ \leq \lim_{k\to\infty} m(\Omega\setminus E_k) + \lim_{k\to\infty} m(\Omega\setminus V_k) =0. \]

Thus,

\[ m\left(\Omega\setminus\bigcup_{k=1}^{\infty}\Omega_k\right)=0. \tag{2.4} \]

Let \(k_0\) be a number starting from which the sets \(\Omega_k\) are nonempty. Fix \(k>k_0\) and a measurable set \(E\) from \(\Omega_k\). Define on the linear manifold \(D_T\) a functional \(F_E\) by the following equality:

\[ F_E^{(k)}f=\int_E (Tf)(s)\,ds,\quad f\in D_T. \tag{2.5} \]

The functional \(F_E^{(k)}\) is, obviously, a linear functional. We shall show that \(F_E^{(k)}\) is a bounded functional on \(D_T\). Let \(f\in D_T\) and \(\|f\|=1\). Since, by assumption, the operator \(T\) has the majorant \(\Lambda(s)\), and for all \(s\in E\subset\Omega_k\), \(\Lambda(s)\leq k\), it follows that

\[ |F_E^{(k)}(f)|\leq \int_E |(Tf)(s)|\,ds \leq kmE. \tag{2.6} \]

As is known, a bounded linear functional defined on an everywhere dense linear manifold can be extended uniquely

extend by continuity to the entire space. We extend the functional defined above by continuity to all of \(L_2(\Omega)\) and retain for the resulting extension the old notation \(F_E^{(k)}\). We note that, by virtue of (2.6) and the density of \(D_T\) in \(L_2(\Omega)\),

\[ \|F_E^{(k)}\| \leq kmE. \tag{2.7} \]

Thus, equality (2.5) defines an abstract set function \(F_E^{(k)}\), which assigns to each measurable set \(E \subset \Omega_k\) a bounded linear functional \(F_E^{(k)}\). We shall show that the set function thus constructed is additive. Indeed, let \(E_1 \subset \Omega_k\), \(E_2 \subset \Omega_k\), and \(E_1 \cap E_2 = \varnothing\). Then for any function \(f \in D_T\),

\[ \begin{aligned} F_{E_1 \cup E_2}^{(k)}(f) &= \int_{E_1 \cup E_2} (Tf)(s)\,ds = \int_{E_1} (Tf)(s)\,ds + \int_{E_2} (Tf)(s)\,ds \\ &= F_{E_1}^{(k)}(f) + F_{E_2}^{(k)}(f) = \bigl(F_{E_1}^{(k)} + F_{E_2}^{(k)}\bigr)(f). \end{aligned} \]

Taking into account that the linear manifold \(D_T\) is everywhere dense in \(L_2\), we obtain from this

\[ F_{E_1 \cup E_2}^{(k)} = F_{E_1}^{(k)} + F_{E_2}^{(k)} . \]

Consequently, the function \(F_E^{(k)}\) is an additive abstract set function taking values in \(L_2(\Omega)\). Since the function \(F_E^{(k)}\) satisfies the estimate (2.7), by the Dunford–Pettis theorem [5] there exists a unique (up to equivalence) measurable abstract function \(\varphi_k(s)\), defined on \(\Omega_k\) and taking values in \(L_2(\Omega)\), such that

\[ F_E^{(k)} = \int_E \varphi_k(s)\,ds,\quad E \subset \Omega_k, \tag{2.8} \]

and, moreover, for all \(f \in L_2(\Omega)\),

\[ F_E^{(k)}(f) = \int_E (f,\varphi_k(s))\,ds. \tag{2.9} \]

Let \(\theta\) be the zero element of the space \(L_2(\Omega)\). Put, for \(s \in \Omega \setminus \Omega_k\), \(\varphi_k(s)=\theta\). Now the functions \(\varphi_k(s)\) are defined on the whole set \(\Omega\), are still measurable, and take values in \(L_2(\Omega)\). Let \(k<l\). Observing that \(F_E^{(k)}=F_E^{(l)}\) for any measurable set \(E\) from \(\Omega_k\), and using the fact that the integrand in (2.8) is determined uniquely (up to equivalence), we obtain that for almost all \(s \in \Omega_k\), \(\varphi_k(s)=\varphi_l(s)\). Hence it follows that the sequence of measurable functions \(\varphi_k(s)\) converges for almost all \(s\) in \(\Omega\) to a certain measurable function \(\varphi(s)\), also taking values in \(L_2(\Omega)\), and, for any \(k\), \(\varphi(s)=\varphi_k(s)\) for almost all \(s \in \Omega_k\). Now it is not difficult to complete the proof of the theorem. By Lemma 3, for the function \(\varphi(s)\) there exists a kernel of Carleman type \(K(s,t)\) such that, for all \(s \in \Omega\),

\[ \overline{K(s,t)}=\varphi(s). \tag{2.10} \]

