UDC 517.91 : 517.944

On Certain Properties of Solutions of Linear Differential Equations with Bounded Coefficients

O. S. Ivankovskaya

Recently a number of papers have appeared in which the solution of an ordinary differential equation on the line or on a half-line is investigated in the following sense: it is assumed that the right-hand side and the solution itself possess a boundedness property in one sense or another, and it is proved that all derivatives of the solution up to the \(n\)-th inclusive (\(n\) is the order of the equation) possess the same property.

The first result of this kind was the well-known lemma of Esclangon, concerning linear differential equations with constant coefficients.

In [3] Yu. I. Lyubich obtained a result analogous to Esclangon’s lemma, replacing the linear differential operator with constant coefficients by an arbitrary polynomial with constant coefficients in a linear, generally speaking unbounded, operator of a very general nature.

M. A. Rutman [2] generalized Esclangon’s lemma and developed it in other directions.

In the present paper the results of Yu. I. Lyubich are extended to polynomials of a linear operator with vector coefficients. We use a method analogous to the analytic technique of M. A. Rutman from [2].

Applications of the theorems proved to certain types of linear differential equations are considered.

1. In the present paper, by \(S\) we shall denote a linear system which is at the same time a (commutative) ring.

Let for any two elements \(x\) and \(y \in S\) there be defined a sequence of scalar products \(\{(x,y)_k\}\) \((k=1,2,\ldots,n,\ldots)\). We assume that the following conditions are satisfied:

\[ \begin{aligned} &1)\quad (\lambda_1 x_1+\lambda_2 x_2,\,y)_k =\lambda_1(x_1,y)_k+\lambda_2(x_2,y)_k;\\ &2)\quad (x,y)_k=\overline{(y,x)}_k;\\ &3)\quad (x,x)_{k+1}\ge (x,x)_k\ge 0,\quad \text{and, beginning with some } k_0=k_0(x),\\ &\qquad (x,x)_k>0 \text{ for all } x\ne 0;\\ &4)\quad (xy,xy)_k\le c\,(x,x)_k\,(y,y)_k,\quad \text{beginning with some } k_1 \\ &\qquad (c \text{ is a constant for the given ring}). \end{aligned} \]

Thus to each element \(x\) of \(S\) there is assigned a monotonically nondecreasing sequence of “quasinorms”:

\[ \|x\|_1,\ \|x\|_2,\ \ldots,\ \|x\|_k,\ \ldots;\qquad \|x\|_k=\sqrt{(x,x)_k}. \]

The limit of this sequence (finite or infinite) will be called the norm of the element \(x\): \(\|x\|=\lim_{k\to\infty}\|x\|_k\). Since the scalar product has properties 1–4, the following holds:

a) \(\|\lambda x\|_k=|\lambda|\,\|x\|_k\) for any \(k\);

b) \(|(x,y)_k|\leqslant \|x\|_k\|y\|_k\) for all \(k>\min\{k_0(x),\,k_0(y)\}\) (the Cauchy—Bunyakovsky inequality);

c) \(\|x+y\|_k\leqslant \|x\|_k+\|y\|_k\) for all \(k>\min\{k_0(x),\,k_0(y)\}\) (the triangle inequality);

d) \(\|xy\|_k\leqslant c\|x\|_k\|y\|_k\) for \(k\geqslant k_1\).

An example of a system \(S\) may be the set whose elements are numerical sequences
\(x=(x_1,x_2,\ldots,x_n,\ldots)\), if the product of two elements
\(x=(x_1,x_2,\ldots,x_n,\ldots)\) and \(y=(y_1,y_2,\ldots,y_n,\ldots)\) is taken to be the element
\(xy=(x_1y_1,x_2y_2,\ldots,x_ny_n,\ldots)\). To each pair of elements
\(x=(x_1,x_2,\ldots,x_n,\ldots)\) and \(y=(y_1,y_2,\ldots,y_n,\ldots)\) we assign the sequence of scalar products
\(\{(xy)_k=\sum_{i=1}^{k}x_i\overline{y_i}\}\). For this sequence, properties 1—4 are valid. The norm of the element
\(x=(x_1,x_2,\ldots,x_n,\ldots)\) is the limit

\[ \|x\|\lim_{k\to\infty}\sqrt{\sum_{i=1}^{k}|x_i|^2}. \]

In a number of important cases, for example for the ring of functions \(f(x)\), defined for
\(-\infty<x<\infty\), bounded and measurable on every finite interval, one can introduce another, more general, system of axioms, which considerably broadens the range of application of the theorems proved below.

