Reports of the Academy of Sciences of the USSR

1. Volume 155, No. 2

MATHEMATICS

Yu. I. Lyubich

ON DENSITY CONDITIONS FOR THE INITIAL MANIFOLD

FOR AN ABSTRACT CAUCHY PROBLEM

(Presented by Academician S. N. Bernstein on 20 XI 1963)

Consider the abstract Cauchy problem

\[ \frac{dx(t)}{dt}=Ax(t)\quad (0\leq t<\infty),\qquad x(0)=x_0, \tag{1} \]

with a linear operator \(A\) in a Banach space. Denote by \(I_A\) the linear manifold of those initial vectors \(x_0\) for which there exists a smooth (in the strong sense) solution of problem (1). We shall be interested in the question under what conditions the manifold \(I_A\) is dense. That this is not always the case is shown by the following simple theorem.

Theorem 1. Let \(A\) be a symmetric operator in a Hilbert space and let \(\operatorname{def} A>0\) in the lower half-plane. Then the manifold \(I_{iA}\) is orthogonal to all eigenvectors of the operator \(A^*\) corresponding to eigenvalues \(\lambda=\alpha+\beta i\) \((\beta>0)\).

Proof. Let \(x_0\in I_{iA}\), and let \(x(t)\) be the corresponding solution. Further, let \(A^*u=\overline{\lambda}u\), \(u\ne0\), \(\operatorname{Im}\lambda<0\). Then

\[ \frac{d}{dt}(x(t),u)=i(Ax(t),u)=i\lambda (x(t),u) \]

and, consequently,

\[ (x(t),u)=(x_0,u)e^{i\lambda t}\quad (0\leq t<\infty). \tag{2} \]

Since \(\|x(t)\|=\mathrm{const}\), by the symmetry of the operator \(A\), it follows from (2), as \(t\to\infty\), that \((x_0,u)=0\).

From this theorem and a theorem of M. G. Krein \((^1)\) there follows

Corollary. If the operator \(A\), satisfying the conditions of Theorem 1, is maximal and simple, then \(I_{iA}=0\).

Denote by \(\Gamma\) a curve in the plane of the complex variable \(\lambda\), given by the equation \(\lambda=\lambda(s)\) \((-\infty<s<\infty)\), where \(\lambda(s)\) is a piecewise smooth function satisfying the following conditions: 1) \(\lambda'(s\pm0)\ne0\) \((-\infty<s<\infty)\); 2) \(\lambda(s_1)\ne\lambda(s_2)\) \((s_1\ne s_2)\); 3) \(|\lambda(s)|\to\infty\) \((|s|\to\infty)\); 4) \(|\lambda(s)|\) is a strictly monotone function in neighborhoods of the values \(s=\pm\infty\); 5)

\[ \sup_{-\infty<s<\infty}\operatorname{Re}\lambda(s)<\infty . \]

Let \(\alpha\) be such that the half-plane \(\operatorname{Re}\lambda<\alpha\) contains the entire curve \(\Gamma\). Denote by \(\Pi_-\) the connected component of the complement of \(\Gamma\) lying in the indicated half-plane. Denote the second component by \(\Pi_+\).

Our constructions are based on the following simple lemmas, the proofs of which we omit.

Lemma 1. Let the operator \(A\) have no spectrum on the line \(\Gamma\), and let a continuous scalar function \(\varphi(\lambda)\) \((\lambda\in\Gamma)\) be such that

\[ \int_{\Gamma}\|\lambda\varphi(\lambda)R_\lambda\|\,|d\lambda|<\infty . \tag{3} \]

where \(R_\lambda\), as usual, is the resolvent of the operator \(A\). If, in addition,

\[ \int_\Gamma \varphi(\lambda)e^{\lambda t}\,d\lambda=0\qquad (0\leq t<\infty), \tag{4} \]

then the range of the operator

\[ S_\varphi=\int_\Gamma \varphi(\lambda)R_\lambda\,d\lambda \]

is contained in the manifold \(I_A\).

