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TWISTED SUMS AND A PROBLEM OF KLEE

arXiv:math/9302205v1 [math.FA] 3 Feb 1993TWISTED SUMS AND A PROBLEM OF KLEEN. T. PeckTo Victor KleeIn [5], Klee asked whether every vector topology τ on a real vector space X isthe supremum of a nearly convex topology τ1 and a trivial dual topology τ2.

Recallthat a vector topology τ1 on X is nearly convex if for every x not in the τ1–closureof {0} there is f in (X, τ1)∗with f(x) ̸= 0; τ2 is trivial dual if (X, τ2)∗= {0}.We do not require that τ1 or τ2 be Hausdorff, even if τ itself is Hausdorff. Thetopology τ is the supremum of τ1 and τ2 if τ1 and τ2 are weaker than τ, and if forevery τ–neighborhood U of the origin 0 there are a τ1–neighborhood V of 0 and aτ2–neighborhood W of 0 such that U ⊃V ∩W.In [5], Klee proved that the usual topology on ℓp, 0 < p < 1, is not the supremumof a locally convex topology and a trivial dual topology; this and other examplesmake the question at the beginning of this paper a natural one.Some relatedquestions on suprema of linear topologies were studied in [7].Given any vector topology τ on X, let K(τ) = ∩{f −1(0) : f ∈(X, τ)∗}.

Itis trivial to answer Klee’s question affirmatively in the case that K(τ) is comple-mented. For in this case, K(τ) must be a trivial dual space in the relative topology;and if L is a complement to K(τ) in X, the relative topology on L is nearly convex.Now simply let τ1 be the product of the trivial topology on K(τ) and the relativeTt bAMS T X

2N. T. PECKtopology on L; and let τ2 be the product of the relative topology on K(τ) and thetrivial topology on L. Then τ = sup(τ1, τ2).So the interesting case is when K(τ) is uncomplemented.

We study the problemwhen (X, τ) is the twisted sum of a separable normed space and the real line. Recallthat a real function F on a normed space E is quasi–linear if(0) (i)F(rx) = rF(x) for all scalars r and all x in E;(ii)|F(x+y)−F(x)−F(y)| ≤C(∥x∥+∥y∥) for all x, y in E and some constantC.Now define the twisted sum of the real line and E (with respect to F) as thevector space XF = R × E equipped with quasi–norm ∥|(r, x)|∥= |r −F(x)| + ∥x∥.It is easy to verify that∥|(r1 + r2, x1 + x2)|∥≤(C + 1)[∥|(r1, x1)|∥+ ∥|(r2, x2)|∥].The space E is said to be a K–space if the subspace R × {0} is complementedin XF for every quasi–linear map F on E. (This is a slight abuse of terminology;strictly speaking, it is the completion of E that is the K–space.) So we are interestedin Klee’s question for the non–K spaces.

The only known non–K spaces are ℓ1–like.The Ribe function is defined on ℓ01, the space of finitely supported elements of ℓ1,byF0(x) =Xixiℓn|xi| − Xixi!ℓn Xixiwith the convention that 0ℓn0 = 0. Ribe [8] proved that F0 is quasi–linear on ℓ01and used F0 to show that ℓ1 is not a K–space.

Closely related functions were usedby Kalton [2] and Roberts [9] to prove the same result. The reflexive space ℓ2(ℓn1)is not a K space and the B convex spaces are K spaces [2] Kalton and Roberts

TWISTED SUMS AND A PROBLEM OF KLEE3[4] showed that c0 and ℓ∞are K–spaces. It is not known whether the James spaceis a K–space.

We are studying Klee’s problem for spaces E and quasi-linear mapsF on E such that R × {0} is not complemented in XF . By Theorem 2.5 of [3],there is no linear map T on E such that |T(x)−F(x)| ≤C∥x∥for all x in E (i.e.

Fdoes not split on E). The corollary to our main theorem implies that none of thespaces above can be a counterexample for Klee’s question, since the F concerneddoes split on an infinite- dimensional subspace.We now state our main result:Main Theorem.

