LINEAR OPERATORS THAT TRANSFORM A NORMAL CONE IN COMPLETELY REGULAR CONES
A. B. NEMETH
The Fredholm resolvents of a wide class of operators, which are sublinear with respect to the ordering induced by the wedge WW in the normed space YY, have the property that transform WW into completely regular cones [6]. These resolvents approximate indefinitely the identity map in the topology of uniform convergence on norm bounded sets. This advantage is associated with the drawback that composing them with convex mappings with values in YY, the resulting operators fail to be convex with respect to the ordering induced by the transformed cone.
The linear operator AA on YY has the property that composed with any WW-convex operator yields a mapping which is A(W)A(W)-convex: The complete regularity of A(W)A(W) remains of a crucial interest for applications. But it appears that when WW isn't regular, the linear operator with this property cannot approximate indefinitely the identity map (Corollary 3). However, some important operators (sec the example in 12) have good properties from this point of view. Hence we devote the present note to investigation of the linear operators AA with the cone range A(W)A(W) being a completely regular cone.
If the linear and positive operator AA maps the closed normal cone CC witi: nonempty interior, contained in the Banach space (B-space), YY, into a compltely regular cone, then any abstract Hammerstein operator AFA F, where FF is CC-convex and continuous, is subdifferentiable at any interior point of the dcmain of FF (sce Proposition 19).
Operators with ompletely regular ene ranges. Let YY be a normed space over the reals and let CC be a cone in YY, i.e., a subset having the propertis C+C sub C,tC sub CC+C \subset C, t C \subset C for any positive real number, tt, and C nn(-C)={0}C \cap(-C)=\{0\}. The cone CC induces a reflexive; transitive and antisymmetric order relation <=\leqslant on YY if we put u <= vu \leqslant v whenever u-u in Cu-u \in C. This order relation relates to the linear structure of YY by the properties: u <= vu \leqslant v implies u+w <= v+wu+w \leqslant v+w for any ww in YY and tu <= tvt u \leqslant t v for any positive real number tt. Since in the sequel we have to do with different cones, we shall call CC-ordering the ordering induced by CC. Similarly, we shall use terms as CC-order bound, CC-monotone etc.
The cone CC is said to be normal if there exists a positive number bb such that ||u|| <= b||v||\|u\| \leqslant b\|v\|, whenever 0 <= u <= v0 \leqslant u \leqslant v.
The cone CC is called completely regular (regular) if any CC-monotone norm bounded ( CC-order bounded) sequence in YY is fundamental. Any regular cone which is complete is normal, and any completely regular cone is normal and hence regular ([3], theorems 1.6 and 1.7).
If AA is a linear operator on YY, then the cone range A(C)A(C) of AA is obvious!y a cone. If A(C)sub CA(C) \subset C, then AA is called positive. For a wide class of cones the positivity of a linear operator implics its continuity. We shall in the present
note ignore this aspect and shall explicitly require in all what follows the continuity of the considered linear and positive operators.
The linear operator AA is said to be of completely regular (regular) type, if its cone range A(C)A(C) is a completely regular (regular) cone. Since any completely regular cone is also regular, any linear operator of completely regular type is also of regular type.
We shall frequently use in the sequel Lemma 4 in [6] and hence we shall state it here in a slightly modified from.
Lemma. The conc CC in YY is completely regular (regular) if it contains no sequence ( y_(1)y_{1} ) having the property that ||y_(i)|| >= d\left\|y_{i}\right\| \geqslant d for any ii and some positive dd and for which the set {sum_(i=1)^(n)y_(i):n inN}\left\{\sum_{i=1}^{n} y_{i}: n \in \mathbf{N}\right\} is norm bounded (C-order bounded).
Proposition. If CC isn't a regular cone of the normed space YY, then no regular type linear and continuous operator (and hence no completely regular type lincar and continuous operator) can have continuous left hand side inverse.
Proof. The linear operator AA has continuous left hand side inverse if and only if there exists a positive bb such that
for any yy in YY (see e.g. V. 4.4. in [4]).
If CC isn't regular, it contains by Iemma 1 a sequence ( y_(i)y_{i} ) with the property that ||y_(i)|| >= d\left\|y_{i}\right\| \geqslant d for any ii and some positive dd, for which the set {sum_(i=1)^(n)y_(i):n in:}in N}\left\{\sum_{i=1}^{n} y_{i}: n \in\right. \in N\} is order bounded. Let yy be a CC-upper bound for this set. Then AyA y will be an A(C)A(C)-upper bound for the set {sum_(i=1)^(n)Ay_(i):n inN}\left\{\sum_{i=1}^{n} A y_{i}: n \in \mathbf{N}\right\}. According (1) and the property of ( y_(t)y_{t} ) it holds
for any ii. Applying once again Lemma 1 we conclude that the cone A(C)A(C) cannot be regular. Q.E.D.
