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UDC 513.88+517.948
MATHEMATICS
E. R. TSEKANOVSKII
ON THE DESCRIPTION OF GENERALIZED EXTENSIONS WITH ONE-DIMENSIONAL IMAGINARY COMPONENT OF THE DIFFERENTIATION OPERATOR WITHOUT SPECTRUM
(Presented by Academician L. S. Pontryagin on 26 XII 1966)
I. Let \(T\) be a closed operator acting in a Hilbert space \(\mathscr G\), for which
\[ T_0 \subset T, \qquad T^* \subset T_0^*, \tag{1} \]
where \(T_0\) is a symmetric operator with dense domain. Consider the Hilbert space \(\mathscr G_+ = D_{T_0^*}\) with scalar product
\[ (f,g)_+ = (T_0^* f, T_0^* g)_0 + (f,g)_0 \qquad (f,g \in \mathscr G_+). \tag{2} \]
Construct a triple of spaces \(\mathscr G_+ \subseteq \mathscr G_0 \subseteq \mathscr G_-\), where \(\mathscr G_-\) is the space of generalized elements generating antilinear functionals on \(\mathscr G_+\) (see \((^2,^4)\)).
In \((^4,^5)\) it was shown that \(T\) and \(T^*\) can always be extended to \(\mathscr G_+ = D_{T_0^*}\) in such a way that the resulting extensions \(T_{\mathscr G_+}\) and \(T_{\mathscr G_+}^{\times}\) are adjoint to one another in the generalized sense; moreover, \(T_{\mathscr G_+}\) was called a generalized extension of the operator \(T\).
M. S. Livšic is responsible for \((^1)\) the well-known theorem that every quasidissipative extension without spectrum of a simple symmetric operator with deficiency index \((1,1)\) is unitarily equivalent to the differentiation operator \(T\)
\[ Tf = \frac{1}{i}\,\frac{df}{dx} \qquad (f(0)=0,\ 0 \leq x \leq \eta), \]
\[ T^* f = \frac{1}{i}\,\frac{df}{dx} \qquad (f(\eta)=0,\ 0 \leq x \leq \eta). \tag{3} \]
It is therefore natural that it is of interest to give a description of all generalized extensions of the differentiation operator (3).
In the present paper we give a description of the generalized extensions \(T_{\mathscr G_+}\) of this operator that have a one-dimensional imaginary component \((T_{\mathscr G_+} - T_{\mathscr G_+}^{\times})/i\), which makes it possible to single out in a natural way an important class of such extensions.
II. Let \(\mathscr G_0 = L_2[0,\eta]\), and let
\[ T_0 f = \frac{1}{i}\,\frac{df}{dx} \qquad (f(0)=f(\eta)=0). \]
It is known \((^3)\) that \(T_0^* f = \dfrac{1}{i}\,\dfrac{df}{dx}\), and \(D_{T_0^*}\) consists of absolutely continuous functions on \([0,\eta]\) whose derivatives belong to \(L_2[0,\eta]\). Obviously, the operators \(T\) and \(T^*\) satisfy condition (1). It follows from (2) that \(\mathscr G_+ = W_2^{(1)}[0,\eta] = D_{T_0^*}\), and the scalar product in \(\mathscr G_+\) is given by the formula
\[ (f,g)_+ = \int_0^\eta f'(x)\,\overline{g'(x)}\,dx + \int_0^\eta f(x)\,\overline{g(x)}\,dx . \]
Let us construct the space \(\mathcal G_-\), so that \(\mathcal G_+\subseteq \mathcal G_0\subseteq \mathcal G_-\).