Let \(f\) be an arbitrary function from \(D_T\). Then, whatever the number \(k\) and the measurable set \(E\) from \(\Omega_k\),

\[ \int_E (Tf)(s)\,ds = F_E^{(k)}(f) = \int_E (f,\varphi_k(s))\,ds = \int_E (f,\varphi(s))\,ds. \]

It follows from this, by virtue of the arbitrariness of the number \(k\) and of the arbitrariness of the set \(E\) (for the given \(k\)), and also by virtue of (2.4), that for almost all \(s\) the equality \((Tf)(s)=(f,\varphi(s))\) holds. Taking (2.10) into account, we finally obtain

\[ (Tf)(s)=(f,\varphi(s))=\int_\Omega K(s,t)f(t)\,dt. \]

This completes the proof of the theorem.

Example. Let \(\Omega=(-\infty,\infty)\) and \((Tf)(s)=sf(s)\). Let

\[ f_{m,n}=\sqrt n\,\chi_{\left[m,\,m+\frac1n\right]},\qquad m=1,2,\ldots \]

It is clear that \(\|f_{m,n}\|=1\). But \(|Tf_{m,n}(s)|>\sqrt n\), \(m=1,2,\ldots,\ n=1,2,\ldots\). Consequently, \(T\) has no majorant and therefore is not an integral operator of Carleman type.

In conclusion of the paragraph we shall prove one important property of a majorant, which we shall use later.

Lemma. Let \(T\) be an integral operator of Carleman type with kernel \(K(s,t)\). Then every majorant \(\Lambda(s)\) of it, for almost all \(s\in\Omega\), satisfies the inequality

\[ K(s)=\left(\int_\Omega |K(s,t)|^2\,dt\right)^{1/2}\leq \Lambda(s). \]

Proof. Using the separability of the space \(L_2(\Omega)\), construct a countable set \(F\) belonging to the unit sphere of the space \(L_2(\Omega)\) and everywhere dense in it. Fix \(f\) from \(F\) and denote by \(\Omega_f\) the set of those points \(s\) for which

\[ (Tf)(s)=\int_\Omega K(s,t)f(t)\,dt \tag{2.11} \]

and

\[ |(Tf)(s)|\leq \Lambda(s). \tag{2.12} \]

Let \(\Omega_0=\bigcup_{f\in F}\Omega_f\). Since the set \(F\) is countable and \(m(\Omega\setminus\Omega_f)=0\), it follows also that \(m(\Omega\setminus\Omega_0)=0\). Let \(s\in\Omega_0\). Taking (2.11) and (2.12) into account, we obtain, for all \(f\in F\) and \(s\in\Omega_0\),

\[ \left|\int_\Omega K(s,t)f(t)\,dt\right|\leq \Lambda(s). \tag{2.13} \]

Since (2.13) holds for all \(f\in F\), and \(F\) is everywhere dense in the unit sphere of the space \(L_2(\Omega)\), we have

\[ K(s)=\left[\int_\Omega |K(s,t)|^2\,dt\right]^{1/2} =\sup_{f\in F}\left|\int_\Omega K(s,t)f(t)\,dt\right|\leq \Lambda(s). \]

The lemma is proved.

§ 3. INTEGRAL OPERATORS WITH CARLEMAN KERNELS

In this section we shall consider Hermitian kernels of Carleman type, i.e., kernels satisfying the equality

\[ K(s,t)=\overline{K(t,s)} \]

for almost all \((s,t)\). According to established terminology, such kernels are called Carleman kernels.

