MATHEMATICS

V. Ya. STETSENKO

(K)-REGULAR CONES

(Presented by Academician P. S. Aleksandrov, October 1, 1960)

In applications of the theory of spaces with a cone to nonlinear problems, an important role is played by the possibility of passing to the limit in monotone bounded sequences. M. A. Krasnosel’skii proposed the following definitions.

Let two cones (K_0) and (K), where (K_0 \subset K), be given in a real Banach space (E). A semi-ordering in (E) is introduced by means of the “wide” cone (K): (x \leq y), if (y - x \in K).

The cone (K_0) is called (K)-regular if every monotone increasing and bounded (all with respect to the cone (K)) sequence of elements of the “narrow” cone (K_0) converges in the space (E), i.e., if from

[
x_1 \leq x_2 \leq \ldots \leq x_n \leq \ldots \quad (x_n \in K_0,\ n = 1,2,\ldots);
\tag{1}
]

[
x_n \leq u_0 \in K_0 \quad (n = 1,2,\ldots)
\tag{2}
]

it follows that (|x_n - x_{n+p}| \to 0) as (n \to \infty).

The cone (K_0) is called completely (K)-regular if every monotone (with respect to the cone (K)) and norm-bounded sequence of elements of (K_0) converges, i.e., if from (1) and

[
|x_n| \leq M \quad (n = 1,2,\ldots)
\tag{3}
]

it follows that the sequence (x_n) is fundamental.

For the case (K = K_0) (in this case (K)-regularity and complete (K)-regularity are called, respectively, regularity and complete regularity), the connection between the notions of regularity, complete regularity, and normality in the sense of M. G. Krein ((^{2})) of a cone was studied in ((^{1})). In particular, it was shown that completely regular cones are regular, and regular cones are normal. There are examples ((^{1})) showing that not every normal cone is regular and not every regular cone is completely regular.

In the present article the case (K_0 \ne K) is considered.

1. It is obvious that regularity and complete regularity of the “wide” cone (K) ensure, respectively, (K)-regularity and complete (K)-regularity of the cone (K_0). It is also obvious that every (K)-regular (completely (K)-regular) cone (K_0) has the property of regularity (complete regularity). However, one can indicate examples of completely regular cones (K_0) which do not have the property of (K)-regularity with respect to some cone (K) (see item 2).

Let us also note that complete regularity of the cone (K_0), together with (K)-regularity of the cone (K_0), does not ensure regularity of the cone (K). The corresponding example will be given below (item 2).

1. According to M. G. Krein ((^{2})), a cone (K) is called normal if there exists a (\delta > 0) such that, for any vectors (e_1, e_2 \in K) ((|e_1| = |e_2| = 1)), the inequality (|e_1 + e_2| \geq \delta) is satisfied. As is known ((^{3})), normality of a cone is equivalent to the existence of such an (M) that from (\theta \leq x \leq y) there follows the inequality (|x| \leq M|y|). In connection with this, the following definition seems natural to us.

The cone (K_0) is called (K)-normal if, for all elements (x,y\in K_0), one can specify an (N) such that from (\theta \leqslant x \leqslant y) there follows the inequality (|x|\leqslant N|y|).

Below, (K_0\langle v,w\rangle) denotes the set of those elements (x\in K_0) for which (v\leqslant x\leqslant w).

Theorem 1. (I. A. Bakhtin ((^{4}))). In order that the cone (K_0) be (K)-normal, it is necessary and sufficient that all sets (K_0\langle \theta,u_0\rangle) ((u_0\in K_0)) be bounded.

From the definition of (K)-normality it follows that every (K)-normal cone (K_0) is normal. The converse assertion is false.

Consider, for example, in the space (l_2) the set (K_0) of numerical sequences
(x=(x_1,x_2,\ldots,x_n,\ldots)) for which (x_1\geqslant 0) and
(0\leqslant x_n\leqslant x_1) ((n=2,3,\ldots)), and as the set (K) take all elements
(y=(y_1,y_2,\ldots,y_n,\ldots)\in l_2) for which
(y_1\geqslant 0,\ -\infty<y_i\leqslant y_1) ((i=2,3,\ldots)). It is not hard to prove that (K_0) and (K) are cones in (l_2), and moreover (K_0\subset K).

The cone (K_0) is part of the perfectly regular cone of nonnegative sequences from (l_2), and therefore (K_0) is perfectly regular and hence also normal. The elements
(x_1=(1,0,0,\ldots)), (x_2=(1,1,0,\ldots)), (x_3=(1,1,1,\ldots),\ldots) belong to (K_0) and are bounded above (with respect to the cone (K)) by the element (u=(1,0,0,\ldots)). However, (|x_n|=\sqrt n=\sqrt n|u|), and the inequality
(|x_n|\leqslant N|u|) is not satisfied for any (N). Thus, (K_0) does not possess the property of (K)-normality.

It is obvious that if the “wide” cone (K) is normal, then the cone (K_0) will be (K)-normal. However, from the (K)-normality of the cone (K_0) there does not always follow the normality of the “wide” cone. For example, if the cone (K_0) is finite-dimensional, then it will be (K)-normal and even (K)-regular with respect to any cone (K).

Theorem 2*. Every (K)-regular cone (K_0) is (K)-normal.