Let E be an ℵ0–dimensional normed space.Assume F is aquasi–linear function on E for which there are a linearly independent sequence (xi)in E and a linear map T on span(xi) such that(1) |T(x) −F(x)| ≤C∥x∥for all x in span(xi) and some constant C.Then there are a trivial dual topology τ2 on R × E, weaker than the quasi–normtopology, and a τ2–neighborhood U of 0 such that if (r, x) ∈U and ∥x∥≤1, then∥|(r, x)|∥< C for some constant C.Before we prove the theorem, we set the framework for the construction withsome auxiliary results. We begin with:Definition.

Suppose (Gi) is a finite or infinite sequence of subsets of E, and (ni)is a sequence of positive integers (of the same length as (Gi)). The (ni)–sum of(Gi) is the set of all finite sumsz = r1z1 + r2z2 + r3z3 .

. .where |ri| ≤1 for all i and z1, .

. .

, zn1 are in G1, zn1+1, . .

. , zn1+n2 are in G2,zn1+n2+1, .

. .

, zn1+n2+n3 are in G3, etc.Note that if |r| ≤1, rz is also in the(n ) sum

4N. T. PECKLemma 1.

Let X be a vector space and let (Un) be a neighborhood base at 0 for apseudo-metrizable vector topology on X, chosen so that Un+1 + Un+1 ⊂Un for alln and [−1, 1]Un ⊂Un for all n. Let (Fn) be a sequence of subsets of X, chosen sothat [−1, 1]Fn ⊂Fn and Fn+1 +Fn+1 ⊂Fn, for all n. Then the sequence (Un +Fn)is a neighborhood base at 0 for a pseudo-metrizable vector topology on X which isweaker than the original topology.Proof. Immediate.□In the next lemma, we specify Fn more closely.Lemma 2.

Let X and (Un) be as in Lemma 1. Let (Gn) be a sequence of subsetsof X. Define subsets Fn of X as follows: for each n in N, Fn is the (2i−n)–sumof the Gi’s for i ≥n.

Then (Fn) satisfies the hypotheses of Lemma 1.Proof. [−1, 1]Fn ⊂Fn as remarked already.

For a typical sum in Fn+1 + Fn+1, atmost 2·2i−(n+1) = 2i−n of the zi’s are in Gi for i ≥n+1, so Fn+1 +Fn+1 ⊂Fn.□Remark. Note that there is an apriori bound on the number of elements of Giappearing in a sum in Fn, for any n: the bound is 2i−1; we use the looser bound2i.In our construction, (Un) is a neighborhood base at 0 for the twisted sum topol-ogy.

The Gi’s of Lemma 2 will be chosen so that (Un + Fn) is a neighborhood baseat 0 for a trivial dual topology τ2; they will also have to be chosen so that τ is thesupremum of τ1 and τ2. The next lemma identifies the topology τ1:Lemma 3.

Let F be a quasi–linear map on a normed space E and let XF =R × E with the quasi–norm ∥|(r, x)|∥= |r −F(x)| + ∥x∥. Assume R × {0} is notcomplemented in XThen the strongest nearly convex topology on R × E which

TWISTED SUMS AND A PROBLEM OF KLEE5is weaker than the quasi–norm topology has a neighborhood base at 0 of sets of theform {(r, x) : ∥x∥< η}.Proof. Sets of the above type are a neighborhood base at 0 for a nearly convextopology weaker than τ, the quasi-norm topology.The closure of {0} for thisweaker topology is R × {0}; and if (r, x) is in XF and x ̸= 0, there is f in E∗withf(x) ̸= 0.

Then f(π(r, x)) ̸= 0, where π is the quotient map of XF onto E.Now suppose ν is a nearly convex topology on R×E, weaker than the quasi-normtopology. Since R × {0} = K(τ) is not complemented, the ν–closure of {0} mustcontain R × {0}.

Let U be ν–open containing 0. Choose V ν–open containing 0,with V + V ⊂U.

