Continuity and Discontinuity of Seminorms on Infinite-Dimensional Vector Spaces. II
aa r X i v : . [ m a t h . F A ] M a r CONTINUITY AND DISCONTINUITY OF SEMINORMSON INFINITE-DIMENSIONAL VECTOR SPACES. II
JACEK CHMIELI ´NSKI
Department of Mathematics, Pedagogical University of Krak´owKrak´ow, PolandE-mail: [email protected]
MOSHE GOLDBERG Department of Mathematics, Technion – Israel Institute of TechnologyHaifa, IsraelE-mail: [email protected]
Abstract.
In this paper we extend our findings in [3] and answer further questions regard-ing continuity and discontinuity of seminorms on infinite-dimensional vector spaces.
Throughout this paper let X be a vector space over a field F , either R or C . As usual,a real-valued function N : X → R is a norm on X if for all x, y ∈ X and α ∈ F , N ( x ) > , x = 0 ,N ( αx ) = | α | N ( x ) ,N ( x + y ) ≤ N ( x ) + N ( y ) . Furthermore, a real-valued function S : X → R is called a seminorm if for all x, y ∈ X and α ∈ F , S ( x ) ≥ ,S ( αx ) = | α | S ( x ) ,S ( x + y ) ≤ S ( x ) + S ( y );hence, a norm is a positive-definite seminorm.Using standard terminology, we say that a seminorm S is proper if S does not vanishidentically and S ( x ) = 0 for some x = 0 or, in other words, ifker S := { x ∈ X : S ( x ) = 0 } , is a nontrivial proper subspace of X . Mathematics Subject Classification.
Key words and phrases. infinite-dimensional vector spaces, Banach spaces, seminorms, norms, norm-topologies, continuity, discontinuity. Corresponding author.
Lastly, just as for norms, we say that seminorms S and S are equivalent on X , if thereexist positive constants β ≤ γ such that for all x ∈ X , βS ( x ) ≤ S ( x ) ≤ γS ( x ) . We recall that if X is finite-dimensional, then all norms on X are equivalent, thus givingrise to a unique norm-topology on X . This well-known fact leads to the following result. Theorem 1 ([6]) . Let S be a seminorm on a finite-dimensional vector space X over F . Then S is continuous with respect to the unique norm-topology on X . Unlike the finite-dimensional case, if X is infinite-dimensional then not all norms on X are equivalent, and accordingly we no longer have a unique norm-topology. Indeed, inour previous paper [3] we explored continuity and discontinuity of seminorms when theassumption of finite-dimensionality was removed. Our two main findings were: Theorem 2 ([3, Theorem 2]) . Let X , an infinite-dimensional vector space over F , be equippedwith a seminorm S and a norm N . Then: (a) S is ubiquitously continuous in X with respect to the topology induced by N if andonly if there exists some point of X at which S is continuous. (b) Similarly, S is ubiquitously discontinuous in X with respect to the above mentionedtopology if and only if there exists some point of X at which S is discontinuous. Theorem 3 ([3, Theorem 3]) . Let S = 0 be a seminorm on an infinite-dimensional vectorspace X over F . Then: (a) There exists a norm with respect to which S is ubiquitously continuous in X. (b)
There exists a norm with respect to which S is ubiquitously discontinuous in X.
As it is, the above results trigger several new questions which are the main business of thepresent paper.Having a second look at Theorem 1, we begin our quest by asking whether there existsan infinite-dimensional normed space X which resembles to some extent the behavior of thefinite-dimensional case in the sense that all seminorms on that space are continuous withrespect to the given norm topology in X . As we shall see next, the answer to this questionis negative: Theorem 4.
