Norm attaining operators which satisfy a Bollob\'as type theorem

In this paper we are interested to study the set $\mathcal{A}_{\|\cdot\|}$ of all norm one linear operators $T$ from $X$ into $Y$ which attain the norm and satisfy the following: given $\epsilon>0$, there exists $\eta$, which depends on $\epsilon$ and $T$, such that if $\|T(x)\|>1 - \eta$, then there is $x_0$ such that $\| x_0 - x\|<\epsilon$ and $T$ itself attains the norm at $x_0$. We show that every norm one functional on $c_0$ which attains the norm belongs to $\mathcal{A} (c_0, \mathbb{K})$. Also, we prove that the analogous result is not true neither for $\mathcal{A}_{\|\cdot\|} (\ell_1, \mathbb{K})$ nor $A (\ell_{\infty}, \mathbb{K})$. Under some assumptions, we show that the sphere of the compact operators belongs to $\mathcal{A}_{\|\cdot\|}$ and that this is no longer true when some of these hypotheses are dropped. We show also that the natural projections on $c_0$ or $\ell_p$, for $1\leq p<\infty$, always belong to this set. The analogous set $\mathcal{A}_{nu}$ for numerical radius of an operator instead of its norm is also defined and studied. We give non trivial examples of operators on infinite dimensional Banach spaces which belong to $\mathcal{A}_{\| \cdot \|}$ but not to $\mathcal{A}_{nu}$ and vice-versa. Finally, we establish some techniques which allow to connect both sets.


Introduction
Over the past few years, some Bishop-Phelps-Bollobás type properties were defined and studied. Some of them are stronger than the Bishop-Phelps-Bollobás property (BPBp, for short), introduced eleven years ago by María Acosta, Richard Aron, Domingo García, and Manuel Maestre (see [3]) in order to study the operator version of the Bollobás' theorem, which, roughly speaking, means that if xx˚, xy « 1, then there exist new elements y˚and y such that y˚« x˚, y « x, and y˚attains the norm at y. One of these properties is the so-called Bishop-Phelps-Bollobás operator property (BPBop, for short), studied in [7,8,11]: a pair of Banach spaces pX, Y q satisfies the BPBop if given ε ą 0, then there is η " ηpεq ą 0 such that whenever an operator T with }T } " 1 from X into Y and a point x satisfy }T pxq} ą 1´η, then there exists a new element x 0 such that }x 0´x } ă ε and T itself attains its norm at x 0 . Notice that it seems to be very restrictive at first glance since, in this case, we do not change the operator which almost attains the norm. Indeed, after some negative results presented in [7] and [11], it was finally proved in [8] that if dimpXq, dimpY q ě 2, then the pair pX, Y q can not satisfy this property. On the other hand, [11,Theorem 2.1] gives a characterization for it in terms of linear functionals. A Banach space X is uniformly convex if and only if the pair pX, Kq satisfies the BPBop.
brings a lot of differences between the local and uniform properties. Now, we have many positive examples about it, not just in the functional case, but also in the operator case (see also [9]).
In this paper, we investigate the set A of all norm one operators T which attain the norm and enjoy the property that for a given ε ą 0, there is η " ηpε, T q ą 0 such that if }T pxq} ą 1´η for some x P S X , then there is x 0 P S X with }x 0´x } ă ε and T attains the norm at x 0 . Let us notice the obvious fact that if a pair of Banach spaces pX, Y q satisfies the (local) BPBop, then every norm one operator T from X into Y which attains the norm belongs to the set A. It is worth mentioning also that if a pair pX, Y q satisfies the (local) BPBop, then the domain space X must be reflexive by James' theorem. However, we present concrete examples of non trivial operators T which satisfies the above property even when X is a non reflexive space. Moreover, we study the equivalent set by considering the numerical radius of T instead of its norm. We will give the precise definition of these sets later. Now, let us restart the introduction with all the necessary definitions and background we need.