Choose ǫ > 0 so that if ∥|(r, x)|∥< ǫ, then (r, x) ∈V .Now, ∥|(F(x), x)|∥= ∥x∥, so if ∥x∥< ǫ, then (F(x), x) ∈V . Also, (r −F(x), 0)is in V since it is in the ν–closure of 0, and so (r, x) is in U.□Notation.

Let Z be a Banach space with a basis (vi). Let (v∗i ) be the coordinatefunctionals on Z.

For n a positive integer and x in Z, set x |[1,n]=nPi=1v∗i (x)vi, andx |(n,∞)=∞Pi=n+1v∗i (x)vi. Say that x is to the right of n if v∗i (x) = 0 for i ≤n.We need two more preliminary results before proving our main theorem:Lemma 4.

Let Z be a Banach space with a monotone basis (vi), let K be a compactsubset of Z, and let ǫ > 0. Then there is n so that if y is to the right of n andx ∈K, then ∥x∥< ∥x + y∥+ ǫ.Proof.

Choose n so that ∥x |(n,∞) ∥< ǫ for every x in K. Now if y is to the rightof n, then ∥x |[1,n] ∥= ∥(x + y) |[1,n] ∥≤∥(x + y)∥, since the basis is monotone, so∥x∥< ∥x + y∥+ ǫ.□

6N. T. PECKLemma 5.

Let yi, 1 ≤i ≤k, be linearly independent elements of a normed spaceE. Define yk+1 = −kPi=1yi, and let η > 0.

Set zi = myi, 1 ≤i ≤k + 1, for somem > 0. Then we can choose m so large that the following condition is satisfied: ifat most k of the ri’s are non–zero, and ifk+1Pi=1rizi < 3, thenk+1Pi=1|ri| < η.Proof.

Choose M > 0 so thatkPi=1|αi| ≤MkPi=1αiyi for all k–tuples (αi). Nowsupposek+1Pi=1rizi < 3, with at most k ri’s non–zero.

If rk+1 = 0, thenkXi=1|ri| ≤3Mm < ηfor m > 3M/η.If rk+1 ̸= 0, then some other ri is 0, r1, say; now,k+1Xi=1rizi = −mrk+1y1 +kXi=2m(ri −rk+1)yi < 3.Therefore|rk+1| +k+1Xi=2|ri −rk+1| < 3Mm ,sok+1Xi=2|ri| ≤|rk+1| +k+1Xi=2|ri −rk+1| + k|rk+1|< 3(k + 1)Mm< ηfor m > 3(k + 1)M/η.□Proof of Main Theorem. We will construct inductively the sets Gn used in Lemma2.

That lemma will give us the sets Fn and then Lemma 1 will provide the topology.We may assume that for xi and T in the theorem, T(xi) = 0 for each i. This ispossible since for each i, there is a scalar αi so that T(x2i−1 + αix2i) = 0; now thesequence x′i = x2i−1 + αix2i also satisfies condition 1) of the Theorem.We can regard E as a subspace of the Banach space Z = C[0, 1], which has amonotone basisAny positive scalar multiple of the quasi norm yields the same

TWISTED SUMS AND A PROBLEM OF KLEE7topology as the quasi–norm, so we can and do assume that the constant C in 0(ii)above is 1. This can be done by multiplying F by a suitable positive constant.Finally, we use ∥∥to refer to the norm on Z and on E. We only calculatenorms of elements of E, but we do use the monotonicity of (vi) in Z.Now we begin the construction of (Gi).Choose 0 < cn ≤2−(n+3) (and thus∞Pn=1cn < 14).

Let (dj) be any sequencewhose linear span is E, and let (ei) be an indexing of (dj) such that each dj occursinfinitely often in (ei). We can assume that ∥dj∥≤1 and that |F(dj)| ≤1 for eachj, by multiplying dj by a positive constant.Assume that finite sets G0, G1 · · · Gn−1 have been constructed, with G0 = {0},satisfying the following conditions:(2) for each 1 ≤i ≤n −1, Gi is a finite set (wi,j : 1 ≤j ≤2i+1), withwi,j = ei + mixℓ(i,j),j ≤2i,wi,2i+1 = ei −2iXj=1mixℓ(i,j);here, (xi) is the sequence in the statement of the theorem.