Our aim is to obtain necessary and sufficient conditions for a linear operator to be an integral operator with a Carleman kernel (or, more briefly, a Carleman operator). The criterion is formulated in Theorem 2. For its proof we shall need the following notation and results from the paper of N. I. Akhiezer [3]:

a) let \(P(s)\) be an arbitrary measurable, nonnegative function, finite almost everywhere, and let \([L_2]_P\) be the linear manifold of all functions \(f\) in \(L_2(\Omega)\) satisfying the inequality

\[ \int_{\Omega} P(s)|f(s)|\,ds<\infty . \]

The linear manifold \([L_2]_P\) is everywhere dense in \(L_2(\Omega)\);

b) let \(K(s,t)\) be a Carleman kernel and

\[ K(s)=\left(\int_{\Omega}|K(s,t)|^2\,dt\right)^{1/2}. \]

The linear manifold \([L_2]_K\) is contained in the linear manifold \(D_B\) of all functions \(f\) in \(L_2(\Omega)\) for which

\[ \int_{\Omega} K(s,t)f(t)\,dt\in L_2(\Omega). \]

Both linear manifolds are everywhere dense in \(L_2(\Omega)\);

c) let \(B\) be the Carleman operator defined on the linear manifold \(D_B\) by the equality

\[ (Bf)(s)=\int_{\Omega}K(s,t)f(t)\,dt, \]

and let \(A\) be its restriction to the linear manifold \([L_2]_K\). The operator \(A\) is symmetric and \(A^*=B\).

We can now formulate our theorem.

Theorem 2. In order that a linear operator \(T\), defined on an everywhere dense linear manifold \(D_T\) in \(L_2(\Omega)\), be an integral operator with a Carleman kernel, it is necessary and sufficient that:

1) the operator \(T\) have a majorant \(\Lambda(s)\);

2) \((T^*f,g)=(f,T^*g)\) for any \(f\) and \(g\) in \([L_2(\Omega)]_\Lambda\).

Necessity. By Theorem 1 the operator \(T\) has a majorant \(\Lambda(s)\). Since \(T\subseteq B\), where \(B\) is the operator defined in c), it follows that \(T^*\supseteq B^*=A^*\supseteq A\). Thus \(D_{T^*}\supseteq D_A=[L_2(\Omega)]_K\). But by the lemma from the preceding section \(K(s)\leq \Lambda(s)\). Consequently, \([L_2]_K\supseteq [L_2]_\Lambda\) and \(D_{T^*}\supseteq [L_2]_\Lambda\). Let us prove the validity of the second assertion of the theorem. We have already shown that \(T^*\supseteq A\) and \(D_A=[L_2]_K\supseteq [L_2]_\Lambda\). But the operator \(A\), as was noted in c), is symmetric; therefore,

\[ (T^*f,g)=(Af,g)=(f,Ag)=(f,T^*g). \]

Sufficiency. Let \(T\) be a linear operator satisfying the conditions of the theorem. Since \(T\) has a majorant, by Theorem 1 \(T\) is an integral operator with a kernel of Carleman type \(K(s,t)\). It remains to show the Hermitian character of the kernel \(K(s,t)\). To this end, consider the linear manifold \(D_{\hat B}\) of all functions \(f\) in \(L_2(\Omega)\) for which

\[ \int_{\Omega}\hat K(s,t)\times \]

\[ \times f(t)\,dt\in L_2(\Omega), \]

and define on \(D_{\hat B}\) the operator \(\hat B\) by the equality

\[ (\hat Bf)(s)=\int_{\Omega}\hat K(s,t)f(t)\,dt . \]

By its very construction the operator \(\hat B\) is an extension of the operator \(T\). Since for all \(f\) in \([L_2]_\Lambda\) and \(g \in L_2(\Omega)\), by virtue of the lemma of § 2,

\[ \int_\Omega\int_\Omega |K(s,t)|\,|f(s)|\,|g(t)|\,dt\,ds \leq \|g\|\int_\Omega\left(\int_\Omega |K(s,t)|^2\,dt\right)^{1/2}|f(s)|\,ds \leq \]

\[ \leq \|g\|\int_\Omega \Lambda(s)|f(s)|\,ds < \infty, \tag{3.1} \]

then

\[ \int_\Omega \overline{K(s,t)} f(s)\,ds \in L_2(\Omega). \]

Consequently, on the everywhere dense in \(L_2(\Omega)\) linear set \([L_2]_\Lambda\) one can define the linear operator \(\hat A\):

\[ (\hat A f)(t)=\int_\Omega \overline{K(s,t)} f(s)\,ds,\quad f\in [L_2]_\Lambda . \]

Inequality (3.1) allows us to apply Fubini’s theorem to the integral

\[ \int_\Omega\int_\Omega \overline{K(s,t)} f(s)\overline{g(t)}\,ds\,dt. \]

