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Let X, Y be intermediate spaces with respect to couples , , respectively. We say that X, Y are (relative) interpolation spaces if there is aH constantK C such that T : implies that T : X Y and H → K → (1.2) T C T . k k L(X;Y ) ≤ k k L(H;K) In the case when C = 1 we speak about exact interpolation. When = and X = Y we simply say that X is an (exact) interpolation space with respectH K to . H Let H be a suitable function of two positive variables and X, Y spaces interme- diate to the couples , , respectively. We say that the spaces X, Y are of type H (relative to , ) ifH forK any positive numbers M , M we have H K 0 1 (1.3) T Mi, i =0, 1 implies T H(M0,M1). k k L(Hi;Ki) ≤ k k L(X;Y ) ≤ The case H(x, y) = max x, y corresponds to exact interpolation, while H(x, y)= x 1−θy θ corresponds to the{ convexity} estimate
(1.4) T T 1−θ T θ . k k L(X;Y ) ≤k k L(H0;K0) k k L(H1;K1) In the situation of (1.4), one says that the interpolation spaces X, Y are of exponent θ with respect to the pairs , . H K 1.2. K-spaces. Given a regular Hilbert couple and a positive Radon measure ̺ on the compactified half-line [0, ] we define anH intermediate quadratic norm by ∞ (1.5) x 2 = x 2 = 1+ t−1 K t, x; d̺(t). k k∗ k k̺ H Z[0,∞] Here the integrand k(t) = 1+ t−1 K(t, x) is defined at the points 0 and by ∞ k(0) = x 2 and k( ) = x 2; we shall write or for the Hilbert space 1 0 ∗ ̺ definedk byk the norm (1.5).∞ k k H H Let T ; and suppose that T Mi; then ∈ L H K k k L(Hi;Ki) ≤ (1.6) K t,Tx ; M 2 K M 2t/M 2, x; , x Σ. K ≤ 0 1 0 H ∈ In particular, Mi 1 for i= 0, 1 implies T x x for all x ̺. It ≤ k k K̺ ≤ k k H̺ ∈ H follows that the spaces , are exact interpolation spaces with respect to , . H̺ K̺ H K Geometric interpolation. When the measure ̺ is given by t−θ π d̺(t)= c dt, c = , 0 <θ< 1, θ 1+ t θ sin θπ we denote the norm (1.5) by ∞ dt (1.7) x 2 := c t−θK (t, x) . k kθ θ t Z0 The corresponding space is easily seen to be of exponent θ with respect to . Hθ H In §3.1, we will recognize θ as the geometric interpolation space which has been studied independently by severalH authors, see [27, 40, 25]. INTERPOLATION BETWEEN HILBERT SPACES 3
2 1.3. Pick functions. Let be a regular Hilbert couple. The squared norm x 1 is a densely defined quadraticH form in , which we represent as k k H0 x 2 = Ax , x = A 1/2x 2 k k1 h i0 k k0 where A is a densely defined, positive, injective (perhaps unbounded) operator in . The domain of the positive square-root A1/2 is ∆. H0 Lemma 1.1. We have in terms of the functional calculus in H0 tA (1.8) K (t, x )= x , x , t> 0. 1+ tA 0 tA In the formula (1.8), we have identified the bounded operator 1+tA with its extension to . H0 Proof. Fix x ∆. By a straightforward convexity argument, there is a unique ∈ decomposition x = x0,t + x1,t which is optimal in the sense that
(1.9) K(t, x)= x 2 + t x 2 . k 0,t k0 k 1,t k1 It follows that x ∆ for i =0, 1. Moreover, for all y ∆ we have i,t ∈ ∈ d x + ǫy 2 + t x ǫy 2 =0, dǫ{k 0,t k0 k 1,t − k1 }|ǫ=0 i.e., A−1/2x tA1/2x , A1/2y =0, y ∆. h 0,t − 1,t i0 ∈ −1/2 1/2 By regularity, we conclude that A x0,t = tA x1,t, whence tA 1 (1.10) x = x and x = x. 0,t 1+ tA 1,t 1+ tA
(Note that the operators in (1.10) extend to bounded operators on 0.) Inserting the relations (1.10) into (1.9), one finishes the proof of the lemma. H
Now fix a positive Radon measure ̺ on [0, ]. The norm in the space ̺ (see (1.5)) can be written ∞ H
(1.11) x 2 = h(A)x , x , k k̺ h i0 where (1 + t)λ (1.12) h(λ)= d̺(t). 1+ tλ Z[0,∞] The class of functions representable in this form for some positive Radon measure ′ ̺ is the class P of Pick functions, positive and regular on R+. Notice that for the definition (1.11) to make sense, we just need h to be defined on σ(A) 0 , where σ(A) is the spectrum of A. (The value h(0) is irrelevant since A is injective).\{ } A calculus exercise shows that for the space (see (1.7)) we have Hθ (1.13) x 2 = Aθx , x . k kθ h i0 4 YACIN AMEUR
1.4. Quadratic interpolation norms. Let ∗ be any quadratic intermediate space relative to . We write H H x 2 = Bx, x k k∗ h i0 where B is a positive injective operator in (the domain of B1/2 is ∆). H0 For a map T ( ) we shall often use the simplified notations ∈ L H T = T , T = T , T = T . k k k k L(H0) k kA k k L(H1) k kB k k L(H∗) The reader can check the identities T = A1/2T A−1/2 and T = B1/2TB−1/2 . k kA k k k kB k k We shall refer to the following lemma as Donoghue’s lemma, cf. [14, Lemma 1].