Let X, Y be Banach spaces over a scalar field K, which can be the real numbers R or the complex numbers C. We denote by B X and S X the closed unit ball and the unit sphere of a Banach space X, respectively. The topological dual space of X is denoted by X˚and the action of an element x˚P X˚at an element x P X is denoted by xx˚, xy. As usual, the set LpX, Y q stands for the bounded linear operators from X into Y and we simply denoted it as LpXq when Y " X. We say that T attains the norm or T is a norm attaining operator when there is x 0 P B X such that }T px 0 q} " }T }. To deal with the numerical radius of an operator, we introduce the set ΠpXq " tpx, x˚q P S XˆSX˚: xx˚, xy " 1u. Given T P LpXq, the numerical radius of T is defined as νpT q :" supt|xx˚, T pxqy| : px, x˚q P ΠpXqu. We say that T attains the numerical radius or T is a numerical radius attaining operator when there is px 0 , x0 q P ΠpXq such that |xx0 , T px 0 qy| " νpT q. For a background on this topic, we refer the reader to [4,5].
Here, we will study the following two sets of bounded linear operators. Definition 1.1. Let X, Y be Banach spaces.
(i) A }¨} pX, Y q stands for the set of all norm attaining operators T P LpX, Y q with }T } " 1 such that if ε ą 0, then there is ηpε, T q ą 0 such that whenever x P S X satisfies }T pxq} ą 1´ηpε, T q, there is x 0 P S X such that }T px 0 q} " }T } " 1 and }x 0´x } ă ε.
In order to show that T P A }¨} pX, Y q (respectively, A nu pXq), we have to show first that T attains the norm (respectively, the numerical radius) and, as we already mentioned, we do not have to assume that X is reflexive anymore as in the BPBop. Obviously, any norm one operator (respectively, numerical radius one operator) which does not attain the norm (respectively, the numerical radius) cannot belong to A }¨} pX, Y q (respectively, to A nu pXq). For that reason, most of the concrete operators defined in the paper are norm one (respectively, numerical radius one) and attain the norm (respectively, the numerical radius).
Let us describe the contents of this paper. In the next section, we show that, as expected, when we are working with finite dimensional spaces, we have a positive result. That is, if dimpXq ă 8, then the set A }¨} pX, Y q coincides with the sphere of LpX, Y q for every Banach space Y and the set A nu pXq coincides with the set of all operators with numerical radius one. As a consequence of it, we get that every norm one functional on c 0 which attains the norm belongs to A }¨} pc 0 , Kq by using the canonical embedding from a finite dimensional Euclidian space pK n , }¨} 8 q (the space K n with the topology induced by the norm of c 0 ) into c 0 . We notice that the same result is also true for norm one functionals on p , when 1 ă p ă 8, or on Hilbert spaces as a consequence of a result in [11]. On the other hand, we present examples of norm one functionals on 1 and 8 which attain the norm but cannot be in A }¨} p 1 , Kq and A }¨} p 8 , Kq, respectively. Next, we show that under some assumptions on the Banach space X, the sphere of the compact operators is contained in A }¨} pX, Y q and also the set of all compact operators T with νpT q " }T } " 1 belongs to A nu pXq. We present some counterexamples that show that this is no longer true by dropping some of these hypothesis. We also present some conditions on X that give the possibility to pass from A }¨} pX, Y q to A }¨} pY˚, X˚q (analogously, from A nu pXq to A nu pX˚q). By observing some examples, we show that this duality is not true in general. Moreover, we prove that if P N is the natural projection on c 0 or p with 1 ď p ă 8, then P N P A }¨} pX, Xq X A nu pXq. Finally, in the last section, we connect these sets by using direct sums and proving that these results are no longer true for general absolute sums.

The results
Recall that in finite dimensional Banach spaces, every operator T attains the norm and numerical radius by compactness. The following result shows that, when X is finite dimensional, the classes A }¨} pX, Y q and A nu pXq are the sets of those operators which have norm one and numerical radius one, respectively. Proposition 2.1. Let X be a finite dimensional Banach space. Then (i) A }¨} pX, Y q " tT P LpX, Y q : }T } " 1u for any Banach space Y , (ii) A nu pXq " tT P LpXq : νpT q " 1u.