Let the system \(S'\) under consideration be such that in it there is still given a sequence of scalar products \(\{(x,y)_k\}\) with conditions 1—3. Properties a—c remain valid for it.

Introduce into \(S'\) still another sequence of quasinorms:
\([x]_1,[x]_2,\ldots,[x]_k,\ldots\), satisfying the conditions:

\(1')\) \([x]_{k+1}\geqslant [x]_k;\quad [x]_k\geqslant 0\), but, starting from some \(k_2\), \([x]_k>0\) for all \(x\ne0\);

\(2')\) \([\lambda x]_k=|\lambda|[x]_k;\)

\(3')\) \([x+y]_k\leqslant [x]_k+[y]_k;\) and a second norm
\([x]=\lim_{k\to\infty}[x]_k\).

At the same time, instead of 4) we assume that the following condition is satisfied:

\(4')\) \(\|xy\|_k\leqslant [x]_k\|y\|_k\).

An example of \(S'\) is the set of functions, defined for \(-\infty<x<\infty\), bounded and measurable on every finite interval.

Put

\[ (f,g)_k=\int_{a_k}^{b_k} f(x)\overline{g(x)}\,dx, \quad \text{where}\quad \lim_{k\to\infty}a_k=-\infty,\quad \lim_{k\to\infty}b_k=+\infty . \]

Next put

\[ [f]_k=\operatorname{Sup}_{a_k\leqslant x\leqslant b_k}|f(x)|. \]

It is easy to verify that properties 1—3 of the scalar product are valid. Properties \(1'—3'\) also, obviously, hold. Let us verify condition \(4'\):

\[ \|fg\|_k= \sqrt{\int_{a_k}^{b_k}|f(x)g(x)|^2\,dx} \leqslant \operatorname{Sup}_{a_k\leqslant x\leqslant b_k}|f(x)| \times \]

\[ \times \sqrt{\int_{a_k}^{b_k}|g(x)|^2\,dx} = [f]_k\|g\|_k. \]

Thus all the conditions are fulfilled, and this set forms a system \(S'\). If
\(\lim_{k\to\infty}\|x\|_k<\infty\), then such an element of the system \(S\) or \(S'\) will be called bounded.

Consider the subset \(M\) of all bounded elements of the ring \(S\) (or \(S'\)). This set is a linear system, since from \(\|x+y\|_k \le \|x\|_k+\|y\|_k\) it follows that if \(x\in M\) and \(y\in M\), then \(x+y\in M\), and, further, from \(\|\lambda x\|_k=|\lambda|\,\|x\|_k\) it follows that if \(x\in M\), then \(\lambda x\in M\) (\(\lambda\) is a scalar).

In the first of the examples considered, the set \(M\) consists of those sequences for which

\[ \lim_{k\to\infty}\sqrt{\sum_{i=1}^{k}|x_i|^2}<\infty, \]

and in the second, of those functions for which

\[ \lim_{k\to\infty}\sqrt{\int_{a_k}^{b_k}|f(x)|^2\,dx}<\infty. \]

The set of elements of the system \(S'\) for which \(\lim_{k\to\infty}[x]_k<\infty\) will be denoted by \(M'\).

\(2^\circ\). Let \(T\) be a homogeneous and additive operator acting in \(S\), and let \(D\) be the domain of definition of \(T\) and of its powers up to the \(n\)-th inclusive \((D\subset S)\).

Theorem 1. Let there be given a polynomial \(Q(\lambda)=\lambda^n+a_1\lambda^{n-1}+\cdots+a_{n-1}\lambda+a_n\), whose coefficients are elements of the set \(M\). If

\[ (Tx,Tx)_k\le |(x,T^2x)_k|+A\|x\|_k\|Tx\|_k+B\|x\|_k^2 \tag{1} \]

for every \(x\) in \(D\) and \(k>k_0(x)\), for some constants \(A\) and \(B\), then from the fact that \(x\in M\) and \(Q(T)x\in M\), it follows that all \(T^m x\in M\) \((m=1,2,\ldots,n)\).

In particular, from the belonging of \(x\) to \(M\) and of \(T^n x\) to \(M\), it follows that \(T^m x\in M\) \((m=1,2,\ldots,n-1)\).