Lemma 2. Let \(\varphi(\lambda)\) \((\lambda\in\Gamma)\) be a summable* scalar function locally satisfying the Lipschitz–Hölder condition. In order that (4) hold, it is necessary that \(\varphi(\lambda)\) admit an extension \(\hat\varphi(\lambda)\) \((\lambda\in\overline{\Pi}_-)\), analytic in \(\Pi_-\) and continuous in \(\overline{\Pi}_-\). This is also sufficient under the additional condition

\[ \lim_{|\lambda|\to\infty}\hat\varphi(\lambda)=0\qquad (\lambda\in\Pi_-). \]

We shall now subject the curve \(\Gamma\) to one more restriction, namely, we shall require that for some \(p>1\)

\[ \int_\Gamma \frac{|d\lambda|}{1+|\lambda|^p}<\infty. \]

Theorem 2. Let the operator \(A\), with dense domain of definition, have no spectrum on the curve \(\Gamma\) and in the domain \(\Pi_+\), and let

\[ \|R_\lambda\|=O(1+|\lambda|^m)\qquad (\lambda\in\overline{\Pi}_+) \]

with some \(m\geq 0\). Then the manifold \(I_A\) is dense.

For the proof it is enough to put

\[ \varphi(\lambda)=\frac{1}{2\pi i(\lambda-a)^{[m+p+2]}}. \]

In this case the operator \(S_\varphi\) is readily computed by the usual methods of the theory of residues and turns out to be equal to \(R_a^{[m+p+2]}\). Consequently, its range coincides with the domain of definition of the operator \(A^{[m+p+2]}\), and the latter is dense (see, for example, \((^2)\), p. 290).

Certain assumptions about the structure of the operator \(A\) make it possible to substantially weaken the growth restriction on the resolvent.

Theorem 3. Let the operator \(A\), with dense domain of definition, have no spectrum on the line \(\Gamma\) and in the domain \(\Pi_+\), and suppose that

\[ \int_\Gamma \frac{\ln_+\|R_\lambda\|}{1+\omega^2(\lambda)}\,|d\omega(\lambda)|<\infty, \tag{5} \]

where \(\omega(\lambda)\) is a function conformally mapping the domain \(\Pi_-\) onto the upper half-plane \(\operatorname{Im}\omega>0\). Suppose that the function \(\omega(\lambda)\) on \(\Gamma\) locally satisfies the Lipschitz–Hölder condition, and that the inverse function \(\lambda(\omega)\) has the same property on the real axis. If the operator \((R_{\lambda_0})^*\), for some \(\lambda_0\in\overline{\Pi}_+\), has no invariant subspaces \(\mathfrak L\ne 0\) on which its spectrum is concentrated at zero, then the manifold \(I_A\) is dense.

We indicate the proof of Theorem 3.

Mapping the domain \(\Pi_-\) onto the upper half-plane and using the Poisson integral, one can, by virtue of (5), construct a function \(\varphi(\lambda)\) \((\lambda\in\overline{\Pi}_-)\), holomorphic in the domain \(\Pi_-\) and continuous in \(\overline{\Pi}_-\), so that condition (3) is fulfilled. Moreover, one can arrange that the following condi-

\[ \text{* This property holds automatically when (3) is fulfilled, since it is not hard to prove that } \lim_{|\lambda|\to\infty}\|\lambda R_\lambda\|\geq 1. \]

conditions: a) \(\varphi(\lambda)\ne 0\) \((\lambda\in \Pi_-)\); b) \(\lim_{|\lambda|\to\infty}\varphi(\lambda)=0\) \((\lambda\in \Pi_-)\); c) the boundary values of the function \(\varphi(\lambda)\) locally satisfy the Lipschitz–Hölder condition.

According to Lemmas 1 and 2, the range of the operator \(S_\varphi\) is contained in \(I_A\). But all essential properties of the function \(\varphi(\lambda)\) are preserved under multiplication by \(e^{k\lambda}\) \((k\ge 0)\). Therefore it remains to prove that the ranges of the operators \(S(k)\equiv S_{\varphi_k}\) \((\varphi_k(\lambda)=\varphi(\lambda)e^{k\lambda})\) form a complete system.