We shall denote with L(Y)\mathscr{L}(Y) the vector space of all linear and bounded operators acting in YY, endowed with the norm topology.
3. Carollary. Let CC be a cone in the BB-space YY that isn't regular. Then the open unit sphere in L(Y)\mathcal{L}(Y) with the centre at the identity map II ean contain no operator of regular (and henece no operator of completely regular) type.
Proof. Any opcrator in the above open sphere has continuous inverse by a theorem of Banach (see e.g. V. 4.5 in [2]). Q.E.D.
4. Remark. In [6] it was shown that the identity map can be indefinitely approximated in the topology of the uniform convergence on norm bounded sets by the Fredholm resolvents of some sublinear operators. These resolvents transform the cone CC whose closure isn't a subspace in some completely regular subcones of its. Restricted to the linear and continuous operators the considered topology is quite the norm topology. Intuitively the above cited result
means that a cone contains subcones ,;arbitrarily close to it" which are completely regular. Although, by Corollary 3, the transformation of a cone that isn't. regular into a such subcone cannot be realised by a linear operator.
5. Proposition. The property of a cone in a normed space to be completely regular is preserved by any linear and boundcd operator with continuous left hand side inverse.
Proof. Let CC be a completely regular cone and assume that A(C)A(C) isn't completely regular for some linear and bounded AA with continuous left hand side inverse. We have for any yy the relation (1) for some positive bb. Invoking Lemma 1, there exists a sequence ( y_(1)y_{1} ) in CC with ||Ay_(1)|| >= d\left\|A y_{1}\right\| \geqslant d for some positive dd and any ii, such that the set {sum_(i=1)^(n)Ay_(i):n inN}\left\{\sum_{i=1}^{n} A y_{i}: n \in \mathbb{N}\right\} is norm bounded. We have for any ii the relation ||y_(i)|| >= d//||A|| > 0\left\|y_{i}\right\| \geqslant d /\|A\|>0, while from (1), ||sum_(i=1)^(n)Ay_(i)|| >= b||sum_(i=1)^(n)y_(i)||\left\|\sum_{i=1}^{n} A y_{i}\right\| \geqslant b\left\|\sum_{i=1}^{n} y_{i}\right\|. Accordingly the set {sum_(i=1)^(n)y_(i):n in(N)}\left\{\sum_{i=1}^{n} y_{i}: n \in \mathrm{~N}\right\} is norm bounded and we have get via Lemma 1 a contradiction with the hypothesis that CC is a completely regular cone. Q.E.D.
6. Remark. Obviously, any linear operator of finite range preserves the complete regularity of a cone. The operators constructed in 12 and 14 furnish other examples having this property. However, there exist linear and compact operators that transform some completely regular cones onto cones without this property (see the example in 17).
Let B\mathscr{B} and C\mathcal{C} be subsets in L(Y)\mathcal{L}(Y). We shall say that C\mathcal{C} is modular over B\mathscr{B}. if for any n inNn \in \mathbf{N}, any B_(1)B_{1} in B\mathscr{B} and any C_(i)C_{i} in C\mathcal{C}, the operator sum_(i=1)^(n)B_(1)C_(i)\sum_{i=1}^{n} B_{1} C_{i} is in CC : If a\mathfrak{a} contains the identity map, then it suffices to restrict nn in the above definition to be >= 2\geqslant 2. It is straightforward to show that if 28 contains all the positive multiples of the identity map (respectively, its multiples with scalars in [0,1][0,1] ), then if C\mathcal{C} is modular over B\mathfrak{B}, it is a convex cone (respectively, a convex set). If C\mathcal{C} is modular over itself we shall say that it is automodular.
We shall need in our next reasonings the
7. Lemma. The sum of a finite number of completely regular subcones of a normal cone is a completely regular cone.
Proof. Let C_(1)C_{1} and C_(2)C_{2} be completely regular subcones of the normal cone CC and assume that C_(1)+C_(2)C_{1}+C_{2} isn't completely regular. By Lemma 1 there exist the sequence (y_(i)^(j))\left(y_{i}^{j}\right) in C_(j),j=1,2C_{j}, j=1,2 and a positive number dd so to have ||y_(i)^(1)++y_(i)^(2)|| >= d\| y_{i}^{1}+ +y_{i}^{2} \| \geqslant d for any ii, while the set
{:(2){sum_(i=1)^(n)(y_(i)^(1)+y_(i)^(2)):n inN}:}\begin{equation*}
\left\{\sum_{i=1}^{n}\left(y_{i}^{1}+y_{i}^{2}\right): n \in \mathbf{N}\right\} \tag{2}
\end{equation*}
is norm bounded. Passing to a subsequence we can assume without loss of the generality that
{:(3)||y_(i_(k))^(1)|| >= d//2","quad k inN:}\begin{equation*}
\left\|y_{i_{k}}^{1}\right\| \geqslant d / 2, \quad k \in \mathbf{N} \tag{3}
\end{equation*}
On the other hand if <=\leqslant denotes the CC-ordering in YY, we have
wherefrom, using the normality of the cone CC and the norm boundendess of (2) it results that the set
{sum_(k=1)^(m)y_(i_(i))^('):k inN}\left\{\sum_{k=1}^{m} y_{i_{i}}^{\prime}: k \in \mathbf{N}\right\}
is norm bounded. But this, together with (3) contradicts the complete regularity of C_(1)C_{1}. Q.E.D.
