Abstract
Full Text
UDC 513.88 : 513.83
MATHEMATICS
Yu. I. LYUBICH, V. A. TKACHENKO
A QUASIANALYTICITY CRITERION
FOR ABSTRACT OPERATORS
(Presented by Academician V. I. Smirnov, January 7, 1969)
The present note is an attempt to extend Carleman’s quasianalyticity criterion \((^{1,2})\) to linear operators in a Banach space. We shall be concerned with conditions under which from inequalities of the form
\[ \|A^n x\|\le m_n\|x\|\qquad (n=0,1,2,\ldots) \tag{1} \]
it follows that \(x=0\). Classical quasianalyticity corresponds to the case when \(A=\frac1i\frac{d}{ds}\) in the space \(C[0,1]\) with the boundary condition \(x(0)=0\). For more general differential operators the problem of quasianalyticity was studied in \((^{3-5})\).
The differentiation operator mentioned above is unbounded and has no spectrum. These properties should be retained in the general formulation of the problem: if on some nonzero invariant subspace \(E\) the operator \(A\) is bounded, then \(\|A^n x\|\le C^n\|x\|\) \((x\in E,\ C=\|A|E\|)\); thus, even with \(m_n=C^n\), inequality (1) will be satisfied by a vector \(x\ne0\).
If the operator \(A\) has no spectrum, then its resolvent \(R(\lambda)\) is an entire function, and it is natural to relate the properties of the operator to the growth of \(R(\lambda)\) as \(|\lambda|\to\infty\). For the differentiation operator the function \(R(\lambda)\) is of exponential type, bounded on the imaginary axis. We shall take this observation as our starting point.
Let \(A\) be an operator without spectrum. The Taylor series of its resolvent has the form
\[ R(\lambda)=\sum_{k=0}^{\infty} A^{-k-1}\lambda^k . \]
Put \(M_A(r)=\max_{|\lambda|=r}\|R(\lambda)\|\) \((r>0)\). The function \(M_A(r)\) as \(r\to\infty\) grows faster than any power of \(r\). Introduce the usual growth characteristic of the sequence \(\{m_n\}\):
\[ T(r)=\sup_n \frac{r^{\,n-1}}{m_n}. \]
Lemma 1. If there exists a vector \(x\ne0\) for which inequalities (1) hold, then
\[ T(r)\le M_A(r). \tag{2} \]
Proof. We take \(\|x\|=1\). Then
\[ 1=\|A^{-n}A^n x\|\le \|A^{-n}\|m_n\le M_A(r)r^{-n+1}m_n \tag{3} \]
by Cauchy’s inequality. Inequality (2) follows from (3).
In view of Lemma 1, from inequalities (1) it will follow that \(x=0\) whenever the function \(T(r)\), in some scale, grows faster than \(M_A(r)\). For example:
Theorem 1. If for some \(\rho>0\)
\[ \int_1^\infty \frac{\ln T(r)}{r^{1+\rho}}\,dr=\infty, \tag{4} \]
while
\[ \int_1^\infty \frac{\ln M_A(r)}{r^{1+\rho}}\,dr<\infty, \tag{5} \]
then from inequality (1) it follows that \(x=0\).
Condition (4) means that the sequence \(\{m_n\}\) satisfies the classical criterion of quasianalyticity “of order \(\rho\)” (see (6), pp. 55–56). Theorem 1 is sharp in the following sense.
Theorem 2. Let \(\psi(t)\) be a positive logarithmically convex function. Suppose that for some \(\rho>0\)
\[ \int_1^\infty \frac{\ln \psi(t)}{t^{1+\rho}}\,dt=\infty. \tag{6} \]
Then there exists an operator \(A\) in a Hilbert space, having no spectrum, and such that: 1) \(\|R(\lambda)\|\leq \psi(|\lambda|)\); 2) for some \(x\ne0\) the sequence \(m_n=\|A^n x\|\) \((n=0,1,2,\ldots)\) satisfies the quasianalyticity condition (4).
The proof is based on the existence of sequences \(\{a_n\}_0^\infty\) and \(\{m_n\}_{-\infty}^\infty\) of positive numbers for which
\[ \sum_{n=0}^\infty a_n^{-1} t^n \leq \psi(t)\quad (t\geq0), \]
\[ a_k m_n \leq m_{k+n+1}\quad (n=0,\pm1,\pm2,\ldots;\ k=0,1,2,\ldots) \]
and \(\{m_n\}_0^\infty\) satisfies condition (4). In the Hilbert space of sequences \(\xi=\{\xi_n\}_{-\infty}^\infty\) with norm
\[ \|\xi\|=\left(\sum_{n=-\infty}^{\infty}|\xi_n|^2 m_n^2\right)^{1/2} \]
the right-shift operator \(A\xi=\{\xi_{n-1}\}_{n=-\infty}^{\infty}\) possesses all the required properties. As \(x\) one should take the coordinate unit vector \(\{\delta_{n0}\}\).