Lemma 1.2. If ∗ is exact interpolation with respect to , then B commutes with every projectionH which commutes with A and B = h(A)Hwhere h is some positive Borel function on σ(A). Proof. For an orthogonal projection E on , the condition E 1 is equivalent H0 k kA ≤ to that EAE A, i.e., that E commutes with A. The hypothesis that ∗ be exact interpolation thus≤ implies that every spectral projection of A commutesH with B. It now follows from von Neumann’s bicommutator theorem that B = h(A) for some positive Borel function h on σ(A).
In view of the lemma, the characterization of the exact quadratic interpolation norms of a given type H reduces to the characterization of functions h : σ(A) R → + such that for all T ∈ L H (1.14) T M and T M T H(M ,M ), k k≤ 0 k kA ≤ 1 ⇒ k kh(A) ≤ 0 1 or alternatively, (1.15) T ∗T M 2 and T ∗AT M 2 A T ∗h(A)T H(M ,M ) 2 h(A). ≤ 0 ≤ 1 ⇒ ≤ 0 1 The set of functions h : σ(A) R satisfying these equivalent conditions forms a → + convex cone CH,A; its elements are called interpolation functions of type H relative to A. In the case when H(x, y) = max x, y we simply write CA for CH,A and speak of exact interpolation functions relative{ } to A.
1.5. Exact Calderón pairs and the K-property. Given two intermediate normed spaces Y , X relative to , , we say that they are (relatively) exact K-monotonic if the conditions H K x0 X and K t,y0; K t, x0; , t> 0 ∈ H ≤ K imply that y0 Y and y0 x0 . ∈ k kY ≤k kX It is easy to see that exact K-monotonicity implies exact interpolation.
Proof of this. If T 1 then x, t: K t,Tx; K t, x; whence k k L(K;H) ≤ ∀ H ≤ K T x x , by exact K-monotonicity. Hence T 1. k kY ≤k kX k k L(X;Y ) ≤ INTERPOLATION BETWEEN HILBERT SPACES 5
Two pairs , are called exact relative Calderón pairs if any two exact interpola- tion (Banach-)H spacesK Y , X are exact K-monotonic. Thus, with respect to to exact Calderón pairs, exact interpolation is equivalent to exact K-monotonicity. The term "Calderón pair" was coined after thorough investigation of A. P. Calderón’s study of the pair (L1,L∞), see [10] and [11]. In our present discussion, it is not convenient to work directly with the definition of exact Calderón pairs. Instead, we shall use the following, closely related notion. We say that a pair of couples , has the relative (exact) K-property if for all H K x0 Σ( ) and y0 Σ( ) such that ∈ K ∈ H (1.16) K t,y0; K t, x0; , t> 0, H ≤ K there exists a map T ( ; ) such that T x0 =y0 and T 1. ∈ L K H k k L(K;H) ≤ Lemma 1.3. If , have the relative K-property , then they are exact relative Calderón pairs. H K Proof. Let Y , X be exact interpolation spaces relative to , and take x0 X and y0 Σ( ) such that (1.16) holds. By the K-property thereH K is T : ∈such that T x∈0 = yH0 and T 1. Then T 1, and so y0 =K →T xH0 k k≤ k k L(X;Y ) ≤ k kY k kY ≤ x0 . We have shown that Y , X are exact K-monotonic. k kX In the diagonal case = , we simply say that is an exact Calderón couple if for intermediate spaces HY,X,K the property of beingH exact interpolation is equivalent to being exact K-monotonic. Likewise, we say that has the K-property if the pair of couples , has that property. H H H Remark 1.4. For an operator T : to be a contraction, it is necessary and sufficient that K → H (1.17) K t,Tx; K t, x; , x Σ( ),t> 0. H ≤ K ∈ K Indeed, the necessity is immediate. To prove the sufficiency it suffices to observe that letting t in (1.17) gives T x 0 x 0, and dividing (1.17) by t, and then letting t →0 ∞, gives that T x k kx ≤. k k → k k1 ≤k k1 2. Mapping properties of Hilbert couples 2.1. Main results. We shall elaborate on the following main result from [2]. Theorem I. Any pair of regular Hilbert couples , has the relative K-property . H K Before we come to the proof of Theorem I, we note some consequences of it. We first have the following corollary, which shows that a strong form of the K-property is true. Corollary 2.1. Let be a regular Hilbert couple and x0,y0 Σ elements such that H ∈ (2.1) K t,y0 M 2 K M 2t/M 2 , x0 , t> 0. ≤ 0 1 0 Then 0 0 (i) There exists a map T such that T x = y and T Mi, ∈ L H k k L(Hi) ≤ i =0, 1. (ii) If x0 X where X is an interpolation space of type H, then ∈ y0 H (M ,M ) x0 . k kX ≤ 0 1 k kX 6 YACIN AMEUR