(3) Set zi,j = wi,j −ei. Then if2i+1Pj=1rjzi,j < 3 with at most 2i rj’s non–zero,2i+1Pj=1|rj| < ci.To define Gn, let K′n be the (2i)–sum of Gi for i ≤n −1, (so K′1 = {0}) and letKn = K′n + [−2n, 2n]en.

Then Kn is a compact subset of E ⊂Z. By Lemma 4,there is an integer sn such that if y is to the right of sn then ∥x∥< ∥x + y∥+ cnfor all x in Kn.

By the linear independence of the sequence (xi), we can choosexℓ(n,1), . .

. , xℓ(n,2n), all to the right of sn, with ℓ(n, j) < ℓ(n, j′) if j < j′.

For easeof notation, put xn,i = xℓ(n,i), 1 ≤i ≤2n, and put xn,2n+1 = −2nP xn,i.

8N. T. PECKBy Lemma 5, we can choose mn so large that if2n+1Xj=1rn,jmnxn,j < 3,with at most 2n of the rn,j non–zero, then2n+1Xj=1|rn,j| < cn.

(4)Finally, for 1 ≤i ≤2n + 1, putwn,i = en + mnxn,iand letGn = (wn,i),1 ≤i ≤2n + 1.Note that since2n+1Pi=1xn,i = 0, en ∈coGn. (We denote the convex hull of A by coA.

)This finishes the construction of (Gi).Now let (Fn) be the subsets of E used in Lemma 1: Fn is the (2i−n) sum of (Gi)for i ≥n. Let (Un) be a neighborhood base at 0 for the quasi–norm topology onR × E, with Un+1 + Un+1 ⊂Un and [−1, 1]Un ⊂Un, for all n; also assume that∥|w|∥< 1 if w ∈U1.

Let τ2 be the topology yielded by Lemma 1.We claim that τ2 is trivial dual. To see this, note that for m ≥n, em ∈co(wm,i) ⊂coFn ⊂co(Fn+Un); since each dj occurs infinitely often in the sequence (em), K(τ2)contains every dj and therefore contains {0}×E.

Also, (1, 0) ∈coUn ⊂co(Un+Fn)for every n, so K(τ2) contains R × {0}. This proves the claim.Now suppose thatx =nXi=12i+1Xj=1ri,j(ei + mixi,j)is in F and that ∥x∥< 1 We will first prove that |F(x)| < 9

TWISTED SUMS AND A PROBLEM OF KLEE9Toward that end: since the xn,j are to the right of sn, the construction of Gnimplies thatn−1Xi=12i+1Xj=1ri,j(ei + mixi,j) +2n+1Xj=1rn,jen < 1 + cn(5)from which, since ∥x∥< 1,2n+1Xj=1rn,jmnxn,j < 2 + cn < 3. (6)Now from (4) and (6), we have2n+1Xj=1|rn,j| < cn;(7)combining this with (5), we haven−1Xi=12i+1Xj=1ri,j(ei + mixi,j) < 1 + 2cn.

(8)For the induction step, assume that for some ℓ,ℓXi=12i+1Xj=1ri,j(ei + mixi,j) < 1 + 2cn · · · + 2cℓ+1. (9)Since the xℓ,j are to the right of sℓ, the construction of Gℓimplies thatℓ−1Xi=12i+1Xj=1ri,j(ei + mixi,j) +2ℓ+1Xj=1rℓ,jeℓ < 1 + 2cn · · · + 2cℓ+1 + cℓ,(10)from which2ℓ+1Xj=1rℓ,jmℓxℓ,j < 2 + 4cn · · · + 4cℓ+1 + cℓ< 3.

(11)Now from (4) and (11), we have2ℓ+1X|rℓ,j| < cℓ;(12)

10N. T. PECKcombining this with (10), we obtainℓ−1Xi=12i+1Xj=1ri,j(ei + mixi,j) < 1 + 2cn · · · + 2cℓ,(13)recalling that ∥ei∥≤1.