We now formulate the main criterion of quasianalyticity for order \(\rho\leq1\).
Theorem 3. Let the operator \(A\) have no spectrum and
\[ \sigma=\varlimsup_{r\to\infty} r^{-\rho}\ln M_A(r)<\infty\quad (0<\rho\leq1). \]
Suppose further that on two rays \(\arg\lambda=\theta_1\), \(\arg\lambda=\theta_2\), \(\theta_2-\theta_1\equiv \pi\rho^{-1}\pmod{2\pi}\), the estimate \(\|R(\lambda)\|\leq\psi(|\lambda|)\) holds, where \(\psi(t)\) \((t\geq0)\) is a positive continuous nondecreasing function for which
\[ \int_1^\infty \frac{\ln \psi(t)}{t^{1+\rho}}\ln t\,dt<\infty. \tag{7} \]
Then, in order that \(x=0\) follow from inequality (1), it is sufficient, and for \(\sigma>0\) necessary, that the quasianalyticity condition (4) be fulfilled.
According to Theorem 2, the quasianalyticity criterion (4) is lost when the condition
\[ \int_1^\infty \frac{\ln \psi(t)}{t^{1+\rho}}\,dt<\infty, \tag{8} \]
weaker than (7). However, we do not know whether condition (7) in Theorem 3 can be replaced by condition (8).
Let us also note that condition (4) in the case \(\sigma=0\) is not necessary. The proof of Theorem 3 uses the fact (see (7)) that, for \(\rho=1,\ \arg\lambda=0\), the growth of the resolvent on the real axis can be absorbed by multiplication by a function of the form
\[ \varphi(\lambda)=\int_0^a e^{i\lambda t}g(t)\,dt \qquad (a>0). \]
Owing to this, one can form functions of \(A\) of the form
\[ \widetilde{\omega}(t;A)=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-i\lambda t}\varphi(\lambda)R(\lambda)\omega(\lambda)\,d\lambda, \]
where \(\omega(\lambda)\) \((-\infty<\lambda<\infty)\) may have arbitrary power growth.
In proving sufficiency (in the case \(\rho=1\)) we take \(\omega(\lambda)=1\) and establish that, by virtue of (1),
\[ \left\|\frac{d^n}{dt^n}\widetilde{\omega}(t;A)x\right\|\leq \operatorname{const}\cdot m_n\|x\|. \]
Then Carleman’s theorem implies \(\widetilde{\omega}(t;A)x=0\) \((t<0)\). Reversing (9), we obtain the representation
\[ R(\lambda)x=\frac{1}{\varphi(\lambda)}\int_0^b e^{i\lambda t}\widetilde{\omega}(t;A)x\,dt, \]
where \(b>0\). By the theorem on addition of indicators ([8], p. 208), the function \(R(\lambda)x\) in the upper half-plane has zero exponential type. An analogous conclusion can be obtained for the lower half-plane. After this, owing to condition (7), the matter reduces to Theorem 1.
In proving necessity, the role of \(\omega(\lambda)\) is played by the Fourier transform of a finite function from the corresponding non-quasianalytic class. The required vector \(x\ne0\), for which the estimates (1) hold, is obtained in the form \(\widetilde{\omega}(t;A)y\) for some \(y\) and some \(t\).
If the operator \(A\) has spectrum, the problem of quasianalyticity disappears, but the problem of completeness arises.
Theorem 4. Let the resolvent \(R(\lambda)\) be meromorphic. Put
\[ m_A(r)=\frac{1}{2\pi}\int_0^{2\pi}\ln\|R(re^{i\theta})\|\,d\theta,\qquad \sigma=\overline{\lim_{r\to\infty}}\, r^{-\rho}m_A(r)\qquad (0<\rho\leq1). \]
Suppose that \(\sigma<\infty\) and that, on the two rays \(\arg\lambda=\theta_1,\ \arg\lambda=\theta_2\), the conditions of Theorem 3 are satisfied. Suppose, further, that the quasianalyticity condition (4) is satisfied. Then the linear span of the root vectors of the operator \(A\) is dense in the set of vectors satisfying inequalities (1).
This theorem is proved by the same method as Theorem 3. We emphasize that the case \(\rho>1\) beyond the scope of Theorems 1, 2 remains uninvestigated.
Kharkov State University
named after A. M. Gorky
Kharkov Polytechnic Institute
named after V. I. Lenin
Received
19 XII 1968
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