8. Proposition. Let CC be a normal cone in the normed space YY. Let BB dchote the subset of CC-positive operators in L(Y)\mathcal{L}(Y) that transform any conpletely regular subcone in CC in completely regular conc. Then BB is an automodular convex cone in L(Y^('))\mathcal{L}\left(Y^{\prime}\right).
Proof. We have obviously B_(1)B_(2)inBB_{1} B_{2} \in \mathscr{B} whenever B_(1)B_{1} and B_(2)B_{2} : are in B\mathscr{B}. Further, by the inclusion
it follows that the cone in the left is completely regular being the subcone of the cone in the right, which is completely regular by Lemma 7. That is, B_(1)++B_(2)B_{1}+ +B_{2} is in E\mathscr{E}. From Proposition 5 we have that II and any positive multiple of its are in so and hence we are done. Q.E.D.
9. Proposition. Let CC be a normal cone in the normed space YY. Let e\mathbb{e} denote the set in L(Y)\mathrm{L}(Y) of the operators that transform CC in completely regular subcones of its. Then E\mathfrak{E} is an automodular convex cone in L(Y)\mathfrak{L}(Y), which is modular over %, where oo\infty is the set in £(Y)£(Y) of CC-positive opcrators transforming the completcly regular subcones of CC in completely regular onces.
Proof. Since C\mathcal{C} is contained in B\mathfrak{B} it suffices to prove that it is modular over 8 . For any B^(')B^{\prime} in B\mathfrak{B} and any C^(')C^{\prime} in C\mathcal{C} the composed operator B^(')C^(')B^{\prime} C^{\prime} is obviously in C\mathcal{C}. Because II is contained in $\$ we have only to prove that B_(1)C_(1)+B_(2)C_(2)B_{1} C_{1}+B_{2} C_{2} is in C\mathcal{C} whenever B_(1)B_{1} and B_(2)B_{2} are in m\mathfrak{m} and C_(1)C_{1} and C_(2)C_{2} are in C\mathcal{C}. But this follows directly from Lemma 7. Q.E.D.
10. Rcmark. From 14 it follows that the cone C\mathcal{C} in general is not closed in the norm topology of the space £(Y)£(Y).
Let AA and BB be in L(Y)\mathcal{L}(Y). We shall put A <= BA \leqslant B if B-AB-A is a CC-positive operator.
11. Proposition. Let CC be a normal cone in YY. Let C\mathcal{C} denote the set of CC-positive operators in f(Y)\boldsymbol{f}(Y) which transform CC in completely yy regular concs. If for some AA and BB in L(Y)\mathcal{L}(Y) there exist the positive scalars alpha\alpha and beta\beta such that
{:(4)alpha B <= A <= beta B",":}\begin{equation*}
\alpha B \leqslant A \leqslant \beta B, \tag{4}
\end{equation*}
then AA is in C\mathcal{C} if and only if BB is in C\mathcal{C}.
Proof. The relation (4) defines in fact an equivalence relation. Hence it is sufficient to show that B inCB \in \mathfrak{C} implies A inCA \in \mathfrak{C}. According the normality of CC, for any sequence ( z_(1)z_{1} ) in CC the norms ||Bz_(1)||\left\|B z_{1}\right\| and ||Az_(1)||,i inN\left\|A z_{1}\right\|, i \in \mathbf{N} are in the same
time bounded from above and respectively, lower bounded by a positive number. Using now Lemma 1 in the way we have done it in the preceeding proofs, we get the required implication. Q.E.D.
2. Examples. The operators of finite range transform a closed cone in completely regular ones. The image by a compact operator of a closed cone is a compactly generated cone and hence the compact operators can be suspected to improve esentially the properties of a cone. Are they of completely regular or of regular type? Unfurtunately they don't. The aim of this paragraph is to show that the property of an operator to be of completely regular as well as of regular type is far to be characterizable with a property like compactness. There are linear and continuous operators of rather general form which are of completely regular type, while some compact operators don't have this property. In the same time we complete the results in the preceeding paragraph.