Consider the family of operators acting from \(\mathcal G_+\) into \(\mathcal G_-\):
\[ T_{\mathcal G_+}(\xi)f=\frac{1}{i}\frac{df}{dx} +if(0)\,[e^{i\xi\eta}\delta(x-\eta)-\delta(x)], \tag{4} \]
\[ \left( \begin{array}{c} -\infty<\xi<+\infty\\ 0\leq x\leq \eta \end{array} \right) \]
\[ T_{\mathcal G_+}^{\times}(\xi)f=\frac{1}{i}\frac{df}{dx} +if(\eta)e^{-i\xi\eta}\,[e^{i\xi\eta}\delta(x-\eta)-\delta(x)]; \tag{5} \]
here \(\delta(x-\eta)\), \(\delta(x)\) are generalized elements (delta functions) generating the functionals
\[ (\delta(x-\eta),f)_0=\overline{f(\eta)},\qquad (\delta(x),f)_0=\overline{f(0)}. \]
It is not hard to see that for each fixed \(\xi\) the operators \(T_{\mathcal G_+}(\xi)\) and \(T_{\mathcal G_+}^{\times}(\xi)\) are extensions of \(T\) and \(T^*\), respectively, and moreover
\[ (T_{\mathcal G_+}(\xi)f,g)_0=(f,T_{\mathcal G_+}^{\times}(\xi)g)_0 \qquad (f,g\in \mathcal G_+). \]
In addition, it follows from (4), (5) that
\[ \frac{T_{\mathcal G_+}(\xi)-T_{\mathcal G_+}^{\times}(\xi)}{i}\,f =(f,[e^{i\xi\eta}\delta(x-\eta)-\delta(x)])_0\,J\,[e^{i\xi\eta}\delta(x-\eta)-\delta(x)] \]
\[ (J=-1). \tag{6} \]
Relation (6) shows that for every \(\xi\) the imaginary component of the family of operators \(T_{\mathcal G_+}(\xi)\) is one-dimensional.
We shall show that the family of operators (4) exhausts all generalized extensions of the operator \(T\) on \(\mathcal G_+=D_{T_0^*}\) that have a one-dimensional imaginary component.
III. In what follows we shall need one theorem, whose proof may be found in \((^5)\) and whose statement we give for convenience.
Theorem 1. Let the operator \(T\) satisfy conditions (1). In order that the operators \(T\) and \(T^*\) be extendable to the whole space
\[ \mathcal G_+=D_{T_0^*}=D_T+D_{T^*} \]
in such a way that the resulting extensions \(T_{\mathcal G_+}\) and \(T_{\mathcal G_+}^{\times}\) are adjoint to each other in the generalized sense, it is necessary and sufficient that there exist linear operators \(P(D_{T^*}\to \mathcal G_-)\) and \(Q(D_T\to \mathcal G_-)\) possessing the following properties:
\[ \begin{aligned} 1)\quad &(Pf_2,g_2)_0=(f_2,T^*g_2)_0,\qquad (Qf_1,g_1)_0=(f_1,Tg_1)_0\\ &\qquad (f_1,g_1\in D_T,\ f_2,g_2\in D_{T^*});\\ 2)\quad &(Pf_2,g_1)_0=(f_2,Qg_1)_0\qquad (g_1\in D_T,\ f_2\in D_{T^*});\\ 3)\quad &P\varphi=T\varphi,\qquad Q\varphi=T^*\varphi\qquad (\varphi\in D_{T_0}=D_T\cap D_{T^*}). \end{aligned} \tag{7} \]
Moreover,
\[ \begin{aligned} T_{\mathcal G_+}f&=Tf_1+Pf_2,\\ T_{\mathcal G_+}^{\times}f&=Qf_1+T^*f_2 \end{aligned} \qquad (f=f_1+f_2,\ f_1\in D_{T^*},\ f\in \mathcal G_+). \tag{8} \]
Theorem 2. There do not exist two distinct generalized extensions \(T_{\mathcal G_+}\) and \(T'_{\mathcal G_+}\) of the operator
\[ Tf=\frac{1}{i}\frac{df}{dx}\qquad (f(0)=0,\ 0\leq x\leq \eta) \]
on \(\mathcal G_+=D_{T_0^*}\), for which the real parts would be extensions of the self-adjoint operator
\[ A_0\psi=\frac{1}{i}\frac{d\psi}{dx}\qquad (\psi(0)+\theta\psi(\eta)=0,\ |\theta|=1). \]
Proof. Let
\[ T_{\mathcal G_+}=A+iB,\qquad T'_{\mathcal G_+}=A'+iB', \tag{9} \]
where \(A=(T_{\mathcal G_+}+T_{\mathcal G_+}^{\times})/2\), \(A'=(T'_{\mathcal G_+}+T_{\mathcal G_+}^{\prime\times})/2\), etc.