Proof. Both items are proved by contradiction. We show (ii). Notice that we just have to prove that tT P LpXq : νpT q " 1u Ă A nu pXq. If this is not the case, then there are ε 0 ą 0 and a numerical radius attaining operator T P LpXq with νpT q " 1 such that for all n P N, there is px n , xnq P ΠpXq with 1 ě |xxn, T px n qy| ě 1´1 n and whenever px, x˚q P ΠpXq satisfies }x´x n } ă ε 0 and }x˚´xn} ă ε 0 , we have |xx˚, T pxqy| ă 1. By compactness, there are subsequences of px n q and pxnq which we denote again by px n q and pxnq, and there are x 0 P B X and x0 P B X˚s uch that x n ÝÑ x 0 and xn ÝÑ x0 as n Ñ 8. Since xxn, x n y " 1 for every n P N, we have that xx0 , x 0 y " 1. This shows that px 0 , x0 q P ΠpXq. Now, since |xxn, T px n qy| ÝÑ 1, we get that |xx0 , T px 0 qy| " 1 , which is a contradiction.
Corollary 2.2. Every norm one functional on c 0 which attains the norm belongs to A }¨} pc 0 , Kq.
On the other hand, concerning p -spaces for 1 ď p ď 8, we have the following result. Recall that every uniformly convex space is reflexive and that every bounded linear functional defined on a reflexive space attains the norm. For simplicity, we denote by NApXq the set of norm attaining functionals on a Banach space X.
Proof. (i). If X is a uniformly convex Banach space, then by [11, Theorem 2.1], every norm one functional on X belongs to the set A }¨} pX, Kq. In particular, every norm one functional on a Hilbert space H or p (1 ă p ă 8) belongs to A }¨} pH, Kq and A }¨} p p , Kq, respectively.
So, ř ně2`1´n´1 n˘| zpnq| " 0. This implies that zpnq " 0 for n ě 2 and then z is of the form z " e iθ e 1 for some θ.
Given ε ą 0, suppose that there is such a ηpε, z˚q ą 0. We take k P N to be such that 1 k ă ηpε, z˚q and then This means that there is z P S 1 such that |xz˚, zy| " 1 and }z´e k } 1 ă ε. By the first part z " e iθ e 1 but }e iθ e 1´ek } 1 " 2, which is a contradiction. So, z˚cannot belong to A }¨} p 1 , Kq although z˚attains its norm.
So, |zpnq| " 1 for every n P N. First assume that there is n 0 ě 1 such that zpn 0 q "´1 (so, we can pick n 0 to be a minimal element). If n 0 ą 1, then which is a contradiction and means that if zp1q " 1 then z should be p1, 1, 1, . . . q. On the other hand, if n 0 " 1, then we consider n 1 P N to be the first index such that zpn 1 q " 1. By arguing at the same way as before, we get again that |xx˚, zy| ă 1.
So, if we assume that such a ηpε, x˚q ą 0 exists, we take k P N with 2 k ηpε, x˚q ą 1, and consider the element e 1`. . .`e k P S 8 . Then, |xx˚, e 1`. . .`e k y| " 1´1 2 k ą 1´ηpε, x˚q.
Let us now consider linear operators instead of functionals. In order to find an operator which does not belong neither to A }¨} pX, Y q nor A nu pXq, it is clear that it is enough to consider any operator which has norm and numerical radius one, but it does not attain neither the norm nor numerical radius. Nevertheless, in Example 2.4, we define an operator which cannot belong neither to A }¨} pX, Xq nor to A nu pXq even though it attains its norm and numerical radius.
Example 2.4. We consider the spaces p and q as p p 2 p q and q p 2 q q, respectively, where 2 p " pK 2 , }¨} p q. For each n P N, define T n : 2 p ÝÑ 2 p by Now, define T : p ÝÑ p as T pzq :" pT n pxpnq, ypnqqq n "ˆˆ1´1 2n˙x pnq, ypnq˙n pz " ppxpnq, ypnqqq n P p q.
In [7, Theorem 2.21.(ii)] it was showed that the operator T attains the norm but T R A }¨} p p , p q.
(i) Every isometry on X belongs to A }¨} pX, Xq.