Suppose the contrary. Then there exist sequences (u_n) and (x_n) (both from (K_0)) such that

[
x_n\leqslant u_n\qquad (n=1,2,\ldots)
\tag{4}
]

and at the same time

[
|x_n|\geqslant n^2|u_n|.
\tag{5}
]

Consider the series (\sum_{n=1}^{\infty}\dfrac{u_n}{|x_n|}). In view of (5), this series converges. Denote its sum by (v_0). Obviously, (v_0\in K_0). From inequality (4) it follows that
[
\frac{x_n}{|x_n|}\leqslant \frac{u_n}{|x_n|},
]
and therefore
[
\sum_{n=1}^{\infty}\frac{x_n}{|x_n|}\leqslant v_0.
]

Consider the sequence
[
y_n=\sum_{k=1}^{n}\frac{x_k}{|x_k|}.
]
Obviously, (y_n\in K_0) ((n=1,2,\ldots)), and
[
y_1\leqslant y_2\leqslant \cdots \leqslant y_n\leqslant \cdots \leqslant v_0.
]
Therefore, by virtue of the (K)-regularity of the cone (K_0), (y_n) converges strongly, which is impossible, since
[
|y_{n+1}-y_n|=1\qquad (n=1,2,\ldots).
]
The contradiction obtained proves the theorem.

From the theorem just proved it follows that the cone (K_0) considered above is not (K)-regular.

3. As was already noted above, from the complete regularity of the cone (K) there follows its regularity. For arbitrary completely (K)-regular cones it has not been possible to prove their (K)-regularity, nor even (K)-normality. We have the following weaker assertions.

* M. A. Krasnosel’skii has another proof of this theorem.

Theorem 3. If the cone (K_0) is completely (K)-regular and (K)-normal, then it is (K)-regular.

Theorem 4. If the cone (K_0) is completely (K)-regular, then it is normal.

Theorem 5. Suppose that for every monotone norm-bounded sequence (x_n\in K):
[
x_1\leq x_2\leq x_3\leq\cdots
]
one can indicate an element (u_0) such that (x_n+u_0\in K_0) ((n=1,2,\ldots)). Then the complete (K)-regularity of the cone (K_0) implies its (K)-regularity.

1. Let us consider the question of under what additional conditions complete (K)-regularity follows from (K)-regularity.

Theorem 6. Let the (K)-regular cone (K_0) be solid. Then (K_0) is a completely (K)-regular cone.

Theorem 7. A (K)-regular cone (K_0) is completely (K)-regular if the cone (K) is normal and the space (E) is weakly complete.

1. We now consider the differential equation
   [
   \frac{dx}{dt}=f(t,x)
   \tag{6}
   ]
   in a Banach space (E), semi-ordered by means of a certain cone (K). The operator (f(t,x)) is assumed to be continuous in the totality of its variables. As Bourbaki showed ((^5)), the existence theorem does not follow from the continuity of (f(t,x)) (i.e., Peano’s theorem in its general form is false for equations in Banach spaces). In this connection there arises the question of additional properties which the operator (f(t,x)) must possess in order for an existence theorem to be valid for equation (6). It turns out that a number of sufficient conditions can be formulated in terms of spaces with a cone, if one uses the notions of regularity or complete regularity of a cone.

As above, (K_0) and (K) are cones in (E), with (K_0\subset K). We shall assume that the operator (f(t,x)) is defined and continuous on the topological product of the interval ([0,a]) and the set (K_0\langle \theta,v_0\rangle). Let (f(t,x)\in K_0) and
[
f(t,x)\leq b(x)v_0
]
for (t\in[0,a]), (x\in K_0\langle \theta,v_0\rangle).

Theorem 8. Suppose that the operator (f(t,x)), for (t\in[0,a]), is monotone on (K_0\langle \theta,v_0\rangle): from (x\leq y) it follows that (f(t,x)\leq f(t,y)). Suppose that the cone (K_0) has the property of (K)-regularity. Then equation (6) has a solution (x(t)), defined on some interval ([0,\delta]\subset[0,a]), and satisfying the initial condition (x(0)=x_0), where (x_0\in K\langle\theta,\gamma v_0\rangle) ((\gamma<1)).

The solution (x(t)) may be obtained by the method of successive approximations:
[
x_{n+1}(t)=x_0+\int_0^t f[s,x_n(s)]\,ds
\quad (t\in[0,\delta],\ n=0,1,2,\ldots),\qquad
x_0(t)\equiv x_0.
]

Theorem 8 remains valid if, instead of requiring (K)-regularity of the cone (K_0), one requires its complete (K)-regularity, and additionally requires of the operator (f(t,x)) that the inequality
[
\sup |f(t,x)|<\infty
\quad (t\in[0,a],\ x\in K_0\langle\theta,v_0\rangle)
]
be satisfied.

By the usual schemes ((^6)), Theorem 8 can be illustrated by applications to integro-differential equations.

The author expresses sincere gratitude to M. A. Krasnosel’skii for posing the problems and for advice.

Voronezh State
University

Received
30 IX 1960

References

1. M. A. Krasnosel’skii, DAN, 135, No. 2 (1960).
2. M. G. Krein, M. A. Rutman, UMN, 3, issue 1 (23) (1948).
3. I. A. Bakhtin, Tr. seminara po funktsional’n. analizu, Voronezhsk. gos. univ., issue 6 (1958).
4. I. A. Bakhtin, Dissertation, Voronezh State Univ., 1958.
5. N. Bourbaki, Elements de Mathématique, part I, book IV, ch. IV, Paris, 1951.
6. M. A. Krasnosel’skii, S. G. Krein, Tr. seminara po funktsional’n. analizu, Voronezhsk. gos. univ., issue 8 (1959).