On Equi-Continuity of Bounded on Order Intervals Families of Semi-Norms

It is well known that order bounded operators from a Banach lattice to a normed lattice are continuous (cf., [1, Theorem 1.31]). The norm completeness of domain is essential (for example, the operator of summation of coordinates in the vector lattice c 00 of all eventually zero real sequences equipped with an arbitrary lattice norm, is bounded on every order interval but is not continuous). In general, it is a difficult task to establish continuity of operators of certain classes (see, e.g., [2–4]). Thus, finding of new equivalent conditions for continuity of operators is of considerable interest.

• ^# The work of the second author was carried out in the framework of the State Task to the Sobolev Institute of Mathematics (Project FWNF-2026-0022).

• 2.    Main Results

(О 2026 Gorokhova, S. G., Emelyanov, E. Y. and Storozhuk, K. V.

It was proved in [5, Theorem 2.1] that collectively order-to-norm bounded sets of operators from a Banach lattice to a normed lattice are equi-continuous. Later, this was extended in [6, Theorem 2.1] to the case of ordered Banach space with a closed generating cone. In the present note, we generalize the above results by proving Theorem 1 which establishes that every equi-bounded on each order interval family of semi-norms on an ordered Banach space with a closed generating cone is equi-continuous. In the end of this note, we also include Theorem 4 extending Theorem 1 further, to families of semi-norms on a completely metrizable ordered vector space with a closed generating positive cone.

We abbreviate a normed (topological, ordered, ordered normed, ordered Banach) vector space by NS (TVS, OVS, ONS, OBS respectively). Throughout the paper, vector spaces are real, operators are linear, L(X,Y ) stands for the space of operators from a vector space X to a vector space Y , B x (a,r) = { x G X : d(x,a) C r } for the closed ball centered at a G X of radius r ^ 0 in a metric space (X, d), and B x for the closed unit ball of an NS X.

Let X be an OVS. If Y is an NS, an operator T G L(X,Y ) is order-to-norm bounded whenever T[a, b] is bounded for each order interval [a, b] of X. More generally, if Y is a TVS, a family T C L(X,Y ) is order-to-topology bounded whenever the set T [a,b] = Ut <=t T[a,b] is topologically bounded for every [a, b] C X.

Recall that a family F of mapping from a metric space (X, d) to a TVS (Y, т ) is equi-continuous on a subset B of X if: for every U G т (0), there exists e >  0, such that for all f G F and all x,y G B , we have f (x) G f (y) + U whenever d(x,y) C e. We say that a family S of semi-norms on a vector space X is equi-bounded on a subset A of X , if SA := Use s s(A) is bounded in R . Clearly, a family S of semi-norms on an NS X is equi-continuous, iff it is equi-bounded on B X . We denote the set of equi-bounded on each order interval of an OVS X semi-norms by S _eb_oi (X).

For further unexplained terminology and notations, we refer to [1, 7, 8].

Recall that a positive cone X ₊ of an OVS X is generating whenever X ₊ — X ₊ = X . The Krein–Smulian theorem [8, Theorem 2.37] tells that, in an OBS X with a closed generating cone, B x П X ₊ — B x П X ₊ is a neighborhood of zero. We extend [5, Theorem 2.1] and [6, Theorem 2.1] to the semi-norm setting as follows.

Theorem 1. If X is an OBS with a closed generating cone then a family S of semi-norms on X is equi-continuous whenever S is equi-bounded on each order interval.

• < 1 Assume the opposite: some S G S _eb_oi (X) is not equi-continuous. By the Krein-Smulian theorem, aB x C B x П X ₊ — B x П X ₊ for some a >  0, and hence S is not equi-continuous on B x П X ₊ . So, there exist sequences (x _n ) in B x П X ₊ and (s _n ) in S , such that s n (x n ) > n ³ for all n. Set the norm limit x := || • ||-£ ^ =1 n ^{- 2} x n . As X ₊ is closed, x G X ₊ and n ^{- 2} x n G [0,x] for each n. Since S G S _eb_oi (X), we have S [0, x] C [0, N] for some N G N . It follows from n ^{- 2} x _n G [0, x] that s _n (nT²x n ) G [0, N] C [0, n] for large enough n. It is absurd, since s n (n ^{- 2} x n ) = n ^{- 2} s _n (x n ) > n for all n. О

An inspection of the proof of Theorem 1 shows: the requirement that X ₊ is generating can be weakened to the requirement that B x П X ₊ — B x П X ₊ is a neighborhood of zero (this is to say, X ₊ gives an open decomposition in X). It is worth mention that the property of open decomposition in topological OVSs was studied in a lot of papers of I. A. Polyrakis and co-authors (see, for example, [9] and references therein).

Remark. Assuming in Theorem 1 the semi-norms consisting the family S to be continuous, one can avoid closedness of X ₊ . Indeed, let us take x G X .It follows from X ₊ —

X ₊ = X that x = x i — X 2 for some x i ,X 2 € X + . As S € S eboi (X), the set S [ - x 2 ,x i ] is bounded. Thus, S x is bounded, because x € [ — x 2 ,x i ]. Since x € X is arbitrary, X = U ~i ris e s s ¹ [0, n], where the sets rs e S ^s 1 [0, n] are closed by continuity of semi-norms in S . The Baire category theorem implies B x (a,r) C Q_{s e} S s ^{- 1} [O,n o ] for some n g € N , r >  0, and a € X .As above, S a C [0, M ] for some M € N . Since s(x) ^ s(x + a) + s(a) for all s € S , we have sup s _e S ,x _e B _X (o ,r ) s(x) ^ n o + M , and hence S is equi-continuous.