From theorem (1) it follows that
\[ T_{\mathfrak G_{+}}f=Tf_1+Pf_2,\qquad T'_{\mathfrak G_{+}}f=Tf_1+P'f_2, \]
\[ T_{\mathfrak G_{+}}^{\times}f=Qf_1+T^*f_2,\qquad T_{\mathfrak G_{+}}^{\prime\times}f=Q'f_1+T^*f_2. \tag{10} \]
From (9) and (10) it follows that
\[ Af=\tfrac12(Tf_1+T^*f_2+Qf_1+Pf_2), \]
\[ (f=f_1+f_2,\ f_1\in D_T,\ f_2\in D_{T^*}), \tag{11} \]
\[ A'f=\tfrac12(Tf_1+T^*f_2+Q'f_1+P'f_2). \]
In (5) it was shown that \(T_0^*f=Tf_1+T^*f_2\) \((f=f_1+f_2,\ f_1\in D_T,\ f_2\in D_{T^*})\). Therefore, taking (11) into account, we obtain
\[ Af=\tfrac12(T_0^*f+Qf_1+Pf_2),\qquad A'f=\tfrac12(T_0^*f+Q'f_1+P'f_2). \tag{12} \]
Suppose now that \(A\) and \(A'\) are extensions of the self-adjoint operator \(A_0\), i.e. \(Af_0=A_0f_0,\ A'f_0=A_0f_0\) \((f_0\in D_{A_0})\). Since \(T_0\subset A_0\subset T_0^*\), we have
\[ A_0f_0=\tfrac12(A_0f_0+Qf_1^0+Pf_2^0), \]
\[ A_0f_0=\tfrac12(A_0f_0+Q'f_1^0+P'f_2^0), \qquad (f_0=f_1^0+f_2^0,\ f_1^0\in D_T,\ f_2^0\in D_{T^*},\ f_0\in D_{A_0}). \]
From the last relations it follows that
\[ (Q-Q')f_1^0=-(P-P')f_2^0. \tag{13} \]
From equalities (7) and (13) it follows that
\[ ((Q-Q')f_1^0,g_1)_0=-((P-P')f_2^0,g_1)_0=0, \]
\[ ((P-P')f_2^0,g_2)_0=-((Q-Q')f_1^0,g_2)_0=0. \]
Hence
\[ ((Q-Q')f_1^0,g_1)_0=0,\qquad ((Q-Q')f_1^0,g_2)_0=0, \]
therefore,
\[ ((Q-Q')f_1^0,g)_0=0 \qquad (g=g_1+g_2,\ g_1\in D_T,\ g_2\in D_{T^*}) \]
and, consequently, \((Q-Q')f_1^0=0\). Similarly, \((P-P')f_2^0=0\). From (8) and (12) it follows that \(T_{\mathfrak G_{+}}f_0=T'_{\mathfrak G_{+}}f_0,\ Af_0=A'f_0\).
Thus, we have proved that the generalized extensions \(T_{\mathfrak G_{+}}\) and \(T'_{\mathfrak G_{+}}\) of the operator \(T\) coincide on the subspace \(D_{A_0}\) under the condition that their real parts are extensions of the self-adjoint operator \(A_0\). We shall now show that \(T_{\mathfrak G_{+}}\) and \(T'_{\mathfrak G_{+}}\) coincide on the whole space \(\mathfrak G_{+}\).
Let \(f_1^0\) be an arbitrary function from \(D_T\). Then, as is easy to see, one can choose a function \(f_2^0\in D_{T^*}\) such that \(f_0=f_1^0+f_2^0\) belongs to the set \(D_{A_0}\). From (9) it follows that \(T_{\mathfrak G_{+}}^{\times}=A-iB,\ T_{\mathfrak G_{+}}^{\prime\times}=A'-iB'\). Therefore
\[ Tf_1^0=Af_1^0+iBf_1^0,\qquad Tf_1^0=A'f_1^0+iB'f_1^0, \]
\[ T^*f_2^0=A'f_2^0-iB'f_2^0,\qquad T^*f_2^0=Af_2^0-iBf_2^0. \tag{14} \]
From relations (14) it follows that
\[ Tf_1^0+T^*f_2^0=A(f_1^0+f_2^0)+iB(f_1^0-f_2^0), \]
\[ Tf_1^0+T^*f_2^0=A'(f_1^0+f_2^0)+iB'(f_1^0-f_2^0); \]
whence it follows immediately that
\[ (A-A')f_0+i(B-B')(f_1^0-f_2^0)=0 \qquad (f_0=f_1^0+f_2^0,\ f_0\in D_{A_0}). \tag{15} \]
Since, as we have shown, \(T_{\mathfrak g_+} f_0=T'_{\mathfrak g_+} f_0\), it already follows easily that \(Bf_0=B'f_0,\ Af_0=A'f_0\). Therefore from (15) \((B-B')(f_1^0-f_2^0)=0\), i.e.,
\[ (B-B')f_1^0=(B-B')f_2^0. \tag{16} \]
Since \(f_1^0+f_2^0=f_0\), we have
\[ (B-B')f_1^0+(B-B')f_2^0=0. \tag{17} \]
From (16) and (17) we obtain that \((B-B')f_1^0=0\) for an arbitrary function \(f_1^0\in D_T\). Similarly one proves that \((B-B')f_2^0=0\) for an arbitrary function \(f_2^0\in D_{T^*}\). Hence it follows that \((B-B')f=0\) \((f\in\mathfrak G_+)\). Since, when the imaginary components coincide, \(B=B'\), the generalized extensions \(T_{\mathfrak g_+}\) and \(T'_{\mathfrak g_+}\), as is known (5), also coincide, it follows that \(T_{\mathfrak g_+}=T'_{\mathfrak g_+}\). The theorem is proved.