Let \(\mathscr L\) be the subspace of those linear functionals \(f\) for which \(f(S(k)x)=0\) for all \(k\ge 0\) and all \(x\), i.e.

\[ \int_\Gamma \varphi(\lambda) f(R_\lambda x)e^{k\lambda}\,d\lambda=0. \]

By Lemma 2, the function \(\varphi(\lambda)f(R_\lambda x)\), and hence also \(f(R_\lambda x)\) for any \(x\), is analytically continued into the domain \(\Pi_-\). But \(f(R_\lambda x)\) is regular also in \(\overline{\Pi}_+\). Therefore \(f(R_\lambda x)\) is an entire function. Thus \((R_\lambda)^* f\) turns out to be an entire function for any \(f\in\mathscr L\). But then the subspace \(\mathscr L\), if it is nonzero, turns out to be endowed with precisely those properties which are forbidden by the formulation of the theorem. Consequently, \(\mathscr L=0\).

Let us note some particular cases of Theorem 3. If \(\Gamma\) is the imaginary axis, then condition (5) takes the form

\[ \int_{-\infty}^{\infty}\frac{\ln_+\|R_{i\omega}\|}{1+\omega^2}\,d\omega<\infty. \]

If \(\Gamma\) is the boundary of the angle
\[ (1-q)\pi\le \arg\lambda\le (1+q)\pi\quad (0<q\le 1/2), \]
then condition (5) takes the form

\[ \int_0^\infty \frac{\ln_+\|R_{-\omega e^{-iq\pi}}\|+\ln_+\|R_{-\omega e^{iq\pi}}\|} {1+\omega^{1+1/2q}}\,d\omega<\infty. \]

Finally, if \(\Gamma\) is the boundary of the half-strip \(|\operatorname{Im}\lambda|\le h,\ \operatorname{Re}\lambda\le 0\), then condition (5) takes the form

\[ \int_0^\infty \ln_+\|R_{-\tau\pm ih}\|\,e^{-\pi\tau/2h}\,d\tau<\infty. \]

This series of conditions shows that the restriction on the growth of the resolvent weakens as the spectrum approaches the negative half-axis. The general character of this observation is clear from the results of S. E. Varshavskii \((^3)\), concerning the asymptotic behavior of conformal mappings.

The following theorem is proved analogously to Theorem 3.

Theorem 4. Suppose that all the conditions of Theorem 3 are satisfied, except the one concerning the operator \((R_\lambda)^*\). If, in addition, the spectrum of the operator \(A\) is nonempty, then \(I_A\ne 0\).

The condition that the spectrum be nonempty is essential, as is shown by the following example, “dual” to the example constructed in \((^4)\).

Let \(A\) be the operator in \(\mathscr L_\rho^2(0,\infty)\) defined by the differentiation operation \(d/ds\) and the boundary condition

\[ \xi(0)=0. \tag{6} \]

Then \(I_A=0\), since if the function \(\xi(s+t)\) \((s,t\ge 0)\) satisfies condition (6) for all \(t\ge 0\), then it is identically equal to zero.

Let us subject the weight \(\rho(s)\) to the conditions: 1) \(\rho(s_1+s_2)\le \rho(s_1)\rho(s_2)\) \((s_1,s_2\ge 0)\); 2) \(\sqrt{\rho(s)}\,e^{ks}\in\mathscr L(0,\infty)\) for all \(k\ge 0\). One may take, for example, \(\rho(s)=e^{-s^2}\). Then the operator \(A\) will have no spectrum.

Incidentally, the resolvent of the operator \(A\) constructed is estimated in the following way:

\[ \|R_\lambda\| \leq \int_0^\infty \sqrt{\rho(s)}\, e^{ks}\, ds \qquad (k=\operatorname{Re}\lambda), \]

whence it is clear that the conditions of Theorem 2 cannot be weakened so that the growth of the resolvent is restricted only on the line \(\Gamma\).

Kharkov State University
named after A. M. Gorky

Received
20 XI 1963

References

\(^{1}\) M. G. Krein, Ukr. Mat. Zh., No. 2, 3 (1949).
\(^{2}\) E. Hille, Functional Analysis and Semigroups, IL, 1951.
\(^{3}\) S. E. Varshavskii, Sbornik of Translations. Mathematics, 2, 4, 67 (1958).
\(^{4}\) Yu. I. Lyubich, DAN, 130, No. 5, 969 (1960).