(ii) There is an isometry on X " 2 such that it does not belong to A nu pXq.
Proof. Notice that it is immediate to prove that every isometry on a Banach space X belongs to the set A }¨} pX, Xq. On the other hand, this does not hold for the set A nu pXq. Consider the right shift operator R : 2 Ñ 2 , that is, Rpxp1q, xp2q, xp3q,¨¨¨q " p0, xp1q, xp2q,¨¨¨q for every x P 2 . It is known that the numerical range W pRq of R is the open unit disk D in the complex plane (see, for example, [13, Example 2]) which implies that νpRq " 1, but |xRx, xy| ă 1 for every x P S 2 .
Recall that a Banach space X satisfies the Kadec-Klee property when the weak and norm topologies coincide on the unit sphere S X . It is well known that every locally uniformly rotund space (LUR, for short) satisfies the Kadec-Klee property (the converse is not true, e.g., 2 1 ). Recall also that, by theŠmulian lemma, the norm of X is Fréchet differentiable at x if and only if pxnq Ă S X˚i s convergent whenever lim n xxn, xy " 1. In the next result, under some assumptions on the involved Banach spaces, we show that some subsets of the space of all compact operators belong to the classes A }¨} and A nu . Proposition 2.6. Let X be a reflexive space which satisfies the Kadec-Klee property. Then, Proof. We prove (ii). Suppose by contradiction that it is not true. Then, there are ε 0 P p0, 1q and a compact operator T P KpXq with νpT q " }T } " 1 such that for every n P N, there is px n , xnq P ΠpXq such that (1) 1 ě |xxn, T px n qy| ě 1´1 n and whenever px, x˚q P ΠpXq satisfies }x´x n } ă ε 0 and }x˚´xn} ă ε 0 , we have |xx˚, T pxqy| ă 1.
Since X is reflexive, B X is weakly compact and then, by Eberlein-Šmulian theorem, there are a subsequence of px n q, which we denote again by px n q, and x 0 P B X such that x n w ÝÑ x 0 . Since T is completely continuous, we have that T px n q ÝÑ T px 0 q in norm. From this and 1 " νpT q " }T } ě }T px n q} ě |xxn, T px n qy| ÝÑ 1, we get that }T px 0 q} " 1. This shows that x 0 P S X . Since w and norm topologies coincide in S X , we have that x n ÝÑ x 0 in norm. Notice now that for each n P N, we have 1 ě |xxn, T px 0 qy| " |xxn, T px n qy|`|xxn, T px 0´xn qy| ě |xxn, T px n qy|´}x 0´xn }.
Since x n converges to x 0 in norm, by using p1q, we get that |xxn, T px 0 qy| ÝÑ 1.
By compactness of B C , there are a subsequence of pxnq, which we denote again by pxnq, and some θ P r0, 2πq such that xxn, T px 0 qy converges to e iθ . Let S P KpXq be the operator defined by S :" e´i θ¨T . One clearly has that Spx 0 q P S X and xxn, Spx 0 qy converges to 1. On account of the Fréchet differentiability of X, byŠmulian lemma, there is x0 P B X˚s uch that xn ÝÑ x0 in norm. Since xxn, x n y " 1 for every n P N, we get that xx0 , x 0 y " 1. So x0 P S X˚a nd so px 0 , x0 q P ΠpXq. Finally, in view of (1) and |xxn, T px n qy| ÝÑ |xx0 , T px 0 qy|, we get that |xx0 , T px 0 qy| " 1. This is a contradiction.
Note that (i) is obtained by using a very similar argument as above (see [15]).
In fact, the above argument shows, under the same assumptions on (ii), that every compact operator T which has norm and numerical radius 1 attains its numerical radius. Notice also that the identity operator always belongs to A nu pXq whereas it is not compact unless X is finite dimensional. So, in the infinite dimensional setting, the inclusion in Proposition 2.6.(ii) must be strict. On the other hand, since every operator from a reflexive space into a space which satisfies the Schur's property is compact and Hilbert spaces satisfy all the hypothesis from Proposition 2.6, we have the following consequence.
Corollary 2.7. Let X be a reflexive Banach space with the Kadec-Klee property and let H be a Hilbert space.