Proof. Let \(x\in M\) and \(Q(T)x\in M\). It is enough to show that \(Tx\in M\). Inequality (1) means that for all \(k\) greater than some \(k_0\),

\[ \|Tx\|_k^2\le \|x\|_k\|T^2x\|_k+A\|x\|_k\|Tx\|_k+B\|x\|_k^2 \]

and

\[ \frac{\|T^2x\|_k}{\|Tx\|_k} \ge \frac{\|Tx\|_k}{\|x\|_k} -A -B\frac{\|x\|_k}{\|Tx\|_k}. \]

If \(x\in M,\ Tx\notin M\), then

\[ \lim_{k\to\infty}\frac{\|Tx\|_k}{\|x\|_k}=\infty. \]

Analogously we obtain that

\[ \lim_{k\to\infty}\frac{\|T^2x\|_k}{\|Tx\|_k}=\infty,\ldots, \lim_{k\to\infty}\frac{\|T^n x\|_k}{\|T^{n-1}x\|_k}=\infty; \]

\[ \lim_{k\to\infty}\frac{\|x\|_k}{\|Tx\|_k}=0,\quad \lim_{k\to\infty}\frac{\|Tx\|_k}{\|T^2x\|_k}=0,\ldots,\quad \lim_{k\to\infty}\frac{\|T^{n-1}x\|_k}{\|T^n x\|_k}=0. \]

Then, a fortiori,

\[ \lim_{k\to\infty}\frac{\|T^i x\|_k}{\|T^n x\|_k}=0 \quad\text{for } i=1,2,\ldots,n-1. \]

Now let

\[ T^n x+a_1T^{n-1}x+\cdots+a_{n-1}Tx+a_nx=y\in M. \]

Then

\[ \|T^n x\|_k \le c\bigl(\|a_1\|_k\|T^{n-1}x\|_k+\cdots+\|a_{n-1}\|_k\|Tx\|_k+\|a_n\|_k\|x\|_k+\|y\|_k\bigr), \]

\[ 1\le c\left( \|a_1\|_k\frac{\|T^{n-1}x\|_k}{\|T^n x\|_k} +\cdots+ \|a_{n-1}\|_k\frac{\|Tx\|_k}{\|T^n x\|_k} +\|a_n\|_k\frac{\|x\|_k}{\|T^n x\|_k} +\frac{\|y\|_k}{\|T^n x\|_k} \right). \]

Passing to the limit in both parts of the inequality as \(k \to \infty\), we find that the left-hand side is equal to 1, while the right-hand side tends to zero, which is impossible. Consequently, \(Tx \in M\), and the theorem is proved.

Suppose the operator \(T\) acts not in the system \(S\), but in the system \(S'\). Then the following is true.

Theorem \(1'\). Let a polynomial
\[ Q(\lambda)=\lambda^n+a_1\lambda^{n-1}+\ldots+a_{n-1}\lambda+a_n \]
be given, whose coefficients are elements of the set \(M'\). If
\[ (Tx,Tx)_k \leq |(x,T^2x)_k|+A\|x\|_k\|Tx\|_k+B\|x\|_k^2 \]
for every \(x\) in \(D\) and \(k>k_0(x)\), for some constants \(A\) and \(B\), then from the fact that \(x \in M\) and \(Q(T)x \in M\) it follows that all \(T^m x \in M\) \((m=1,2,\ldots,n)\).

In particular, from the membership of \(x\) in \(M\) and \(T^n x\) in \(M\) it follows that
\(T^m x \in M\) \((m=1,2,\ldots,n-1)\).

Proof. Since all \(a_i \in M'\), there exists \(c>0\) such that \([a_i]_k<c\) \((i=1,2,\ldots,n)\);
\[ T^n x=y-a_1T^{n-1}x-a_2T^{n-2}x-\ldots-a_{n-1}Tx-a_nx, \]
\[ \|T^nx\|_k \leq \|y\|_k+\|a_1T^{n-1}x\|_k+\|a_2T^{n-2}x\|_k+\ldots+\|a_{n-1}Tx\|_k+\|a_nx\|_k \leq \]
\[ \leq \|y\|_k+c\|T^{n-1}x\|_k+c\|T^{n-2}x\|_k+\ldots+c\|Tx\|_k+c\|x\|_k; \]
\[ 1\leq \frac{\|y\|_k}{\|T^nx\|_k} +c\left( \frac{\|T^{n-1}x\|_k}{\|T^nx\|_k} +\frac{\|T^{n-2}x\|_k}{\|T^nx\|_k} +\ldots+ \frac{\|Tx\|_k}{\|T^nx\|_k} +\frac{\|x\|_k}{\|T^nx\|_k} \right). \]
Passing to the limit in both parts of the inequality as \(k\to\infty\), we obtain the contradiction: \(1\leq 0\).