Let C[0,1]C[0,1] denote the space of continuous real valued functions defined on [ 0,1 ] endowed with the uniform norm and ordered by the cone CC of nonnegative functions. This cone is closed and normal.
12. The linear operators in C[0,1]C[0,1] with the representing kernels bounded rom above and from below by positive multiples of a measure function, which represents a lincar and positive functional, are of completely regular type.
Let AA be a linear and positive operator in C[0,1]C[0,1] and assume that the represcuting kernel KK of its (see e.g. VI. 9.46 in [1]) satisfies the following conditions :
(i) There exist a normalized function gg of bounded variation on [0,1][0,1] and a positive real beta\beta, such that
for any ss and tt in [0,1][0,1];
(ii) There exist an s_(0)s_{0} in [0,1][0,1] and a positive scalar alpha\alpha such that
alpha g(dt) <= K(s_(0),dt)\alpha g(d t) \leqslant K\left(s_{0}, d t\right)
for any tt in [0,1][0,1].
For any yy in the cone CC it holds by (i)
||Ay||=s u p_(s)int_(0)^(1)y(t)K(s,dt) <= betaint_(0)^(1)y(t)g(dt)\|A y\|=\sup _{s} \int_{0}^{1} y(t) K(s, d t) \leqslant \beta \int_{0}^{1} y(t) g(d t)
If ||Ay|| >= d\|A y\| \geqslant d, then this relation together with (ii) yields
{:(5)alpha d//beta <= int_(0)^(1)y(t)K(s_(0),dt)=(Ay)(s_(0)):}\begin{equation*}
\alpha d / \beta \leqslant \int_{0}^{1} y(t) K\left(s_{0}, d t\right)=(A y)\left(s_{0}\right) \tag{5}
\end{equation*}
If ( x_(i)x_{i} ) is a sequence in A(C)A(C) with the property that ||z_(d)|| >= d\left\|z_{d}\right\| \geqslant d for some positive dd and any ii, then we have by (5) the inequality
z_(t)(s_(0)) >= alpha d//betaz_{t}\left(s_{0}\right) \geqslant \alpha d / \beta
and hence
||sum_(i=1)^(n)z_(1)|| >= n alpha d//beta;\left\|\sum_{i=1}^{n} z_{1}\right\| \geqslant n \alpha d / \beta ;
that is, the set {sum_(i=1)^(n)z_(i):n inN}\left\{\sum_{i=1}^{n} z_{i}: n \in \mathbf{N}\right\} cannot be norm bounded. Thus by Lemma 1 , A(C)A(C) is a completely regular cone.
We have in particular, that if the representing kernel KK of the positive and compact operator AA satisfies the condition
for any tt in [0,1][0,1] and some s_(0)s_{0} in this interval, then AA is of completely regular type.
Indeed, we have then
K(s_(0),t)dt >= alpha dt,K\left(s_{0}, t\right) d t \geqslant \alpha d t,
and
K(s,t)dt <= beta dtK(s, t) d t \leqslant \beta d t
for any ss and tt since KK is continuous and hence bounded.
13. Example of a positive integral operator with continuous kernel acting in C[0,1]C[0,1] that isn't of completely regular type.
Consider the increasing sequence ( a_(n)a_{n} ) of distinct real numbers in ( 0,1//20,1 / 2 ). Let we construct the functions k_(n)k_{n} by putting for any n inNn \in \mathbb{N}
is non-negative and continuous on [0,1]xx[0,1][0,1] \times[0,1].
We shall show that the integral operator AA defined by the relation
(Ay)(s)=int_(0)^(1)K(s,t)y(t)dt(A y)(s)=\int_{0}^{1} K(s, t) y(t) d t
isn't of completely regulat type with respect to the cone CC of the non-negative
functions in C[0,1]C[0,1]. To this end, we consider the sequence ( y_(n)y_{n} ) in CC defined by
y_(n)(t)=max{0,c_(n)((a_(n)-a_(n+1))^(2)-(t-a_(n)-a_(n+1))^(2))},quad n inN,y_{n}(t)=\max \left\{0, c_{n}\left(\left(a_{n}-a_{n+1}\right)^{2}-\left(t-a_{n}-a_{n+1}\right)^{2}\right)\right\}, \quad n \in \mathbf{N},
where
{:(6)c_(n)=(int_(2a_(n))^(2a_(n+1))((a_(n)-a_(n+1))^(2)-(t-a_(n)-a_(n+1))^(2))^(2)dt)^(-1):}\begin{equation*}
c_{n}=\left(\int_{2 a_{n}}^{2 a_{n+1}}\left(\left(a_{n}-a_{n+1}\right)^{2}-\left(t-a_{n}-a_{n+1}\right)^{2}\right)^{2} d t\right)^{-1} \tag{6}
\end{equation*}
Then we have the properties
(a) The function z_(n)(s)=int_(0)^(1)k_(n)(s,t)y_(n)(t)dtz_{n}(s)=\int_{0}^{1} k_{n}(s, t) y_{n}(t) d t vanish outside the interval [2a_(4),2a_(n+1)]\left[2 a_{4}, 2 a_{n+1}\right];
(b) ||z_(n)||=1\left\|z_{n}\right\|=1;
(c) int_(0)^(1)k_(m)(s,t)y_(n)(t)dt=0\int_{0}^{1} k_{\mathrm{m}}(s, t) y_{\mathrm{n}}(t) d t=0 for any ss, whenever m!=nm \neq n.