Let N be a norm on an infinite-dimensional vector space X over F . Thenthere exist a norm N ′ as well as a proper seminorm S which are ubiquitously discontinuousin X with respect to the norm topology induced by N .Proof. Let B N := { x ∈ X : N ( x ) ≤ } be the unit ball in our normed space. Sincespan B N = X , we consult Corollary 4.2.2 in [7] which ensures us that B N contains a Hamelbasis H for X . Now, fix a sequence { h , h , h , . . . } of distinct elements in H and set g n = h n n , n = 1 , , , . . . . Then B := { g , g , g , . . . } ∪ ( H r { h , h , h , . . . } ) ONTINUITY AND DISCONTINUITY OF SEMINORMS 3 is a Hamel basis of X as well. So every x in X assumes a unique representation of the form x = X b ∈ B α b ( x ) b, α b ( x ) ∈ F , where { b ∈ B : α b ( x ) = 0 } is a finite set.With this representation at our disposal, we can easily confirm that the real-valued func-tions N ′ ( x ) := X b ∈ B | α b ( x ) | , x ∈ X, and S ( x ) := X b ∈ B r { g } | α b ( x ) | , x ∈ X, are a norm and a proper seminorm on X , respectively.We get, however, that N ( g n ) = 1 n N ( h n ) ≤ n , n = 1 , , , . . . , whereas, N ′ ( g n ) = 1 , n = 1 , , , . . . , and S ( g n ) = 1 , n = 2 , , , . . . . Consequently, N ′ and S are discontinuous at zero with respect to N ; so by Theorem 2(b),both N ′ and S are ubiquitously discontinuous in X with respect to the norm topologyinduced by N and our assertion follows. (cid:3) In view of Theorem 4, it seems natural to ask whether, given a norm N on X , there existsanother norm or a proper seminorm which are ubiquitously continuous in X with respectto N . Surely, every norm which is equivalent to N provides a positive answer to the abovequestion. If, however, we look for a non-equivalent norm which is ubiquitously continuouswith respect to the topology induced by N , we have nothing to say, unless X is completewith respect to both norms—a case discussed in Corollary 1 later on. Turning to properseminorms, the answer to our question is positive if somewhat involved: Theorem 5.
Let N be a norm on an infinite-dimensional vector space X over F . Thenthere exists a proper seminorm S which is ubiquitously continuous in X with respect to N .Proof. Let V be a nontrivial finite-dimensional subspace of X , and let U be another subspaceof X such that X = U ⊕ V . Furthermore, let N ′ : U → R be the real-valued function thatmeasures, with respect to N , the distance from elements in U to the subspace V ; that is, N ′ ( u ) := dist N ( u, V ) = inf v ∈ V N ( u − v ) , u ∈ U. (1)Since V is finite-dimensional, the infimum in (1) is attained, so we can write N ′ ( u ) = min v ∈ V N ( u − v ) , u ∈ U. We shall now show that N ′ is a norm on U . Indeed, if N ′ ( u ) = 0 for some u ∈ U , thenthe closedness of V implies that u ∈ V ; thus u = 0, and it follows that N ′ ( u ) > u = 0 . JACEK CHMIELI ´NSKI AND MOSHE GOLDBERG
Moreover, for u ∈ U and λ ∈ F r { } we get N ′ ( λu ) = min v ∈ V N ( λu − v ) = min v ∈ V N ( λu − λv ) = | λ | min v ∈ V N ( u − v ) = | λ | N ′ ( u ) . And since N ′ (0) = 0, we infer that N ′ ( λu ) = | λ | N ′ ( u ) for all u ∈ U and λ ∈ F. Finally, select u , u ∈ U . Then there exist elements v , v ∈ V such that N ′ ( u ) = N ( u − v ) and N ′ ( u ) = N ( u − v ) . Therefore, N ′ ( u + u ) = min v ∈ V N ( u + u − v ) ≤ N ( u + u − ( v + v )) ≤ N ( u − v ) + N ( u − v ) = N ′ ( u ) + N ′ ( u ) , and the fact that N ′ is a norm on U is in the bag.Next, for any x ∈ X and its unique decomposition x = u + v where u ∈ U and v ∈ V , weput S ( x ) := N ′ ( u ) . Since N ′ is a norm on U , it is easily verified that the mapping S : X → R is a seminorm on X . Furthermore, since ker S = V and V is a nontrivial proper subspace of X , we concludethat S is a proper seminorm on X.Lastly, for any x = u + v in X with u ∈ U and v ∈ V , we obtain S ( x ) = N ′ ( u ) = min v ′ ∈ V N ( u − v ′ ) ≤ N ( u + v ) = N ( x ) . Hence, S is majorized by N on X , implying the continuity of S at zero; so Theorem 2(a)forces the desired result. (cid:3) Falling back on Theorem 3(a), we remember that every seminorm on an infinite-dimen-sional vector space X is ubiquitously continuous with respect to some norm on X . Assumingthat a given seminorm is ubiquitously continuous with respect to two norms, we ask whetherboth norms are necessarily equivalent . The following example answers this question in thenegative:
Example 1.