Remark. In the proof of Theorems 1 and \(1'\) we used, for the given \(x\), the following inequalities:
\[ (Tx,Tx)_k \leq |(x,T^2x)_k|+A\|Tx\|_k\|x\|_k+B\|x\|_k^2; \]
\[ (T^2x,T^2x)_k \leq |(Tx,T^3x)_k|+A\|T^2x\|_k\|Tx\|_k+B\|Tx\|_k^2; \]
\[ (T^{n-1}x,T^{n-1}x)_k \leq |(T^{n-2}x,T^nx)_k|+A\|T^{n-1}x\|_k\|T^{n-2}x\|_k+B\|T^{n-2}x\|_k^2 \]
for some constants \(A\) and \(B\).

Consequently, if these inequalities are fulfilled for some element \(x\) of \(D\), then the assertions of the theorem are valid for this element \(x\), even if the condition
\((Ty,Ty)_k \leq |(y,T^2y)_k|+A\|Ty\|_k\|y\|_k+B\|y\|_k^2\)
is not fulfilled for every \(y\in D\).

Theorem 2. Let the operator \(T=R+H\), where \(H\) is defined on the whole space \(S\) \((HS\subset S)\), and let \(D\) \((D\subset S)\) be the domain of definition of the operator \(R\) and of its powers up to the \(n\)-th inclusive. Suppose further that the operator \(R\) satisfies the condition
\[ (y,Rx)_k=\pm(Ry,x)_k+F_k(x,y), \]
where \(|F_k(x,y)|\leq A\|y\|_k\|x\|_k\) for all \(k>k_0\), and \(H\) is a linear operator uniformly bounded with respect to any of the quasinorms, i.e. such that there exists a constant \(c_0\) for which
\((Hx,Hx)_k\leq c_0^2\|x\|_k^2\) for any \(k\) and any \(x\in S\). Then from the membership of \(x\) in \(M\) and \(Q(T)x\) in \(M\) it follows that
\(T^m x\in M\) \((m=1,2,\ldots,n)\).

Proof. Let
\[ (y,Rx)_k=(Ry,x)_k+F_k(x,y). \]

(The case \((y, Rx)_k = -(Ry, x)_k + F_k(x, y)\) is considered analogously). Since the operator \(H\) is bounded, the operator \(H^*\), defined by the equality \((Hx, y)_k = (x, H^*y)_k\) for any \(k\), is also bounded and \(\|H\|_k = \|H^*\|_k\). Consequently,

\[ \left|(H^2x, x)_k\right| = \left|(Hx, H^*x)_k\right| \leq c_0^2 \|x\|_k^2. \]

\[ \begin{aligned} (Tx, Tx)_k &= (\{R+H\}x, \{R+H\}x)_k = (Rx, Rx)_k + (Rx, Hx)_k + \\ &\quad + (Hx, Rx)_k + (Hx, Hx)_k = (R^2x, x)_k + (RHx, x)_k + (HRx, x)_k + (H^2x, x)_k - \\ &\quad - (HRx, x)_k - (H^2x, x)_k + (Rx, Hx)_k + (Hx, Hx)_k + \\ &\quad + F_k(Rx, x) + F_k(Hx, x); \end{aligned} \]

\[ \|Tx\|_k^2 \leq \|T^2x\|_k \|x\|_k + 2c_0 \|Tx\|_k \|x\|_k + \]

\[ + A \|Rx\|_k \|x\|_k + Ac_0 \|x\|_k^2. \]

Let now \(x \in M\), and \(Tx \notin M\). Then, for all \(x \ne 0\), beginning with a certain \(k\),

\[ \frac{\|Tx\|_k}{\|x\|_k} \leq \frac{\|T^2x\|_k}{\|Tx\|_k} + 2c_0 + Ac_0 \frac{\|x\|_k}{\|Tx\|_k} + A \frac{\|Rx\|_k}{\|Tx\|_k}. \]

We shall show that

\[ \lim_{k \to \infty} \frac{\|Rx\|_k}{\|Tx\|_k} \leq 1. \]