Remark 2.8. We notice that there is a norm one compact operator T on 1 (which is non reflexive and has Schur property) such that then }T } " 1 but T never attains its norm. On the other hand, in [2, Example 1.8], the author constructed a non compact operator S on a separable Hilbert space such that νpSq " }S} " 1 but does not attains its numerical radius. So, this S cannot belong to A nu pHq.
In the next example, we present a numerical radius attaining compact operator S R A nu with νpSq " }S} " 1 defined on a Banach space X which is not reflexive, its norm is nowhere Fréchet differentiable and satisfies the Schur's property (and, in particular, the Kadec-Klee property).
Example 2.9. Let c 0 be the real space of all sequences which converge to zero. Consider the operator T : c 0 ÝÑ c 0 to be defined as (2) pT pxqqp1q " 8 ÿ j"1 1 2 j xpjq and pT pxqqpkq " 0 pk ě 2q px " pxpjqq 8 j"1 P c 0 q.
It is proved in [2, Proposition 2.8] that }T } " νpT q " 1 but T does not attain neither the norm nor the numerical radius. In particular, T cannot belong neither to A }¨} pc 0 , c 0 q nor to A nu pc 0 q. We claim that S " T˚is an operator which is compact numerical radius attaining operator with νpSq " }S} " 1 but does not belong to A nu . Notice that S : 1 ÝÑ 1 is given by Moreover, notice also that νpSq " νpT q " 1, xz, e 1 y " 1 where z " p1, 1, 1, . . . q P S 8 , and that xz, Se 1 y " which implies that S attains the numerical radius (and the norm).
But this is not possible since ε ą }z´z 0 } 8 ě |zpn 0`1 q´z 0 pn 0`1 q| ě 1, which is a contradiction. This shows that S R A nu p 1 q.
Let us recall that in Corollary 2.7, we proved that if a compact operator T defined on a Hilbert space is such that νpT q " }T } " 1, then T must belong to the set A nu pHq. However, the following example shows that it can happen that 1 " νpT q ă }T } and even so T P A nu pHq.
Example 2.10. Let H be a separable infinite dimensional real Hilbert space. We will define an operator T on H such that T is compact, attains its numerical radius, 1 " νpT q ă }T } and T P A nu pHq. Let 0 ă α ď 1 and tα n u be a sequence such that α 1 ą 1, 0 ă α n ă 1 for n ě 2, and α n Ñ 0 as n Ñ 8. Let tJ 1 , J 2 , J 3 u be a partition of N such that |J 1 | " |J 2 | " ℵ 0 , |J 3 | " ă 8. Write the subsets J 1 , J 2 as J 1 " tn k : k ě 1u, J 2 " tm k : k ě 1u where n 1 ď n 2 ď . . . , m 1 ď m 2 ď . . . and each n k corresponds to m k via an one-to-one correspondence between J 1 and J 2 . Define T : H Ñ H by T pe n k q "´α k e m k pk P Nq, T pe m k q " α k e n k pk P Nq, T pe n q " αe n pn P J 3 q, where te n : n ě 1u is an orthonormal basis of H.
We claim that T is a compact operator. Indeed, note first that for every x P H, we have xx, e n yT pe n q " ÿ kPN nPJ3´α k xx, e n k ye m k`α k xx, e m k ye n k`α xx, e n ye n . Now since J 3 is finite, we may take j sufficiently large so that j R J 3 and j P J 1 Y J 2 . Assume j P J 1 . If j " n k for some k ě 1, then }T e j } " |α k |. Since α n Ñ 0 as n Ñ 8, we have }T e j } Ñ 0 as j Ñ 8. For a given ε ą 0, choose j 0 such that }T e j } ă ε for j ě j 0 . For each n ě 1, set T n :" ř n j"1 x¨e j yT pe j q. Let us observe that for every x P S H and n ě m ě j 0 (j 0 is large enough so that j 0 R J 3 ), we have that since tT e j u n j"m are orthogonal. This shows that pT n q is a Cauchy sequence in LpHq which converges to T . Since each T n is finite-rank, it follows that T is compact. Now we calculate the norm and numerical radius of T . Note first that, by using (3), we have the following equalities: (i) xT pxq, e n k y " α k xx, e m k y, (ii) xT pxq, e m k y "´α k xx, e n k y and (iii) xT pxq, e n y " αxx, e n y, for n k P J 1 , m k P J 2 and n P J 3 . Then, for each x P S H , we have xT pxq, xy " α k xe n k , xyxx, e m k y´α k xe m k , xyxx, e n k y`αxe n , xyxx, e n y.