Let c denote the familiar space of all infinite F -valued sequences with finitesupport, i.e., c = { x = { ξ , ξ , ξ , . . . } : ξ i ∈ F , ∃ i such that ξ i = 0 for all i ≥ i } . Consider the proper seminorm S ( x ) := | ξ | , x = { ξ , ξ , ξ , . . . } ∈ c , and the two norms N ( x ) := ∞ X i =1 | ξ i | , x = { ξ , ξ , ξ , . . . } ∈ c ,N ∞ ( x ) := max i | ξ i | , x = { ξ , ξ , ξ , . . . } ∈ c . Surely, S ( x ) ≤ N ∞ ( x ) ≤ N ( x ) for all x ∈ c . (2) ONTINUITY AND DISCONTINUITY OF SEMINORMS 5
Now let { x n } ∞ n =1 be an arbitrary sequence in c . If either N ( x n ) → N ∞ ( x n ) → n → ∞ then, by (2), S ( x n ) →
0, hence S is continuous at zero with respect to N and N ∞ ;so by Theorem 2(a), S is ubiquitously continuous with respect to both norms.As it is, however, N and N ∞ are non-equivalent. To justify this statement, set x n = { n , . . . , n | {z } n times , , , , . . . } , n = 1 , , , . . . . Then N ( x n ) = 1 for all n , whereas N ∞ ( x n ) = n → n → ∞ , and we are done.We appeal now to the second part of Theorem 3 which tells us that every seminorm on X is ubiquitously discontinuous with respect to some norm on X . In analogy to our previousquestion, we assume that a given seminorm is ubiquitously discontinuous with respect to twonorms on X , and ask whether these norms are necessarily equivalent . Again, the followingexample furnishes a negative answer: Example 2.
Let us resort to the space c and to the non-equivalent norms N and N ∞ in Example 1. By Theorem 4, there exists a proper seminorm S which is ubiquitouslydiscontinuous in X with respect to N . In particular, S is discontinuous at zero, so for somesequence { x n } ∞ n =1 in c we have N ( x n ) → S ( x n )
0. Further, since N ∞ ≤ N on c , we get N ∞ ( x n ) → S ( x n )
0, implies the discontinuity of S at zero with respect to N ∞ . It thus follows that S is ubiquitously discontinuous in c withrespect to N and N ∞ , and our goal is achieved.With Examples 1 and 2 in store, we assert that the space c is incomplete with respect toeither of the corresponding norms N and N ∞ .To substantiate our claim, consider for example the sequence { x n } ∞ n =1 in c where x n = { , , . . . , n , , , , . . . } , n = 1 , , , . . . . For any positive integers n > m we get N ( x n − x m ) = n X i = m +1 i ≤ m and N ∞ ( x n − x m ) = 12 m +1 , which renders { x n } a Cauchy sequence with respect to both N and N ∞ .Now fix an arbitrary element y = { η , η , . . . , η k , , , , . . . } in c . Then, for all n , n > k ,we obtain N ( x n − y ) ≥ k +1 and N ∞ ( x n − y ) ≥ k +1 . Whence y may not be the limit of { x n } with respect to either N or N ∞ , and our assertionis validated.In fact, c cannot be made into a Banach space, for it is incomplete with respect to anynorm . To verify this statement we summon the canonical basis { e n } ∞ n =1 of c where e n isthe vector whose n -th entry is 1 and all others vanish. Evidently, { e n } is a countable Hamelbasis for c , while it is known (e.g., [2], [8]) that a Hamel basis of a (separable or not)infinite-dimensional Banach space is uncountable. JACEK CHMIELI ´NSKI AND MOSHE GOLDBERG
With the above observation in mind, we attend now to the case where X is a completespace with respect to certain norms. Following standard nomenclature, we shall henceforthcall a norm N complete if X is complete with respect to N . Theorem 6.