for any nn in N\mathbf{N}, wherefrom via Lemma 1 we conclude that A(C)A(C) isn't a completely regular cone.
14. The integral operator AA constructed in 13 can be indefinitely approximatcd in the norm topology by positive integral opcrators of completely regular type.
We refer for the notations to the preceeding point. Consider the function
wherefrom A_(m)A_{m} converges in the norm to AA when m in oom \in \infty.
We have to check that A_(m)A_{m} is for any mm of completely regular type. We observe first that for any yy in CC the function A_(m)yA_{m} y attains its local maxima at the points a_(1)+a_(2),dots,a_(m)+a_(m+1)a_{1}+a_{2}, \ldots, a_{m}+a_{m+1}. Indeed, suppose that ss is in the interval [2a_(j),2a_(j+1)](j=1,dots,m)\left[2 a_{j}, 2 a_{j+1}\right](j=1, \ldots, m). Then by the property (i) of k_(j)k_{j},
(A_(m)y)(s)=int_(0)^(1)K_(m)(s,t)y(t)dt=int_(0)^(1)k_(1)(s,t)y(t)dt,quad(s in[2a_(1),2a_(j+1)])\left(A_{m} y\right)(s)=\int_{0}^{1} K_{m}(s, t) y(t) d t=\int_{0}^{1} k_{1}(s, t) y(t) d t, \quad\left(s \in\left[2 a_{1}, 2 a_{j+1}\right]\right)
Now, by the property (ii) of k_(j)k_{j},
int_(0)^(1)k_(j)(s,t)y(t)dt <= int_(0)^(1)k_(j)(a_(j)+a_(j+1),t)y(t)dt=(Ay)(a_(j)+a_(j+1))\int_{0}^{1} k_{j}(s, t) y(t) d t \leqslant \int_{0}^{1} k_{j}\left(a_{j}+a_{j+1}, t\right) y(t) d t=(A y)\left(a_{j}+a_{j+1}\right)
that is, for ss in [2a_(j),2a_(j+1)]\left[2 a_{j}, 2 a_{j+1}\right],
Consider now an arbitrary sequence (z_(1))\left(z_{1}\right) in A_(m)(C)A_{m}(C) with the property that ||z_(d)|| >= d\left\|z_{d}\right\| \geqslant d for some positive dd and for any ii. We have
z_(1)=A_(n)y,i inN,z_{1}=A_{n} y, i \in \mathbf{N},
for some y_(i)y_{i} in CC. According to the property (i) of k_(j)k_{j} it follows that z_(i)(s)=0z_{i}(s)=0 for ss in [0,1]\\[2a_(1),2a_(m+1)][0,1] \backslash\left[2 a_{1}, 2 a_{m+1}\right]. By the relation (7) we have that the maximum of zz must be attained on some point a_(1)+a_(j+1,)j=1,dots,ma_{1}+a_{j+1,} j=1, \ldots, m. That is, since ||z_(d)|| >= d\left\|z_{d}\right\| \geqslant d, there exists at least a j(1 <= j <= m)j(1 \leqslant j \leqslant m) so to have
z(a_(1)+a_(j+1)) >= dz\left(a_{1}+a_{j+1}\right) \geqslant d
Because jj can have a finite number of values, it follows that there exists an index h(1 <= h <= m)h(1 \leqslant h \leqslant m) and a subserquence (uz_(i_(i)))\left(u z_{i_{i}}\right) of (z_(j))\left(z_{j}\right) such that
z_(i_(1))(a_(n)+a_(n+1)) >= dz_{i_{1}}\left(a_{n}+a_{n+1}\right) \geqslant d
for any ll in N\mathbf{N}. This means that
sum_(i=1)^(n)z_(i)(a_(n)+a_(h+1)) >= rd\sum_{i=1}^{n} z_{i}\left(a_{n}+a_{h+1}\right) \geqslant r d
and hence the set
{sum_(j=1)^(n)z_(j):n inN}\left\{\sum_{j=1}^{n} z_{j}: n \in \mathbf{N}\right\}
cannot be norm bounded. From Lemma 1 we have then that A_(-)(C)A_{-}(C) is a completely regular cone.