Applying this theorem to the indicated integral, we obtain

\[ \int_\Omega\left(\int_\Omega \overline{K(s,t)} f(s)\,ds\right)\overline{g(t)}\,dt = \int_\Omega f(s)\left(\int_\Omega \overline{K(s,t)}\,\overline{g(t)}\,dt\right)\,ds, \tag{3.2} \]

or, what is the same,

\[ (\hat A f,g)=(f,Bg), \]

i.e. \(\hat B \subset \hat A^*\). But \(T \subset \hat B\), hence \(\hat A \subset \hat A^{**}\subset T^*\). Using this, the condition of the theorem can be written for all \(f\) and \(g\) from \(D_{\hat A}=[L_2]_\Lambda\) in the following form:

\[ (\hat A f,g)=(f,\hat A g), \]

or, in detailed notation,

\[ \int_\Omega\left(\int_\Omega \overline{K(s,t)} f(s)\,ds\right)\overline{g(t)}\,dt = \int_\Omega f(s)\left(\int_\Omega \overline{K(t,s)}\,\overline{g(t)}\,dt\right)\,ds. \]

Taking (3.2) into account, we further obtain

\[ \int_\Omega f(s)\left[\int_\Omega \bigl(\overline{K(s,t)}-K(t,s)\bigr)\overline{g(t)}\,dt\right]\,ds=0. \]

Since \(f\) and \(g\) run through the set \([L_2]_\Lambda\), everywhere dense in \(L_2(\Omega)\), it follows from the last equality that for almost every \(s\in\Omega\), \(\overline{K(s,t)}=K(t,s)\) for almost all \(t\in\Omega\). The theorem is proved.

An almost obvious, but important, consequence of Theorem 2 is the following

Theorem 3. A self-adjoint linear operator is an integral operator with a Carleman kernel if and only if it has a majorant.

Proof. Necessity is a direct consequence of Theorem 1. To prove sufficiency we shall use the proved

in Theorem 2 by the relation \(\hat A\subseteq T^*\). Since \(T\) is a self-adjoint operator, for all \(f,g\) from \([L_2]_\Lambda\subseteq D_{T^*}=D_T\),

\[ (T^*f,g)=(f,T^*g). \]

Thus the conditions of Theorem 2 are satisfied; consequently, the operator \(T\) is an integral operator with a Carleman kernel.

§ 4. OPERATORS OF A SPECIAL FORM

Let \(T\) be a linear operator acting in \(L_2(\Omega)\) and defined on a linear manifold \(D_T\) everywhere dense in \(L_2(\Omega)\). Suppose that there are a numerical sequence \(\lambda_k\), an orthonormal sequence \(\varphi_k\), and some sequence of elements \(\psi_k\) from \(L_2\) such that, for every function \(f\) from \(L_2\), the series \(\sum_{k=1}^{\infty}\lambda_k(f,\varphi_k)\psi_k\) converges in the norm of the space \(L_2(\Omega)\) to \(Tf\). Thus, let

\[ Tf=\sum_{k=1}^{\infty}\lambda_k(f,\varphi_k)\psi_k. \tag{4.1} \]

Additional information about the operator \(T\), contained in the equality (4.1), makes it possible to reduce the clarification of the question of the existence of a majorant for the operator \(T\) to checking a certain inequality relating \(\lambda_k\) and \(\psi_k\).

Theorem 4. In order that the linear operator \(T\), defined by the equality (4.1), be an integral operator of Carleman type, it is necessary and sufficient that, for almost all \(s\in\Omega\),

\[ \lambda(s)=\left(\sum_{k=1}^{\infty}|\lambda_k|^2|\psi_k(s)|^2\right)^{1/2}<\infty . \tag{4.2} \]

Necessity. Let \(T\) be an integral operator of Carleman type with kernel \(K(s,t)\). Let \(\Omega_0\) be the set of those points \(s\in\Omega\) for which

\[ K^2(s)=\int_{\Omega}|K(s,t)|^2\,dt<\infty . \tag{4.3} \]

Denote by \(\Omega_k^{(1)}\) the set of all \(s\in\Omega\) for which

\[ \int_{\Omega} K(s,t)\varphi_k(t)\,dt=T\varphi_k(s), \tag{4.4} \]

and by \(\Omega_k^{(2)}\) the set of all \(s\in\Omega\) for which

\[ T\varphi_k(s)=\chi_k\varphi_k(s). \tag{4.5} \]