Theorem 3. In order that a generalized extension \(T_{\mathfrak g_+}\) of the operator
\[ Tf=\frac{1}{i}\frac{df}{dx}\quad (f(0)=0,\ 0\leq x\leq \eta) \]
have a one-dimensional imaginary component, it is necessary and sufficient that the real part \((T_{\mathfrak g_+}+T_{\mathfrak g_+}^{\times})/2\) be an extension of some self-adjoint operator \(A_0\) which is an extension of the operator
\[ T_0f=\frac{1}{i}\frac{df}{dx}\quad (f(0)=f(\eta)=0,\ 0\leq x\leq \eta). \]
Proof. It is known (3) that all self-adjoint extensions of the operator \(T_0\) have the form
\[ T(\theta)f=\frac{1}{i}\frac{df}{dx}\quad (f(0)+\theta f(\eta)=0,\ |\theta|=1,\ 0\leq x\leq \eta). \tag{18} \]
Let now \(T_{\mathfrak g_+}\) be some generalized extension of the operator \(T\), for which \((T_{\mathfrak g_+}+T_{\mathfrak g_+}^{\times})/2\) is an extension of the operator \(T(\theta)\) for fixed \(\theta\) \((|\theta|=1)\). From (4) it follows that \(\xi\) can be chosen so that \((T_{\mathfrak g_+}(\xi)+T_{\mathfrak g_+}^{\times}(\xi))/2\) will also be an extension of \(T(\theta)\) for the same \(\theta\). By virtue of Theorem 2, \(T_{\mathfrak g_+}\) and \(T_{\mathfrak g_+}^{\times}(\xi)\) must coincide. From (6) it follows that \(T_{\mathfrak g_+}\) has a one-dimensional imaginary component. Sufficiency is proved.
Let now \(T_{\mathfrak g_+}\) be a generalized extension of the operator \(T\) having a one-dimensional imaginary component, i.e.,
\[ T_{\mathfrak g_+}=A+i(\,\cdot\,,e_0)_0Je_0\quad (A=A^{\times},\ J=\pm1). \tag{19} \]
From (19) it follows that
\[ T f_1=Af_1+i(f_1,e_0)_0Je_0,\quad T^* f_2=Af_2-i(f_2,e_0)_0Je_0,\quad (f_1\in D_T,\ f_2\in D_{T^*}), \]
hence
\[ Af=T_0^*f-i(f_1-f_2,e_0)_0Je_0\quad (f=f_1+f_2,\ f_1\in D_T,\ f_2\in D_{T^*}). \tag{20} \]
It is not difficult to show that for any \(f_1\in D_T\) one can indicate such an \(f_2\in D_{T^*}\) that \((f_1,e_0)_0=(f_2,e_0)_0\). Therefore the operator \(A\), as is seen from (20), will act on vectors \(f=f_1+f_2\) \(((f_1,e_0)_0=(f_2,e_0)_0)\) as an operator in the space \(\mathfrak G_0\) and will be a symmetric extension of \(T_0\). Consequently, as is known (3), it must coincide with one of the operators \(T(\theta)\) (see (18)). The theorem is proved.
Corollary. The totality of generalized extensions with one-dimensional imaginary component of the differentiation operator \(T\) (3) consists of the family (4).
Let us note in conclusion that all generalized extensions (4) in the space \(\mathfrak G\) also have no spectrum in the finite part of the plane (see (5)).
Donetsk State University
Received
6 XII 1966
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