Indeed,

\[ \begin{aligned} \|Rx\|_k^2 &= (Rx, Rx)_k = (\{R+H\}x, \{R+H\}x)_k - (Rx, Hx)_k - \\ &\quad - (Hx, Rx)_k - (Hx, Hx)_k = (Tx, Tx)_k - (Tx, Hx)_k - (Hx, Tx)_k + \\ &\quad + (Hx, Hx)_k \leq \|Tx\|_k^2 + 2c_0 \|Tx\|_k \|x\|_k + c_0^2 \|x\|_k^2; \end{aligned} \]

\[ \lim_{k \to \infty} \frac{\|Rx\|_k^2}{\|Tx\|_k^2} \leq 1 + \lim_{k \to \infty} \frac{2c_0 \|x\|_k}{\|Tx\|_k} + \lim_{k \to \infty} \frac{c_0^2 \|x\|_k^2}{\|Tx\|_k}; \]

\[ \lim_{k \to \infty} \frac{\|Rx\|_k^2}{\|Tx\|_k^2} \leq 1; \qquad \lim_{k \to \infty} \frac{\|Rx\|_k}{\|Tx\|_k} \leq 1. \]

Therefore

\[ \lim_{k \to \infty} \frac{\|T^2x\|_k}{\|Tx\|_k} \geq \lim_{k \to \infty} \frac{\|Tx\|_k}{\|x\|_k} - 2c_0 - A = \infty. \]

The remainder of the proof follows from Theorem 1.

Theorem \(2'\). Let the operator \(T = R + H\), where \(H\) is defined on all of the space \(S'\) \((HS' \subseteq S')\), and let \(D\) \((D \subseteq S')\) be the domain of definition of the operator \(R\) and of its powers up to the \(n\)-th inclusive. Suppose, further, that the operator \(R\) satisfies the condition

\[ (y, Rx)_k = \pm (Ry, x)_k + F_k(x, y), \]

where \(|F_k(x, y)| \leq A \|y\|_k \|x\|_k\) for all \(k > k_0\), and \(H\) is a linear operator uniformly bounded in any of the quasinorms. Then from the membership of \(x\) in \(M\) and \(Q(T)x\) in \(M\) (the coefficients of the polynomial \(Q(T)\) belong to \(M'\)) it follows that

\[ T^m x \in M \qquad (m = 1, 2, \ldots, n). \]

Exactly as in Theorem 2, we prove that

\[ \lim_{k \to \infty} \frac{\|T^2x\|_k}{\|Tx\|_k} = \infty, \quad \text{if } x \in M,\ \text{and } Tx \notin M. \]

The remainder of the proof follows from Theorem \(1'\).

Theorem 3. If all the conditions of Theorem 1 (or Theorem \(1'\)) are satisfied and, moreover, \(A=0,\ B=0\), i.e.

\[ (Tx,Tx)_k \leq |(x,T^2x)_k|, \]

then

\[ \|T^m x\|_k \leq \|T^n x\|_k^{\frac{m}{n}}\|x\|_k^{\frac{n-m}{n}} \quad (m=1,2,\ldots,n-1). \]

Proof. By the hypothesis of the theorem,

\[ (Tx,Tx)_k \leq |(x,T^2x)_k|. \]

\[ \|Tx\|_k^2 \leq \|T^2x\|_k\|x\|_k;\qquad \|Tx\|_k \leq \|T^2x\|_k^{1/2}\|x\|_k^{1/2}; \]

\[ \|T^2x\|_k^2 \leq \|T^3x\|_k\|Tx\|_k \leq \|T^3x\|_k\|T^2x\|_k^{1/2}\|x\|_k^{1/2}; \]

\[ \|T^2x\|_k^{3/2} \leq \|T^3x\|_k\|x\|_k^{1/2};\qquad \|T^2x\|_k \leq \|T^3x\|_k^{2/3}\|x\|_k^{1/3}. \]

Further:

\[ \|T^3x\|_k^2 \leq \|T^4x\|_k\|T^2x\|_k \leq \|T^4x\|_k\|T^3x\|_k^{2/3}\|x\|_k^{1/3}; \]

\[ \|T^3x\|_k^{4/3} \leq \|T^4x\|_k\|x\|_k^{1/3};\qquad \|T^3x\|_k \leq \|T^4x\|_k^{3/4}\|x\|_k^{1/4}. \]

Suppose now that the relation

\[ \|T^i x\|_k \leq \|T^{i+1}x\|_k^{\frac{i}{i+1}}\|x\|_k^{\frac{1}{i+1}} \]

is valid. We shall show that then the following is also valid:

\[ \|T^{i+1}x\|_k \leq \|T^{i+2}x\|_k^{\frac{i+1}{i+2}}\|x\|_k^{\frac{1}{i+2}}. \]