Put \(\hat\Omega_0\) to be the aggregate of all points \(s\) from \(\Omega_0\) belonging simultaneously to all the sets \(\Omega_k^{(1)}\) and \(\Omega_k^{(2)}\), \(k=1,2,\ldots\). It is clear that \(m(\Omega\setminus\hat\Omega_0)=0\). Let \(s\in\hat\Omega_0\). Then, by virtue of (4.3)—(4.5) and Bessel’s inequality,

\[ \lambda^2(s)\sum_{k=1}^{\infty}|\lambda_k|^2|\psi_k(s)|^2 = \sum_{k=}^{\infty}|(T\varphi_k)(s)|^2 = \tag{4.6} \]

\[ = \sum_{k=1}^{\infty}\left|(K(s,t),\overline{\varphi_k(t)})\right|^2 \leq \int_{\Omega}|K(s,t)|^2\,dt=K^2(s)<\infty . \]

Sufficiency. Let \(f\in D_T\) and \(\|f\|=1\). In view of (4.1) and (4.2), the series
\[ \sum_{k=1}^{\infty}\lambda_k(f,\varphi_k)\psi_k(s) \]
converges absolutely for almost all \(s\in\Omega\) to \((Tf)(s)\). But then
\[ |(Tf)(s)|=\left|\sum_{k=1}^{\infty}\lambda_k(f,\varphi_k)\psi_k(s)\right| \leq \left(\sum_{k=1}^{\infty}|\lambda_k|^2|\psi_k(s)|^2\right)^{1/2} =\lambda(s). \tag{4.7} \]

The function \(\lambda(s)\) is, obviously, measurable, nonnegative, and, by the hypothesis, finite almost everywhere. Thus, \(\lambda(s)\) is a majorant of the operator \(T\), and, consequently, the operator \(T\) is an integral operator of Carleman type.

Corollary. Let \(T\) be an integral operator of Carleman type with kernel \(k(s,t)\). Then for almost all \(s\)
\[ K(s)=\lambda(s). \tag{4.8} \]

Proof. By the lemma of § 2, for almost all \(s\in\Omega\), \(K(s)\leq \lambda(s)\). Hence, and from (4.6), the validity of equality (4.7) follows for almost all \(s\).

Example. The fractional integration operator
\[ T_a f=\sum_{n=-\infty}^{\infty}(in)^{-\alpha}(f,e^{ins})e^{ins}\quad (n\ne 0) \]
is an integral operator of Carleman type if \(\frac12<\alpha\), and is not such if \(0<\alpha\leq \frac12\). Indeed, in this case
\[ \lambda(s)=\sum_{(n\ne 0)} n^{-2\alpha}, \]
and the series converges if \(1<2\alpha\), and diverges if \(0<2\alpha\leq 1\).

Theorem 5. In order that the linear operator \(T\), defined by equality (4.1), be an integral operator with Carleman kernel, it is necessary and sufficient that the function \(\lambda(s)\) be finite almost everywhere and that, for any \(f\) and \(g\) from \([L_2]\),
\[ (T^*f,g)=(f,T^*g). \]

Necessity. The proof is carried out by a verbatim repetition of the arguments given in the proof of necessity in Theorem 2, with the sole difference that in our case, in view of (4.8),
\[ \Lambda(s)=\lambda(s)+K(s). \]

Sufficiency follows immediately from Theorem 3, since, as we showed above (see (4.7)), \(\lambda(s)\) is a majorant of the operator \(T\).

A simple consequence of Theorem 3 is also the following

Theorem 6. In order that the self-adjoint operator
\[ Tf=\sum_{k=1}^{\infty}\lambda_k(f,\varphi_k)\varphi_k \]

was an integral operator with a Carleman kernel, it is necessary and sufficient that, for almost all \(s\),

\[ \lambda(s)=\sum_{k=1}^{\infty}|\lambda_k|^2|\varphi_k(s)|^2<\infty . \]

Let us note that all the results of this article are also valid for operators acting in a real Hilbert space. In this case the kernels under consideration should be regarded as real. Let us consider one more example. Let \(\Omega=[0,1]\), \(\chi_n^k\), \(n=1,2,\ldots,\ k=1,2,\ldots,2^n\), be the Haar functions, \(T\chi_n^k=\alpha_n^k\chi_n^k\), where \(\alpha_n^k\) are real numbers. Let

\[ a_n=\min_k|\alpha_n^k|,\qquad A_n=\max_k|\alpha_n^k|. \]