The first two terms are canceled out because H is real and then (4) xT pxq, xy " α ÿ nPJ3 |xx, e n y| 2 for x P S H which implies that νpT q ď α. Since |xT e n , e n y| " α for every n P J 3 , we have that T attains its numerical radius and νpT q " α.
It follows that T attains its numerical radius at }π 3 px 0 q}´1π 3 px 0 q P S H . Moreover, This shows that for a given ε ą 0, there is ηpε, T q ą 0 such that whenever x 0 P S H satisfies xT px 0 q, x 0 y ą 1´ηpε, T q, there is x 1 P S H such that νpT q " xT px 1 q, x 1 y " 1 and }x 1´x0 } ă ε.
Example 2.10 gives, in particular, a non trivial example of an operator which belongs to A nu but not to A }¨} . On the other hand, notice that we have proved in Example 2.9 that the operator T˚belongs to A }¨} but it cannot belong to A nu although it attains its numerical radius and }T˚} " νpT˚q " 1. Let us present next a non trivial example of an operator S on an infinite dimensional Banach space such that S P A }¨} X A nu and S˚P A }¨} X A nu .
Proposition 2.12. Let X, Y be Banach spaces and T P LpX, Y q.
Proof. Note that (ii) is just a consequence of (i) since, in this case, X is, in particular, reflexive. Let us prove (i). Let Y be a uniformly smooth Banach space. Let T P A }¨} pX, Y q. Then, }T˚} " }T } " 1 and T˚is also norm attaining. In order to prove that T˚P A }¨} pY˚, X˚q, let ε P p0, 1q be given and consider ηpε, T q ą 0. Set ηpε, T˚q :" min where ε Þ Ñ δ Y˚p εq stands for the modulus of convexity of Y˚. Pick y1 P S Y˚t o satisfy }T˚py1 q} ą 1´ηpε, T˚q.
Now we prove (iii). Since X is reflexive, we just have to prove one direction. Assume T P A nu pXq. Note that T˚P LpX˚q also attains its numerical radius. Now let ε ą 0 be given and set ηpε, T˚q :" ηpε, T q ą 0. Let px1 , x˚1 q P ΠpX˚q be such that |xx˚1 , T˚px1 qy| ą 1´ηpε, T˚q.
In Proposition 2.12, if we drop off the hypothesis on Banach spaces X and Y , it is possible to construct non trivial operators which do not satisfy the conclusion of that result. Recall that, in Example 2.9, we constructed the operator T defined on the non reflexive space c 0 such that T˚P A }¨} p 1 , 1 q but T R A }¨} pc 0 , c 0 q. The next example shows that the adjoint operator T˚of the same T implies the existence of an operator S defined on a non reflexive space X such that S P A }¨} pX, Xq but S˚R A }¨} pX˚, X˚q.
This shows that |z 0 pjq| " 1 for all j P N.
For a given ε P p0, 1q, suppose that there is ηpε, T˚˚q ą 0. Let n 0 P N be such that 2 n ηpε, T˚˚q ą 1 for every n ě n 0 . Consider the vector z P S 8 defined as z " p1, 1,¨¨¨, 1 lo omo on n0-th , 0, 0,¨¨¨q. Then, However, we have observed that if }T˚˚pz 0 q} " 1 then |z 0 pjq| " 1 for all j P N. Thus the vector z cannot be close to norming points of T˚˚. This shows that T˚˚R A }.} p 8 , 8 q.
On the other hand, by changing slightly the definition of the operator in Example 2.9, we define a new operator T on a non reflexive space X such that T P A nu pXq but T˚R A nu pXq.