15. The operator AA constructed in 13 is of regular type. We have to show in accordance with Lemma 1 , that if (z_(i))\left(z_{i}\right) is a sequence in A(C)A(C) with the property that there exists a positive dd such that ||z_(d)|| >= d\left\|z_{d}\right\| \geqslant d for any ii, then the set
{:(8){sum_(i=1)^(n)z_(i):n inN}:}\begin{equation*}
\left\{\sum_{i=1}^{n} z_{i}: n \in \mathbf{N}\right\} \tag{8}
\end{equation*}
cannot be A(C)A(C)-order bounded (by any element in A(C)A(C) ).
Let aa be the limit of the sequence (a_(1))\left(a_{1}\right). Then K(2a,t)=0K(2 a, t)=0 for any tt in [0,1][0,1]. Hence z(2a)=0z(2 a)=0 for any zz in A(C)A(C). Assume that zz is an element in A(C)A(C) which is a CC-order bound for the set (8). This means that
sum_(i=1)^(n)z_(i)(s) <= z(s),quad s in[0,1],n inN.\sum_{i=1}^{n} z_{i}(s) \leqslant z(s), \quad s \in[0,1], n \in \mathbf{N} .
Since zz is continuous, z(a,+a_(j+1))rarr z(2a)=0z\left(a,+a_{j+1}\right) \rightarrow z(2 a)=0. Assume that h inNh \in \mathbf{N} has the property that z(a_(1)+a_(j+1)) < d//2z\left(a_{1}+a_{j+1}\right)<d / 2 for any j >= hj \geqslant h. Since z_(s)(s) <= z(s)z_{s}(s) \leqslant z(s) for any ii and since ||z_(d)|| >= d\left\|z_{d}\right\| \geqslant d, it follows that the maximum of any element z_(1)z_{1} must be attained at a point s < a_(h)+a_(h+1)s<a_{h}+a_{h+1}. According the reasonings in the point 14, an ss with this property must be one of the points a_(j)+a_(j+1)a_{j}+a_{j+1} for j <= hj \leqslant h. Hence we get a contradiction as in the above point with the norm boundendess of the set (8) which follows from the CC-order boundendess of it. Now, if the set (8) would be A(C)A(C)-order bounded by some element in A(C)A(C), then it would be also CC-order bounded by the same clement. But this contradicts, as we have seen above, the hypothesis that ||z_(i)|| >= d\left\|z_{i}\right\| \geqslant d for any ii. Thus A(C)A(C) must be regular.
16. Example of a positive integral operator in C[0,1]C[0,1] with continuous kernel, which isn't of regular type.
We shall use the constructions in the example 13, restricting the terms fo the sequence (a_(1))\left(a_{1}\right) to satisfy 1//4 < a_(1) < 1//2,i in N1 / 4<a_{1}<1 / 2, i \in N. Let be a_(0)=1//4a_{0}=1 / 4 and put
is continuous and non-negative and have the property that
int_(0)^(1)K^(1)(s,t)y_(0)(t)dt=1\int_{0}^{1} K^{1}(s, t) y_{0}(t) d t=1
for any ss in [0,1][0,1].
The elements
z_(n)(s)=int_(0)^(1)K^(1)(s,t)y_(n)(t)dt,n_(i)inNz_{n}(s)=\int_{0}^{1} K^{1}(s, t) y_{n}(t) d t, n_{i} \in \mathbf{N}
are of the norm 1 and have the property that ||sum_(n=1)^(m)z_(n)||=1\left\|\sum_{n=1}^{m} z_{n}\right\|=1 for any mm in N. Let us denote by c(s)c(s) the function identically 1 on [0,1][0,1], and let consider the difference
This is for any mm a non-negative function of norm <= 1\leqslant 1.
Consider the sequence (b_(i))\left(b_{i}\right), where b_(i)=a_(i)-1//4,i inNb_{i}=a_{i}-1 / 4, i \in \mathbf{N}, and put
{:h_(n)(s,t)=u_(n)(s)max{b_(n)-b_(n+1))^(2)-(t-b_(n)-b_(n+1))^(2)},n inN.\left.h_{n}(s, t)=u_{n}(s) \max \left\{b_{n}-b_{n+1}\right)^{2}-\left(t-b_{n}-b_{n+1}\right)^{2}\right\}, n \in \mathbf{N} .
h_(a)h_{\mathrm{a}} is a non-negative continuous function vanishing outside the strip [0,1]xx xx[2b_(n),2b_(n+1)][0,1] \times \times\left[2 b_{n}, 2 b_{n+1}\right], satisfying the inequality h_(n)(s,t) <= (b_(n)-b_(n+1))^(2)h_{n}(s, t) \leqslant\left(b_{n}-b_{n+1}\right)^{2}. Hence
with c_(n)c_{n} given by (6). We shall show that the compact operator AA defined by
(Ay)(s)=int_(0)^(1)K(s,t)y(t)dt(A y)(s)=\int_{0}^{1} K(s, t) y(t) d t
where K=K^(1)+K^(2)K=K^{1}+K^{2} isn't of regular type.