Since

\[ \|T^{i+1}x\|_k^2 \leq \|T^{i+2}x\|_k\|T^i x\|_k, \]

we have

\[ \|T^{i+1}x\|_k \leq \|T^{i+2}x\|_k^{1/2}\|T^i x\|_k^{1/2} \leq \|T^{i+2}x\|_k^{1/2} \|T^{i+1}x\|_k^{\frac{i}{2i+1}} \|x\|_k^{\frac{i}{2(i+1)}}. \]

\[ \|T^{i+1}x\|_k^{\frac{i+2}{2(i+1)}} \leq \|T^{i+2}x\|_k^{1/2} \|x\|_k^{\frac{1}{2(i+1)}} ; \]

\[ \|T^{i+1}x\|_k \leq \|T^{i+2}x\|_k^{\frac{i+1}{i+2}} \|x\|_k^{\frac{1}{i+2}}. \]

Consequently,

\[ \|T^{n-1}x\|_k^{\frac{n}{n-1}} \leq \|T^n x\|_k\|x\|_k^{\frac{1}{n-1}}; \qquad \|T^{n-1}x\|_k \leq \|T^n x\|_k^{\frac{n-1}{n}}\|x\|_k^{\frac{1}{n}}. \]

\[ \|T^{n-2}x\|_k \leq \|T^{n-1}x\|_k^{\frac{n-2}{n-1}}\|x\|_k^{\frac{1}{n-1}} \leq \|T^n x\|_k^{\frac{n-2}{n}} \|x\|_k^{\frac{1}{n}\cdot\frac{n-2}{n-1}} \|x\|_k^{\frac{1}{n-1}} = \]

\[ = \|T^n x\|_k^{\frac{n-2}{n}}\|x\|_k^{\frac{2}{n}}. \]

\[ \|T^{n-3}x\|_k \leq \|T^{n-2}x\|_k^{\frac{n-3}{n-2}}\|x\|_k^{\frac{1}{n-2}} \leq \left[ \|T^n x\|_k^{\frac{n-2}{n}} \|x\|_k^{\frac{2}{n}} \right]^{\frac{n-3}{n-2}} \|x\|_k^{\frac{1}{n-2}} = \]

\[ = \|T^n x\|_k^{\frac{n-3}{n}}\|x\|_k^{\frac{3}{n}}. \]

Suppose that for \(m_1\) it is true that

\[ \|T^{\,n-m_1}x\|_k \leq \|T^n x\|_k^{\frac{n-m_1}{n}}\|x\|_k^{\frac{m_1}{n}}. \]

We shall show that the same relation is also valid for \((m_1+1)\):

\[ \begin{aligned} \|T^{\,n-(m_1+1)}x\|_k &\leq \|T^{\,n-m_1}x\|_k^{\frac{n-m_1-1}{\,n-m_1\,}} \|x\|_k^{\frac{1}{\,n-m_1\,}} \\ &\leq \|T^n x\|_k^{\frac{n-m_1}{n}\cdot\frac{n-m_1-1}{\,n-m_1\,}} \|x\|_k^{\frac{m_1}{n}\cdot\frac{n-m_1-1}{\,n-m_1\,}} \|x\|_k^{\frac{1}{\,n-m_1\,}} \\ &\leq \|T^n x\|_k^{\frac{n-(m_1+1)}{n}} \|x\|_k^{\frac{m_1+1}{n}} . \end{aligned} \]

Consequently, the inequality

\[ \|T^{\,n-m_1}x\|_k \leq \|T^n x\|_k^{\frac{n-m_1}{n}} \|x\|_k^{\frac{m_1}{n}} \]

is valid for all \(m_1 \leq n-1\). Denote \(n-m_1=m\). Then

\[ \|T^m x\|_k \leq \|T^n x\|_k^{\frac{m}{n}} \|x\|_k^{\frac{n-m_1}{n}} \qquad (m=1,2,\ldots,n-1). \]

Remark. It is obvious that if, in the theorem just proved, the condition \((Tx,Tx)_k \leq |(x,T^2x)_k|\) is replaced by the condition
\(\lim_{k\to\infty}(Tx,Tx)_k \leq \lim_{k\to\infty}|(x,T^2x)_k|\), then the result obtained is the following:

\[ \lim_{k\to\infty}\|T^m x\|_k \leq \lim_{k\to\infty}\|T^n x\|_k^{\frac{m}{n}} \cdot \lim_{k\to\infty}\|x\|_k^{\frac{n-m}{n}} \qquad (m=1,2,\ldots,n-1). \]

1. In what follows we shall show that the hypotheses of Theorems 1–3 are satisfied, in particular, for the differentiation operator. Here the following lemmas will play an essential role; their validity is easily verified.