Corollary 1. If X is an OBS with a closed generating cone, then every bounded on each order interval semi-norm on X is continuous.

The assumption on X to be Banach space is essential in the above Corollary. Indeed, let us consider a semi-norm s(x) = ^2^ =1 \ x k | on c gg . Then s is equi-bounded on each order interval of c gg , yet not continuous. To see the condition X ₊ — X ₊ = X is essential, take any infinite-dimensional Banach space X , any discontinuous semi-norm s on X , and set a trivial cone X ₊ = { 0 } in X. Then s is bounded on each order interval of c gg . The closedness of X ₊ is also essential. To see this, take any infinite dimensional Banach space X , make it an OBS, and set a semi-norm s(x) = ||Tx||, where the operator T : X ^ X is as in [6, Example 2.12b)].

The key result of [6] turns a consequence of Theorem 1 as follows.

Theorem 2 [6, Theorem 2.1] . Let X be an OBS with a closed generating cone, Y be an NS, and T be an order-to-norm bounded subset of L(X,Y ) . Then T is bounded in the operator norm.

• < 1 For each T € T define a semi-norm s t (x) = | Tx | on X. Since T is order-to-norm bounded, S T = { s T : T € T } € S _eboi (X). By Theorem 1, S T is equi-continuous. So, S T is bounded on B x , and hence the set T is norm bounded. >

Recall that a positive cone X ₊ of an ONS X is normal whenever order intervals of X are bounded. It is worth noting that not every bounded operator in the conditions of Theorem 2 is bounded on order intervals. As an example, consider the identity operator I X on the OBS X = C 1 [0,1], and observe that X ₊ is closed and generating (but not normal).

Corollary 2 [10, Theorem 1] . Every order bounded operator from an OBS with a closed generating cone to an ONS with a normal cone is continuous.

The next theorem should be compared with [11, Theorem 2.5].

Theorem 3. Let X be an OBS with a closed generating cone and (Y, т ) be a locally convex space. Then every order-to-topology bounded subset of L(X, Y ) is equi-continuous.

• <    Let T be an order-to-topology bounded subset of L(X,Y ). Take a base U of т at zero consisting of convex circled open sets and denote by f U the Minkowski functional of U € U . For every T € T and U € U , we define a semi-norm s U on X by s U (x) = \ f u (Tx) \ . Since T is order-to-topology bounded, the set T [a,b] is т -bounded in Y for each order interval [ a, b ] C X . Therefore, S_T ^U = { s _T ^U : T € T } € S _eboi ( X ) for each U € U . Theorem 1 implies that the corresponding set of operators is bounded on B x for each U € U , say sup T _e t . x e B X \ f u (Tx) = sup T _e T . x e B X s U (x) <  N u € N . It follows T{N - ^Bx) C U for every U € U . Since U is a base of т at zero, the set T is equi-continuous. >

Corollary 3. Every order-to-topology bounded operator from an OBS with a closed generating cone to a locally convex space is continuous.

We conclude the paper with the following extension of Theorem 1.

Theorem 4. Let (X, т ) be a completely metrizable OVS with a т -closed generating positive cone. Then every family of semi-norms on X equi-bounded on each order interval is equi-continuous.

• < 1 Suppose some S G S eboi (X) is not equi-continuous. Since т is completely metrizable, there exist sequences (x n ) in X and (s n ) in S , such that x n -4 0 and s n (x n ) ^ n ² for all n. By [8, Lemma 2.30], there exist a subsequence (x n k ) and u G X + , such that for each k we have n k x n k G [ — u,u]. Since S G S eboi (X), then S [ — u,u] C [0,N] for some N G N . It follows from n k x n k G [ — u, u] that s n _k (n k ^{- 1} x n _k ) G [0, N] C [0, k] for large enough k. It is absurd, since s n k (n k -1 x n _k ) > n k ^ k for all k. >

Remark. Note that under the conditions of Theorem 4, the conclusion of the Krein– Smulian theorem also holds [8, Corollary 2.11]. However, the proof of Theorem 1 does not directly transfer to the current situation. On the other hand, if X + has a nonempty interior, the metrizability of τ and τ -closedness of X + are not necessary in Theorem 4. Indeed, if u is an interior point of X + then [ — u, u] is a neighborhood of zero [8, Lemma 2.5]. The rest is obvious.

Corollary 4. If the cone of a completely metrizable OVS X is closed and generating, then every family of linear functionals on X equi-bounded on each order interval of X is equi-continuous.

• <    Let F be an order-to-norm bounded subset of L(X, R ). For each f G F , define a semi-norm S f (x) = | f (x) | on X. Since F is equi-bounded on each order interval of X, then { s f } f ^ f G S eboi (X). Theorem 4 implies that { s f } f _E F is equi-continuous, and hence F is equi-continuous. >

In particular, we have the following extension of the result of G. Ya. Lozanovsky (cf., [8, Theorem 2.34]).

Corollary 5. If the cone of a completely metrizable OVS X is closed and generating, then every order bounded linear functional on X is continuous.