We observe first that ee and the sequence (z_(n))\left(z_{n}\right) are in A(C)A(C). Further, we have
{:[int_(0)^(1)K(s","t)v_(n)(t)dt=int_(2b_(n))^(2b_(n+1))h_(n)(s","t)v_(n)(t)dt=],[=c_(n)u_(n)(s)int_(2b_(n))^(2b_(n+1))((b_(n)-b_(n+1))^(2)-(t-b_(n)-b_(n+1))^(2))^(2)dt=u_(n)(s)]:}\begin{gathered}
\int_{0}^{1} K(s, t) v_{n}(t) d t=\int_{2 b_{n}}^{2 b_{n+1}} h_{n}(s, t) v_{n}(t) d t= \\
=c_{n} u_{n}(s) \int_{2 b_{n}}^{2 b_{n+1}}\left(\left(b_{n}-b_{n+1}\right)^{2}-\left(t-b_{n}-b_{n+1}\right)^{2}\right)^{2} d t=u_{n}(s)
\end{gathered}
by the definition of the sequence ( b_(n)b_{n} ) and of the numbers c_(n),n inNc_{n}, n \in \mathbf{N}.
The obtained relation shows that u_(n)u_{n} is in A(C)A(C) and that the set
{sum_(n=1)^(m)z_(n):m inN}\left\{\sum_{n=1}^{m} z_{n}: m \in \mathbf{N}\right\}
is A(C)A(C)-order bounded by the element ee of A(C)A(C). But ||z_(n)||=1\left\|z_{n}\right\|=1 for any nn in N , and invoking Lemma 1 again we conclude that AA isn't of regular type.
17. Example of a linear, positive and compact operator in cc that transforms a completely regular cone in a cone that isn't completely regular.
Denote by cc the space of convergent sequences of real numbers, endowed with the usual norm. Let CC be the cone of the sequences in cc with non-negative terms. The subcome C_(1)C_{1} in CC of the nondecreasing sequences is completely regular. Indeed, if yy is in C_(1),y=(y^(')),y^(')inRC_{1}, y=\left(y^{\prime}\right), y^{\prime} \in \mathbf{R}, then ||y||=limy^(')\|y\|=\lim y^{\prime}. Accordingly, for y_(1)y_{1} and y_(2)y_{2} in C_(1)C_{1} we have ||y_(1)+y_(2)||=||y_(1)||+||y_(2)||\left\|y_{1}+y_{2}\right\|=\left\|y_{1}\right\|+\left\|y_{2}\right\| and hence there cainot exist any sequeuce (y_(1))\left(y_{1}\right) of elements in C_(1)C_{1} such that ||y_(1)|| >= d\left\|y_{1}\right\| \geqslant d for some positive dd and any ii, for which {sum_(i=1)^(n)y_(i):n in(N)}\left\{\sum_{i=1}^{n} y_{i}: n \in \mathrm{~N}\right\} is a norm bounded set. That is, C_(1)C_{1} is completely regular by Lemma 1.
Let us consider the infinite matrix of real numbers denoted by AA,
with delta_(i)^(j)\delta_{i}^{j} standing for the Kronecker symbol. If we define AyA y for some yy in cc as to be the multiplication of AA by the (column) vector yy, then AA can be interpreted as a linear operator in cc. It is straightforward to see that AA is compact.
Define the sequence ( y_(6)y_{6} ) of the elements in the completely regular conc C_(1)C_{1} by putting
for any nn in N\mathbf{N}. That is, A(C_(1))A\left(C_{1}\right) isn't completely regular by Lemma 1.