Lemma 1. Let \(f(x)\) \((-\infty<x<\infty)\) be a bounded function, \(\sup |f(x)|=M\), for which there exist and are bounded

\[ f'(x),\ f''(x)\ldots f^{(n)}(x); \qquad \sup |f^{(j)}(x)|=M_j \quad (j=1,2,\ldots,n). \]

Then there exists a \(c>0\), depending only on \(M\) and \(M_j\), such that, beginning with some \(k\),

\[ |f^{(j)}(x)| \leq c\sqrt{\int_{a_k}^{b_k}|f(t)|^2\,dt} \qquad (j=0,1,\ldots,n) \]

for all \(x\in [a_k,b_k]\) \((a_k\to-\infty,\ b_k\to+\infty)\).

Lemma 2. If \(f(x,y)\) \((-\infty<x,y<\infty)\) is a bounded function, \(\sup |f(x,y)|=M\), for which there exist and are bounded

\[ \frac{\partial^{i+j}f}{\partial x^i\,\partial y^j} \quad (i=1,2,\ldots,n;\ j=1,2,\ldots,m); \qquad \sup\left|\frac{\partial^{i+j}f}{\partial x^i\,\partial y^j}\right| =M_{ij}, \]

then there exists a \(c>0\), depending only on \(M\) and \(M_{ij}\), such that

\[ \left| \int_{a_k}^{b_k} \frac{\partial^{i+j}f(x,y_0)}{\partial x^i\,\partial x^j} \frac{\partial^{p+l}f(x,y_0)}{\partial x^p\,\partial y^l} \,dx \right| \leq c \int_{a_k}^{b_k}\int_{d_k}^{l_k} |f(x,y)|^2\,dx\,dy \]

for all \(y_0\in[d_k,l_k]\), \(0\leq i+p\leq n\); \(0\leq j+l\leq m\).

1. Consider the set of continuous bounded functions \(f(x)\) \((|f(x)|\leq L)\) on the axis \((-\infty<x<\infty)\) or on the half-axis \((0\leq x<\infty)\). As was shown above, this set forms a system \(S'\).

Theorem 4. If \(f(x)\) is a bounded solution of the linear differential equation

\[ \frac{d^{n}f}{dx^{n}}+a_{1}(x)\frac{d^{n-1}f}{dx^{n-1}}+a_{2}(x)\frac{d^{n-2}f}{dx^{n-2}}+\cdots+a_{n}(x)f(x)=\varphi(x) \]

\[ [\,|\varphi|\le L\,] \]

with coefficients bounded on the axis \(-\infty<x<\infty\) or on the half-axis \(0\le x<\infty\), then from the convergence of the integrals

\[ \int_{-\infty}^{+\infty}|\varphi(x)|^{2}\,dx \]

and

\[ \int_{-\infty}^{+\infty}|f(x)|^{2}\,dx \]

there follows the convergence of the integrals

\[ \int_{-\infty}^{+\infty}|f'(x)|^{2}\,dx,\ldots,\int_{-\infty}^{+\infty}|f^{(n)}(x)|^{2}\,dx. \]

Proof. It is known that if \(f(x)\) is a bounded solution of a linear differential equation of order \(n\) with bounded coefficients and right-hand side, then all derivatives of this solution up to the \(n\)-th inclusive are also bounded [2]. Therefore, using Lemma 1, we obtain the following.

Let the operator

\[ T=\frac{d}{dx}. \]

Then

\[ (Tf,Tf)_{k}=\int_{a_{k}}^{b_{k}} f'(x)\overline{f'(x)}\,dx =f'(b_{k})\overline{f(b_{k})}-f'(a_{k})\overline{f(a_{k})} -\int_{a_{k}}^{b_{k}} f(x)\overline{f''(x)}\,dx \]

\[ \le c_{1}\sqrt{\int_{a_{k}}^{b_{k}} |f(x)|^{2}\,dx}\, \sqrt{\int_{a_{k}}^{b_{k}} |f'(x)|^{2}\,dx} -\int_{a_{k}}^{b_{k}} f(x)\overline{f''(x)}\,dx =c_{1}\|f(x)\|_{k}\|Tf(x)\|_{k}-(f,T^{2}f)_{k} \]