This finishes the induction step.The above argument has yielded that2i+1Xj=1ri,jei < ci(14)for each i; from this and ∥x∥< 1, we havenXi=12i+1Xj=1ri,jmixi,j < 1 +∞Xn=1cn < 2. (15)¿From (15) and (1), recalling T(xi,j) = 0,FnXi=12i+1Xj=1ri,jmixi,j < 2.To estimateFnXi=12i+1Xj=1ri,jei,recall that |F(ei)| ≤1 for each i, soF2i+1Xj=1ri,jei < 2−i.ThereforeFnXi=12i+1Xj=1ri,jei ≤nXi=12−i +nXi=1i2i+1Xj=1ri,jei< 1 +nXi=1i · 2−i < 4(usingF(P ui) ≤P |F(ui)| + P i∥ui∥).

Finally,|F(x)| ≤2 + 4 +nXi=12i+1Xj=1ri,jei+nXi=12i+1Xj=1ri,jmixi,j<2 + 4 + 1 + 29

TWISTED SUMS AND A PROBLEM OF KLEE11To complete the proof of the theorem, suppose (r, x) ∈U1 + F1 and ∥x∥≤1.Write (r, x) = (r, y)+(0, z), with (r, y) ∈U1 and z ∈F1. Then |r −F(y)|+∥y∥≤1;from this and ∥x∥≤1 follows ∥z∥≤2.

Now since z ∈F1, the preceding paragraphimplies |F(z)| < 18. At last,|r −F(x)| ≤|r −F(y)| + |F(y) −F(x)|≤1 + |F(y) −F(x)|≤1 + |F(z)| + ∥z∥+ ∥x∥< 22,so ∥|(r, x)|∥< 23.

The proof is complete.□Corollary. Let E be a separable normed space and let E0 be an ℵ0-dimensionalsubspace of E which is dense in E. Assume that there is a quasi-linear map F onE0 which splits on an infinite-dimensional subspace of E0.

Then the twisted sumtopology on R ⊗F E is the supremum of a trivial dual topology and a nearly exotictopology.Proof. Let q denote the quotient map of R⊗F ˜E onto ˜E, where ˜E is the completionof E. (For x ∈E0, q(r, x) = x.) The subspace E0 satisfies the hypotheses of themain theorem.

Therefore there are a trivial dual topology τ2 on R × E0, weakerthan the twisted sum topology; a τ2- neighborhood V of 0; and a constant C sothat if x ∈E0, (r, x) ∈V , and ∥x∥< 1, then ∥|(r, x)|∥< C.We can assume that V contains a τ2-neighborhood U of 0 of the form Bα + Fn,where Fn ⊂E0 is as constructed as in the proof of the main theorem, and for anyβ > 0,B{(r x) ∈R × E : ∥|(r x)|∥< β}

12N. T. PECKSets of the form Bβ + q−1(Fm), whereBβ = {w ∈R ⊗F E : ∥|w|∥< β},obviously form a neighborhood base at the origin for a vector topology τ 2 on R⊗F E,weaker than the twisted sum topology.

The topology τ 2 is trivial dual since itsrestriction to the dense subspace R × E0 is trivial dual.Now choose 0 < γ < 1/2 so that Bγ + Bγ ⊂Bα, and assume that w ∈Bγ +q−1(Fn) and ∥q(w)∥< 1/2. Choose w0 ∈R × E0 so that ∥|w −w0|∥< γ. Then∥q(w) −q(w0)∥< γ, so ∥q(w0)∥< γ + 1/2 < 1.

Clearly, w0 ∈Bα + q−1(Fn), andso from our assumption, ∥|w0|∥< C. Now, ∥|w|∥< (α/γ)(C + 1), and the proof iscomplete.□The theorem and corollary apply to several spaces which are either not K–spacesor for which it is not known whether they are K–spaces:Theorem. For the following pairs of normed spaces E and quasi–linear maps Fon E, the twisted sum topology on XF = R×E is the supremum of a nearly convextopology and a trivial dual topology:(a) E is any infinite–dimensional subspace of ℓ01 (whether or not it is a K–space), F is the Ribe function F0;(b) E is the linear span of the usual unit vector basis for the James space, underthe James norm; F is any quasi–linear function on E;(c) E is the span of the usual unit vector basis in ℓp(ℓn1), for 1 < p < ∞(thisis a reflexive non–K space); F will be described below.Proof.