It is natural to ask whether the canonical projections P N on a Banach space X with Schauder basis are elements of the set A }¨} pX, Xq or A nu pXq. The following result answers this question positively when X is a classical sequence space c 0 or p with 1 ď p ă 8.
Proposition 2.15. Let X " c 0 or p with 1 ď p ă 8. Then, P N P A }¨} pX, Xq and P N P A nu pXq.
Then, since ř N j"1 x8pjqx 8 pjq " 1, we have that xx˚, x n y " 1. In particular, px n , x˚q P Πpc 0 q. Moreover, (i) |xx˚, P N px n qy| " 1, Now, to get a contradiction, we consider n large enough to make }x˚´xn} 1 and }x n´xn } 8 small. So, P N P A nu pc 0 q as desired. The proof that P N P A nu p 1 q is very similar to the last one and we omit it.

Connecting the sets A }¨} and A nu
In this section, we introduce a natural approach to connect the sets A }¨} and A nu through direct sums. If we have an operator T P LpW, Zq, then there is the simplest way to define r T : W ' Z Ñ W ' Z: consider r T :" ι 2˝T˝P1 , that is, r T pw, zq " p0, T wq for every pw, zq P W ' Z. Conversely, if we have an operator S P LpW ' Zq, then we can consider q S : W Ñ Z defined as q S :" P 2˝S˝ι1 , that is, q Spwq " pP 2˝S qpw, 0q for every w P W . We start with the following result that shows that under some assumptions one can obtain an operator S in A nu pW ' Zq whenever T belongs to A }¨} pW, Zq where S is generated by T , i.e., S " r T .
Remark 3.2. Proposition 3.1 is no longer true if we consider general Banach spaces instead of uniformly smooth ones. Indeed, consider the real Banach space 1 . Example 2.9 provides an operator that belongs to A }¨} p 1 , 1 q but not to A nu p 1 q. We will show that this operator does not satisfy the property stated in Proposition 3.1. Indeed, let T : 1 ÝÑ 1 be defined as 2 j e j , for all x P 1 , and consider r T P Lp 1 ' 1 1 q. Note that if ppx, yq, px˚, y˚qq P Πp 1 ' 1 1 q satisfies (14) |xpx˚, y˚q, r T px, yqy| " |xy˚, T pxqy| "ˇˇˇˇ8 ÿ j"1 y˚pjqxp1q 2 jˇ" 1, then, since |y˚pjqxp1q| ď 1 for all j P N, one gets easily that y˚pjqxp1q has to be equal to either 1 or´1 for all j P N. From here and the fact that ppx, yq, px˚, y˚qq P Πp 1 ' 1 1 q, we get that the only possibilities have the form x " se 1 , y " 0, x˚" ps, x˚p2q, x˚p3q, . . .q, and y˚" pr, r, r, . . .q with |x˚pjq| ď 1 for all j ą 1, where s, r P t´1, 1u. Now, suppose by contradiction that for a given ε P p0, 1q, there is ηpε, r T q ą 0. Let n 0 P N be such that n0 ÿ j"1 1 2 j ą 1´ηpε, r T q, and set w " e 1 , z " 0, w˚" e1 , and z˚" e1`. . .`en 0 . It is immediate to check that ppw, zq, pw˚, z˚qq P Πp 1 ' 1 1 q and also that |xpw˚, z˚q, r T pw, zqy| ą 1´ηpε, r T q. Then, there must be some ppx, yq, px˚, y˚qq P Πp 1 ' 1 1 q satisfying (14) and such that }pw, zq´px, yq} 1 ă ε and }pw˚, z˚q´px˚, y˚q} 8 ă ε. But this is already a contradiction, since ε ą }px˚´w˚, y˚´z˚q} 8 ě }y˚´z˚} 8 ě |y˚pn 0`1 q´z˚pn 0`1 q| ě 1.
Therefore r T R A nu p 1 ' 1 1 q as desired, even though T P A }¨} p 1 , 1 q.