The self-adjoint operator \(T\) is Carleman if the series \(\sum a_n^2 2^n<\infty\), and is not Carleman if the series \(\sum A_n^2 2^n\) diverges. Indeed, in this case

\[ \sum a_n^2 2^n\leq \lambda^2(s)\leq \sum A_n^2 2^n . \]

The class of operators considered in this paragraph includes, in particular, completely continuous operators, since every completely continuous operator \(T\) can be represented in the form

\[ Af=\sum_{k=1}^{\infty}\lambda_k(f,\varphi_k)\psi_k, \]

where \(\lambda_k\geq 0\) are the singular values of the operator, and \(\{\varphi_k\}\) and \(\{\psi_k\}\) are orthonormal sequences of the so-called associated Schmidt elements. Thus, all the results of this paragraph are also valid for completely continuous operators.

In conclusion, let us consider the question of complete continuity of integral operators with kernels of Carleman type.

Definition. Let \(T\) be a bounded linear operator in \(L_2(\Omega)\), and let \(\Omega\) be a bounded set. Let \(E\) be a measurable subset of the set \(\Omega\). Define the bounded operator \(P_E\) by the equality

\[ P_E f=f\chi_E . \]

We shall call an operator \(T\) an absolutely continuous operator if, for every \(\varepsilon>0\), there exists \(\delta>0\) such that for all \(E\) with \(mE<\delta\),

\[ \|P_E T\|<\varepsilon . \tag{4.9} \]

The following is true.

Theorem 7. In order that an integral operator of Carleman type be completely continuous, it is sufficient that it be absolutely continuous.

Proof. Let \(T\) be an absolutely continuous operator of Carleman type. From the results of the first paragraph it follows that the operator \(T\) can be represented in the form

\[ (Tf)(s)=(f,\varphi(s)),\qquad f\in L_2(\Omega), \]

where \(\varphi(s)\) is a measurable abstract function, defined on \(\Omega\) and taking values in \(L_2(\Omega)\). Let \(\varepsilon>0\) be given. Since \(\Omega\) is a bounded set, for the given \(\varepsilon\) there exists \(E\) with \(mE<\delta\) and a sequence of finite-valued functions \(\omega_k(s)\), converging for all \(s\in\Omega\setminus E\)

uniformly. Since the range of values of each function \(\omega_k\) is compact, it follows from uniform convergence that the set of functions \(\varphi(s)\chi_{\Omega\setminus E}(s)\) is also compact. Note that

\[ Tf=P_E Tf+P_{\Omega\setminus E}Tf . \]

We shall show that the operator \(P_{\Omega\setminus E}T\) is completely continuous. Indeed,

\[ (P_{\Omega\setminus E}T)f=Tf(s)\chi_{\Omega\setminus E}=(f,\varphi(s))\chi_{\Omega\setminus E}= \]

\[ =(f,\varphi(s)\chi_{\Omega\setminus E}). \]

Let \(f_n\) converge weakly to zero. Since the set of values of the function \(\varphi(s)\chi_{\Omega\setminus E}\) is compact, the sequence \(P_{\Omega\setminus E}Tf_n=(f_n,\varphi(s)\chi_{\Omega\setminus E})\) converges to zero uniformly for all \(s\). Hence, and from boundedness of the set, it follows that the sequence \(P_{\Omega\setminus E}Tf_n\) converges to zero in the norm of \(L_2(\Omega)\). Thus, the operator \(P_{\Omega\setminus E}T\) maps every sequence weakly converging to zero into a sequence strongly converging to zero, and therefore is completely continuous. Since the operator \(T\) is absolutely continuous and \(mE<\delta\), we have

\[ \|T-P_{\Omega\setminus E}T\|=\|P_E T\|<\varepsilon . \]

By the arbitrariness of \(\varepsilon\), it follows that the operator \(T\) is completely continuous.

References

1. Neumann J. Charakterisierung des Spektrums eines Integraloperators. Collected works, Vol. IV, 38—55, 1962.
2. Carleman T. Sur les équations intégrales singulières a noyau réel et Symmetrique. Uppsala, 1923.
3. Ахиезер Н. И. УМН, 2, вып. 5(21), 91—132, 1947.
4. Hille E., Phillips R. Functional Analysis and Semi-Groups. IL, Moscow, 1962.
5. Dunford N., Schwartz J. T. Linear Operators. General Theory. IL, Moscow, 1962.

Received by the editors
June 30, 1965.

Institute of Mathematics
Siberian Branch, Academy of Sciences of the USSR