For (a), let H = {x ∈E : Pixi = 0}. Note that if x, y ∈H and x andy have disjoint supports F (x + y)F (x) + F (y) Since H has codimension at

TWISTED SUMS AND A PROBLEM OF KLEE13most 1 in E and E is infinite dimensional, there is a sequence of non–zero elements(xi) in H satisfying sup (support xi) < inf (support xi+1) for all i.As remarked above, F0 is linear on span(xi), so if we define T(xi) = F(xi), thelinear function T certainly satisfies hypothesis (1) of the theorem. Therefore thetheorem applies to E.For (b), it is known that the even unit vectors e2n span a pre–Hilbert subspaceof the James space (see [1]).The B–convexity of span(e2n) and Theorems 2.6of [2] and 2.5 of [3] imply that there is a linear map T on span(e2n) such that|T(x) −F(x)| ≤C∥x∥for all x in span(e2n).

Therefore the theorem applies. (c) For each n let (ei,n) be the usual unit vector basis of ℓn1, and let E be thespan of the ei,n in ℓp(ℓn1).

Let (cn) be any sequence in ℓq, 1p + 1q = 1. Let F0 be theRibe function and define F on E byF((xn)) =XncnF0(xn).We claim that F is quasi–linear.

For this, if (xn) and (yn) are in E, the sequences(∥xn∥1) and (∥yn∥1) are ℓp sequences, and for each n,|cnF0(xn + yn) −cnF0(xn) −cnF0(yn)|≤cn(∥xn∥1 + ∥yn∥1).¿From H¨older’s inequality,|F((xn + yn)) −F((xn)) −F((yn))|≤∥(cn)∥q(∥(xn)∥p + ∥(xn)∥p).Theorem 4.7 of [2] gives that E is not a K–space. The F just defined provesthis directly for suppose there is a linear T on E with |T(x)F(x)| ≤C∥x∥for

14N. T. PECKall x in E. Then since F(ei,n) = 0 for all i, n, |T(ei,n)| ≤C for all i, n. ButF 1nnPi=1ei,n= −cn log n, a contradiction if we choose cn so that (cn log n) isunbounded.Finally, our theorem applies in this situation.

To show this, for each n pick a unitvector xn in ℓn1. The sequence (xn) is equivalent to the usual basis of ℓp, which isB–convex; the results already mentioned imply that there is a linear T on span(xn)such that |T(x) −F(x)| ≤C∥x∥for all x in span(xn).

This finishes the proof.□Note that, because of the separability, the corollary applies to the completionsof the twisted sums in (a)–(c) above.References1. R. C. James, A non–reflexive Banach space isometric with its second conjugate space, Proc.Nat.

Acad. Sci.

37 (1951), 174–177.2. N. J. Kalton, The three space problem for locally bounded F –spaces, Compositio Math.

37(1978), 243–276.3. N. J. Kalton and N. T. Peck, Twisted sums of sequence spaces and the three–space problem,Trans.

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Soc. 255 (1979), 1-30.4.

N. J. Kalton and James W. Roberts, Uniformly exhaustive submeasures and nearly additiveset functions, Trans. Amer.

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Symp. on Gen. Topology and its Relationsto Modern Analysis and Algebra, Academic Press, 1962, pp.

238–249.6. J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces I, Ergeb.

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u. ihrer Gren-zgeb., Springer–Verlag, 1977, p. 92.7. N. T. Peck and H. Porta, Linear topologies which are suprema of exotic topologies, StudiaMath.

47 (1973), 63-73.8. M. Ribe, Examples for the non locally convex three space problem, Proc.

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Soc.237 (1979), 351–355.9. James W. Roberts, A nonlocally convex F –space with the Hahn–Banach approximation prop-erty, Springer Lecture Notes 604, Springer–Verlag, 1977, pp.

76–81.University of Illinois, Urbana, IL 61801

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