3. The subdifferentiability of some Hammerstein type operators. A totally ordered subset of the ordered vector space YY is said to be a chain. The space YY is said to be chain complete if any chain that is bounded from below (from above) has an infimum (a supremum) in YY. If YY is a regular space ordered by a closed cone, then the limit of any monotonically decreasing (increasing) sequence is also the infimum (supremum) of this sequence (see II.3.2 in [8]). Hence isn't difficult to show (see for example the reasoning in the proof of Proposition 2 in [7]), that a space with this property is chain complete. Thus for it we have the conditions used in [5] in order to prove the existence of the
subgradients for convex mappings. Ths operator FF from the vector space XX to the ordered vector space (Y, <= )(Y, \leqslant) is said to be convex if
F(tx_(1)+(1-t)x_(2)) <= tF(x_(1))+(1-t)F(x_(2))F\left(t x_{1}+(1-t) x_{2}\right) \leqslant t F\left(x_{1}\right)+(1-t) F\left(x_{2}\right)
for any x_(1)x_{1} and x_(2)x_{2} in XX and any tt in [0,1][0,1]. The linear operator AA from XX to YY is said to be a subgradient of FF at xx if
F(x+u)-F(x) >= AuF(x+u)-F(x) \geqslant A u
for any uu in XX.
Suppose that XX and YY are B-spaces and that YY is ordered by a closed, normal cone with nonempty interior. Then if FF is a continuous convex operator from XX to YY, then from the existence of a subgradient of FF, it follows its continuity. There exist examples (see e.g. [4]) showing that even for rather nice convex operators there are points in the domain of them at which no subgradient exists. We shall use in this paragraph the results we have estabilished in order to give some sufficient conditions for the existence of subgradients. First of all we prove the following preparatory result:
18. Lemma. The closure of any completely regular cone is completely regular too.
Proof. Assume that CC is completely regular and bar(C)\bar{C} isn't. Then there exist a d > 0d>0 and a sequence ( y_(i)y_{i} ) in bar(C)\bar{C} with the property that ||y_(i)|| >= d\left\|y_{i}\right\| \geqslant d for any ii, so to {sum_(i=1)^(n)y_(i):n inN}\left\{\sum_{i=1}^{n} y_{i}: n \in \mathbf{N}\right\} be a norm bounded set (Lemma 1). Suppose that ||sum_(i=1)^(n)y_(i)||⩽⩽alpha\left\|\sum_{i=1}^{n} y_{i}\right\| \leqslant \leqslant \alpha for any nn. Let z_(i)z_{i} be elements in CC which satisfy the conditions ||z_(i)||⩾⩾d//2\left\|z_{i}\right\| \geqslant \geqslant d / 2 and ||z_(i)-y_(i)|| < 2^(-i)\left\|z_{i}-y_{i}\right\|<2^{-i} for any ii. Then
for any nn. That is, the set {sum_(i=1)^(n)z_(i):n inN}\left\{\sum_{i=1}^{n} z_{i}: n \in \mathbf{N}\right\} is norm bounded. Thus we have get a contradiction via Lemma 1 with the hypothesis that CC is a completely regular cone. Q.E.D.
19. Proposition. Let YY be a BB-space ordered by a closcd normal conc CC with nonemply interior and let FF be a continuous convex mapping from the BB-spacc XX to YY. If AA is a positive operator in YY of completely regular type, then the abstract Hammerstein operator AFA F has continuous subgradients at any point of XX.
Proof. From Lemma 18, bar(A(C))\overline{A(C)} will be a closed completely regular cone. The operator AFA F will be convex with respect to the bar(A(C))\overline{A(C)}-ordering in YY and hence it will have bar(A(C))\overline{A(C)}-subgradients in any point of XX. Since bar(A(C))sub C\overline{A(C)} \subset C, these subgradients will be CC-subgradients too. Hence they will. be continuous opcrators by our comments at the beginning of this paragraph. Q.E.D.
20. Corollary. Let the space C[0,1]C[0,1] be ordered by the cone of non-negative functions and let FF be a continuous convex operator acting in it. Consider the Hammerstein operator defined by
G(x)(s)=int_(0)^(1)F(x(t))K(s,dt)G(x)(s)=\int_{0}^{1} F(x(t)) K(s, d t)
where the kernel KK satisfics the conditions in 12. Then GG has continuous subgradients in each point of C[0,1]C[0,1].
Proof. The positive cone in C[0,1]C[0,1] is closed, normal and has nonempty interior. The linear operator defined by
(Ay)(s)=int_(0)^(1)y(t)K(s,dt)(A y)(s)=\int_{0}^{1} y(t) K(s, d t)
is of completely regular type by 12 . Now, G=AFG=A F and hence we are in the conditions of Proposition 19. Q.E.D.
(Received March 13, 1981)
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OPERATORI LINIARI CE TRANSFORMA UN CON NORMAL IN CONURI COMPLET REGULARE
(Rezumat)
In lucrare sint studiați operatorii liniari și continui care transformă un con normal in conuri complet regulare. Se dau conditii suficiente pentru ca un operator liniar şi continuu din spatiul funcțiilor continue definite pe un interval compact de pe axa reală, să aibă aceastà proprietate. Se construiesc operatori liniari şi compacţi definiți în acest spațiu, care nu transformă conul funcţilor pozitive într-un con regular.