\[ \le c\|f(x)\|_{k}\|Tf(x)\|_{k}+\|f\|_{k}\|T^{2}f\|_{k}, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ (T^{n-1}f,T^{n-1}f)_{k} =\int_{a_{k}}^{b_{k}} f^{(n-1)}(x)\overline{f^{(n-1)}(x)}\,dx =f^{(n-1)}(b_{k})\overline{f^{(n-2)}(b_{k})} \]

\[ -f^{(n-1)}(a_{k})\overline{f^{(n-2)}(a_{k})} -\int_{a_{k}}^{b_{k}} f^{(n)}(x)\overline{f^{(n-1)}(x)}\,dx \le \]

\[ \le c_{n-1}\|f^{(n-1)}(x)\|_{k}\|f^{(n-2)}(x)\|_{k} +\|f^{(n-1)}(x)\|_{k}\|f^{(n)}(x)\|_{k} \le \]

\[ \le c\|T^{n-1}f\|_{k}\|T^{n-2}f\|_{k} +\|T^{n-2}f\|_{k}\|T^{n}f\|_{k}, \]

where \(c=\max(c_{1},c_{2},\ldots,c_{n-1})\), and on the basis of the remark to Theorem 1,

\[ T^{m}f\in M \]

for all \(m=1,2,\ldots,n\), i.e., the integrals

\[ \int_{-\infty}^{+\infty}|f^{(m)}(x)|^{2}\,dx \]

converge for all \(m=1,2,\ldots,n\).

The theorem is proved.

Let us now consider the set of bounded continuous functions of two variables \(f(x,y)\) \((-\infty<x,y<\infty)\). We define the sequence of scalar products of two functions \(f(x,y)\) and \(g(x,y)\) in the following way:

\[ (f,g)_{k}=\int_{a_{k}}^{b_{k}}\int_{d_{k}}^{l_{k}} f(x,y)\overline{g(x,y)}\,dx\,dy; \qquad \lim_{k\to\infty} a_{k},d_{k}=-\infty;\qquad \lim_{k\to\infty} b_{k},l_{k}=+\infty, \]

The norm of the function \(f(x,y)\) is defined as the limit

\[ \|f(x,y)\|=\lim_{k\to\infty}\sqrt{\int_{a_k}^{b_k}\int_{d_k}^{l_k}|f(x,y)|^2\,dx\,dy}. \]

Theorem 5. If \(f(x,y)\) is a solution of the differential equation

\[ \frac{\partial^{2n} f}{\partial x^n \partial y^n} +a_1(x,y)\frac{\partial^{2(n-1)} f}{\partial x^{n-1}\partial y^{n-1}} +\cdots +a_{n-1}\frac{\partial^2 f}{\partial x\partial y} +a_n f=\varphi, \]

where \(|a_i(x,y)|\le A\);

\[ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}|\varphi(x,y)|^2\,dx\,dy<\infty, \]

and \(f(x,y)\) is bounded together with its derivatives up to

\[ \frac{\partial^{2n} f}{\partial x^n\partial y^n}, \]

then from the convergence of the integral

\[ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}|f(x,y)|^2\,dx\,dy \]

there follows the convergence of the integrals

\[ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \left| \frac{\partial^{2m} f}{\partial x^m\partial y^m} \right|^2\,dx\,dy \qquad (m=1,2,\ldots,n). \]

The proof is analogous to the proof of Theorem 4.

Similar theorems can also be formulated for certain other differential equations, for example for the following:

\[ \left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right)^n f(x,y) +a_1\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right)^{n-1} f(x,y) +\cdots+ \]

\[ +a_{n-1}\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right)f(x,y) +a_n f(x,y)=\varphi(x,y), \]

\[ \left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)^n f(x,y) +a_1\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)^{n-1} f(x,y) +\cdots+ \]

\[ +a_{n-1}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)f(x,y) +a_n f(x,y)=\varphi(x,y), \]

\[ \frac{\partial^{n(i+j)} f(x,y)}{\partial x^{ni}\partial y^{nj}} +a_1\frac{\partial^{(n-1)(i+j)} f(x,y)} {\partial x^{(n-1)i}\partial y^{(n-1)j}} +\cdots+ \]

\[ +a_{n-1}\frac{\partial^{i+j} f(x,y)}{\partial x^i\partial y^j} +a_n f(x,y)=\varphi(x,y). \]

References

1. Esclanqon E. CR., 160, 1915, 475.
2. Rutman M. A. In: Studies on Contemporary Problems of the Constructive Theory of Functions. Moscow, Fizmatgiz, 1961, pp. 294.
3. Lyubich Yu. I. DAN SSSR, 102, No. 5, 881, 1955.

Received by the editors
22 November 1965

Odessa State University