Let N be a norm on an infinite-dimensional vector space X over F , such that N is ubiquitously continuous with respect to two complete norms N and N . Then N and N are equivalent.Proof. Consider the norm N ′ ( x ) := max { N ( x ) , N ( x ) } , x ∈ X, and let us prove that N ′ is complete. Select an arbitrary Cauchy sequence { x n } ∞ n =1 withrespect to N ′ so that N ′ ( x n − x m ) → n, m → ∞ . Since N ′ majorizes both N and N , it follows that { x n } is a Cauchy sequence with respectto N and N as well.Furthermore, since N and N are complete, we may exhibit elements x ′ , x ′′ ∈ X such that N ( x n − x ′ ) → N ( x n − x ′′ ) → , n → ∞ ; (3)so by the continuity of N with respect to N and N , we conclude that N ( x n − x ′ ) → N ( x n − x ′′ ) → n → ∞ . Consequently, N ( x ′ − x ′′ ) ≤ N ( x ′ − x n ) + N ( x n − x ′′ ) → n → ∞ ;thus N ( x ′ − x ′′ ) = 0, and we infer that x ′ = x ′′ = ¯ x for some ¯ x ∈ X . By (3) therefore, N ( x n − ¯ x ) → N ( x n − ¯ x ) →
0; hence N ′ ( x n − ¯ x ) → N ′ issecured.Finally, we invoke the well-known Banach inverse mapping theorem (e.g., [5, Corollary10.9]), which implies, [5, Corollary 10.10], that two complete norms on X such that onemajorizes the other, must be equivalent .Indeed, since N , N and N ′ are complete, and as N ( x ) ≤ N ′ ( x ) , N ( x ) ≤ N ′ ( x ) , x ∈ X, we deduce that N and N are each equivalent to N ′ . Whence N and N are equivalent,and the proof is at hand. (cid:3) Reflecting on Theorem 6, we maintain that the assumption of completeness of N and N cannot be dropped . Indeed, revisit the space c and the norms N and N ∞ in Example 1.Putting N = N ∞ and employing the fact that N ∞ ≤ N , we ascertain that N is ubiquitouslycontinuous with respect to both N and N ∞ ; yet, as shown in Example 1, N and N ∞ arenon-equivalent. ONTINUITY AND DISCONTINUITY OF SEMINORMS 7
With Theorem 6 fresh in our mind, we may record the following simple proposition:
Corollary 1.
Let N and N be two complete norms on an infinite-dimensional vector space X over F . Then: (a) N is ubiquitously continuous with respect to N if and only if N and N are equiv-alent. (b) N is ubiquitously discontinuous with respect to N if and only if N and N arenon-equivalent. (c) N is ubiquitously continuous with respect to N if and only if N is ubiquitouslycontinuous with respect to N . (d) N is ubiquitously discontinuous with respect to N if and only if N is ubiquitouslydiscontinuous with respect to N .Proof. To prove (a), set N = N . Then Theorem 6 forces the equivalence of N and N since N and N are complete, and N is ubiquitously continuous with respect to both norms. Parts(b)–(d) are obtained from part (a) without much difficulty, and the corollary follows. (cid:3) Part (a) of Corollary 1 tells us that a Banach space may not admit another non-equivalentcomplete norm which is ubiquitously continuous with respect to the original norm.