We recall that an absolute norm |¨| a is a norm in R 2 which satisfies |p1, 0q| a " |p0, 1q| a " 1 and |pu, vq| a " |p|u|, |v|q| a for every u, v P R. The absolute sum of two Banach spaces W and Z with respect to |¨| a is the Banach space WˆZ endowed with the norm }pw, zq} a " |p}w}, }z}q| a for every w P W and every z P Z. We denote it as W ' a Z. We say that |¨| a is of type 1 if the vector p1, 0q is a vertex of B pR 2 ,}¨}aq . Equivalently, |¨| a is of type 1 if and only there is a positive number K ą 0 such that |u|`K|v| ď |pu, vq| a for every u, v P R (see [14,Propositions 5.5 and 5.6]). Note that the 1 -norm is of type 1.

Remark 3.3.
It is natural to ask if we can replace 1 -sum in Proposition 3.1 by an absolute sum of type 1: if T P A }¨} pW, Zq, then it is true that r T P A nu pW ' a Zq, where |¨| a is of type 1? We show that such a generalization cannot be done by presenting a concrete example. Consider the norm |||¨||| on R 2 defined as |||pp, qq||| " |p|`p1{2q|q| for every pp, qq P R 2 . Then the norm |||¨||| is of type 1 (with constant K " 1{2).
Suppose that T belongs to A }¨} pW, Zq. To see that r T belongs to A nu pW ' a Zq, where }¨} a is the type 1 absolute norm induced from the norm |||¨|||, we have to check that r T has its numerical radius 1 and attains its numerical radius. Notice that }pw, zq} a " }w}`p1{2q}z} for every pw, zq P W ' a Z and it is easy to observe that }pw˚, z˚q} a˚" maxt}w˚}, 2}z˚}u for every pw˚, z˚q P W˚' a˚Z˚. Sinceˇˇˇx pw˚, z˚q, r T pw, zqyˇˇ" |xpw˚, z˚q, p0, T wqy| " |xz˚pT wqy| ď }z˚} ď 1 2 }pw˚, z˚q} a˚" 1 2 for every ppw, zq, pw˚, z˚qq P ΠpW ' a Zq, we have that νp r T q ď 1{2. This shows that r T cannot belong to A nu pW ' a Zq.
Next we prove the analogous result for 8 -sum but under different hypothesis on the underlying spaces.
Proposition 3.5. Let Z be a uniformly convex and uniformly smooth Banach space. Let W be an arbitrary Banach space. If T P A }¨} pW, Zq, then r T P A nu pW ' 8 Zq.
Remark 3. 6. Similar to what happened on Proposition 3.1, Proposition 3.5 is not true in general. Indeed, consider the real Banach space 1 . Like we did in Remark 3.2, we will show that the operator introduced in Example 2.9 does not satisfy the property stated in Proposition 3.5. Let T : 1 ÝÑ 1 be defined as T pxq :" 2 j e j , for all x P 1 , and let r T P Lp 1 ' 8 1 q be defined accordingly. First note that if ppx, yq, px˚, y˚qq P Πp 1 ' 8 1 q, we have: (i) }px, yq} " maxt}x} 1 , }y} 1 u " 1.
Therefore r T R A nu p 1 ' 8 1 q as desired, even though T P A }¨} p 1 , 1 q.
We say that an absolute norm |¨| a in R 2 is of type 8 if the vector p1, 0q is not an extreme point of B pR 2 ,}¨}aq . Equivalently, |¨| a is of type 8 if and only there exists b 0 ą 0 such that |p1, b 0 q| a " 1.
Suppose that T belongs to A }¨} pW, Zq. We claim that r T does not belong to A nu pW ' a Zq, where }¨} a is the type 8 absolute norm induced from the norm |||¨|||. Note that }pw, zq} a " maxt}w}, p1{2q}z}u for every pw, zq P W ' a Z and it is plain to observe that }pw˚, z˚q} a˚" }w˚}`2}z˚} for every pw˚, z˚q P W˚' a˚Z˚. As we calculated in Remark 3.3,ˇˇx pw˚, z˚q, r T pw, zqyˇˇď 1 2 }pw˚, z˚q} a˚" 1 2 for every ppw, zq, pw˚, z˚qq P ΠpW ' a Zq, we have that νp r T q ď 1{2. This proves our claim.