Thequestion whether a Banach space may admit an incomplete norm which is non-equivalent aswell as ubiquitously continuous with respect to the original one , is answered affirmatively bythe following example.
Example 3.
Consider the space l ∞ of all infinite F -valued bounded sequences with the usualnorm N ∞ ( x ) := sup i | ξ i | , x = { ξ , ξ , ξ , . . . } ∈ l ∞ . It is well known (e.g., [4, Proposition 1.16]) that equipped with the above norm, l ∞ is aBanach space over F .Define a second norm on l ∞ by N ′ ( x ) := ∞ X i =1 − i | ξ i | , x = { ξ , ξ , ξ , . . . } ∈ l ∞ . Obviously, N ′ ( x ) ≤ N ∞ ( x ) , x ∈ l ∞ ;so N ′ is ubiquitously continuous with respect to N ∞ .Recall now the sequence { e n } ∞ n =1 where, as before, e n is the vector whose n -th entry is 1and all others vanish. Since N ′ ( e n ) = 2 − n → n → ∞ , and N ∞ ( e n ) = 1 for all n = 1 , , , . . . , the possibility that our two norms are equivalent shutters. Finally, we observe that N ′ isincomplete on l ∞ for otherwise, Theorem 6 would imply the equivalence of N ′ and N ∞ , acontradiction. JACEK CHMIELI ´NSKI AND MOSHE GOLDBERG
Encouraged by the above example, we ask whether every Banach space admits an incom-plete norm which is non-equivalent as well as ubiquitously continuous with respect to theoriginal norm . The answer to this question remains open.Bringing up Theorem 6 again, we shall show now that the role of the norm N may notbe replaced by a proper seminorm. More precisely, the following example will confirm that if S is a proper seminorm on an infinite-dimensional vector space X over F , such that S isubiquitously continuous with respect to two complete norms N and N , then N and N arenot necessarily equivalent . Example 4.
Let l be the space of all absolutely summable F -valued sequences with thefamiliar norm, N ( x ) := ∞ X i =1 | ξ i | , x = { ξ , ξ , ξ , . . . } ∈ l , and let c be the space of all F -valued sequences that converge to zero with N ∞ ( x ) := sup i | ξ i | , x = { ξ , ξ , ξ , . . . } ∈ c . (4)It is well known that furnished with the above norms, both l and c are separable Banachspaces (e.g., [4, Propositions 1.16 and 1.42]). Hence (see [8]), choosing Hamel bases, B for l and B ′ for c , these bases must be of the same cardinality c ; so there exists a bijection f from B onto B ′ . Now, as in the proof of Theorem 4, each x in l takes on a uniquerepresentation of the form x = X b ∈ B α b ( x ) b, α b ( x ) ∈ F , where { b ∈ B : α b ( x ) = 0 } is a finite set. Thus f ( x ) := X b ∈ B α b ( x ) f ( b ) , x ∈ l , is a linear mapping from l into c ; and since f is a bijection from B onto B ′ , we leave it tothe reader to verify that f is a linear bijection from l onto c .Next, for every x = { ξ , ξ , ξ , . . . } ∈ c we define three auxiliary mappings: The left shift L ( x ) := { ξ , ξ , ξ , . . . } , the right shift R ( x ) := { , ξ , ξ , ξ , . . . } , and the truncation T ( x ) := { ξ , , , , . . . } . Obviously, L , R , and T are linear mappings from c into c . Further, since l is a linearsubspace of c , we observe that L , R , and T are linear from l into l .Aided by the above mappings, we define yet another map, F : l → c , by F ( x ) := R ( f ( L ( x ))) + T ( x ) , x ∈ l . ONTINUITY AND DISCONTINUITY OF SEMINORMS 9
Clearly, F is linear. Moreover, if F ( x ) = 0 then, by the definitions of R and T , we notethat both R ( f ( L ( x ))) = 0 and T ( x ) = 0. In addition, the first of these conditions im-plies that f ( L ( x )) = 0; and since f is a bijection, it follows that L ( x ) = 0. This, to-gether with T ( x ) = 0, ensures that x = 0. Thus we have shown that F ( x ) = 0 implies x = 0, so F is injective. Furthermore, selecting an arbitrary element y ∈ c , and setting x := R ( f − ( L ( y ))) + T ( y ), it is quite straightforward to confirm that x belongs to l and F ( x ) = y ; hence F is surjective, and we conclude that F is a linear bijection from l onto c .Introduce now the real-valued function N : l → R , defined by N ( x ) := N ∞ ( F ( x )) , x ∈ l , where N ∞ is the norm in (4). Since F is a linear bijection and N ∞ is a complete norm on c , it is a routine matter, if somewhat tedious, to verify that N , just like N , is a completenorm on l , a task left to the reader.Next, consider a seminorm on l defined by S ( x ) := | ξ | , x = { ξ , ξ , ξ , . . . } ∈ l . Evidently, S is majorized by N on l . Further, we point out that for any x = { ξ , ξ , ξ , . . . } ∈ l , the first entry of the sequence F ( x ) is ξ , so S is majorized by N as well. It thus followsthat S is continuous at zero, hence ubiquitously in l with respect to N and N .As it stands now, we have displayed on l a seminorm S and two complete norms, N and N , such that S is ubiquitously continuous with respect to both norms. Hence, in order toattain our goal, it suffices to prove that N and N are non-equivalent.Indeed, suppose to the contrary that N and N are equivalent, so that for some positiveconstants β ≤ γ , βN ( x ) ≤ N ( x ) ≤ γN ( x ) , x ∈ l ;that is, βN ( x ) ≤ N ∞ ( F ( x )) ≤ γN ( x ) , x ∈ l , or in other words, since F ( x − x ′ ) = F ( x ) − F ( x ′ ) for all x and x ′ in l , βN ( x − x ′ ) ≤ N ∞ ( F ( x ) − F ( x ′ )) ≤ γN ( x − x ′ ) , x, x ′ ∈ l . (5)As we shall see, however, (5) will imply that F is an isomorphism between the Banachspaces l and c . And this, in turn, will lead to a contradiction since it is known (e.g., [1,Corollary 2.1.6]) that l and c are non-isomorphic.We recall that a mapping from one Banach space to another is an isomorphism if itis a linear continuous bijection with a continuous inverse. Hence to prove that F is anisomorphism between l and c , it remains to show that F and its inverse F − are continuous.To this end, let us first consider an arbitrary sequence { x n } ∞ n =1 in l such that x n → x ′ forsome x ′ ∈ l . Referring to the right inequality in (5), we get N ∞ ( F ( x n ) − F ( x ′ )) ≤ γN ( x n − x ′ ) → n → ∞ . Hence F ( x n ) → F ( x ′ ), and it follows that F is ubiquitously continuous in l .Conversely, let { y n } ∞ n =1 be an arbitrary sequence in c such that y n → y ′ for some y ′ ∈ c .Since F is a bijection from l onto c , we can find a sequence { x n } ∞ n =1 and an element x ′ in l so that y n = F ( x n ) and y ′ = F ( x ′ ). Whence, the left inequality in (5) yields, βN ( x n − x ′ ) ≤ N ∞ ( F ( x n ) − F ( x ′ )) → n → ∞ . Thus x n → x ′ or, otherwise put, F − ( y n ) → F − ( y ′ ); so F − is ubiquitously continuous in c , and Example 4 is in our grasp.As our paper draws to its end, we consider for the last time a nontrivial seminorm S onan infinite-dimensional vector space X . Surely, the class of all norms on X is the unionof two distinct classes, C and D , where C consists of all norms with respect to which S iscontinuous, and D is the set of all norms with respect to which S is discontinuous. Whilethe characterization of these two classes seems to be of interest, we have no clue on how toapproach this job. Hence we leave it in the good hands of the reader as an open problem. References [1] Fernando Albiac and Nigel J. Kalton,
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