Proof. (i) Introduce a new couple by letting x = Mi x H . The relation K k kKi k k i (2.1) then says that K t,y0; K t, x0; , t> 0. H ≤ K By Theorem I there is a contraction T : such that T x0 = y0. It now suffices K → H to note that T = Mi T ; (ii) then follows from Lemma 1.3. k k L(Hi) k k L(Ki;Hi) We next mention some equivalent versions of Theorem I, which uses the families of functionals K and E defined (for p 1 and t,s > 0) via p p ≥ p p Kp(t)= Kp(t, x)= Kp t, x; = inf x0 0 + t x1 1 H x=x0+x1 {k k k k } (2.2) p Ep(s)= Ep(s, x)= Ep s, x; = infp x x0 1 . H k x0 k0 ≤s {k − k } 1/p p Note that K = K2 and that Ep(s) = E1 s ; the E-functionals are used in approximation theory. One has that E is decreasing and convex on R and that p + Kp(t) = inf s + tEp(s) , s>0 { } which means that Kp is a kind of Legendre transform of Ep. The inverse Legendre transformation takes the form K (t) s E (s) = sup p . p t − t t>0 It is now immediate that, for all x Σ and y Σ , we have ∈ K ∈ H (2.3) K (t,y) K (t, x), t> 0 E (s,y) E (s, x), s> 0. p ≤ p ⇔ p ≤ p 2/p p/2 Since moreover Ep(s)= E2 s , the conditions in (2.3) are equivalent to that K(t,y) K(t, x) for all t> 0. We have shown the following result. ≤ Corollary 2.2. In Theorem I, one can substitute the K-functional for any of the functionals Kp or Ep. Define an exact interpolation norm relative to by k·k̺,p H x p = 1+ t−1 K (t, x) d̺(t) k k̺,p p Z[0,∞] where ̺ is a positive Radon measure on [0, ]. This norm is non-quadratic when p =2, but is of course equivalent to the quadratic∞ norm corresponding to p =2. 6 2.2. Reduction to the diagonal case. It is not hard to reduce the discussion of Theorem I to a diagonal situation. Lemma 2.3. If the K-property holds for regular Hilbert couples in the diagonal case = , then it holds in general. H K Proof. Fix elements y0 Σ( ) and x0 Σ( ) such that the inequality (1.16) ∈ H ∈ K holds. We must construct a map T : such that T x0 = y0 and T 1. K → H k k≤ To do this, we form the direct sum = ( 0 0, 1 1). It is clear that + = ( + ) ( + ), and thatS H ⊕ K H ⊕ K S0 S1 H0 H1 ⊕ K0 K1 K t, x y; = K t, x; + K t,y; . ⊕ S H K Then K t, 0 y0; K t, x0 0; . ⊕ S ≤ ⊕ S INTERPOLATION BETWEEN HILBERT SPACES 7
Hence assuming that the couple has the K-property, we can assert the existence of a map S ( ) such that SS(x0 0) = 0 y0 and S 1. Letting P : ∈ L S ⊕ ⊕ k k ≤ 0 + 1 0 + 1 be the orthogonal projection, the assignment T x = PS(x 0) Snow definesS → K a mapK such that T x0 = y0 and T 1. ⊕ k k L(H;K) ≤ 2.3. The principal case. The core content of Theorem I is contained in the fol- lowing statement. Theorem 2.4. Suppose that a regular Hilbert couple is finite dimensional and that all eigenvalues of the corresponding operator A areH of unit multiplicity. Then has the K-property . H We shall settle for proving Lemma 2.4 in this section, postponing to Section 5 the general case of Theorem I. To prepare for the proof, we write the eigenvalues λi of A in increasing order, σ(A)= λ n where 0 < λ < < λ . { i}1 1 ··· n Let e be corresponding eigenvectors of unit length for the norm of . Then for a i H0 vector x = xiei we have n n P x 2 = x 2 , x 2 = λ x 2. k k0 | i| k k1 i| i| 1 1 X X Working in the coordinate system (e ), the couple becomes identified with the i H n-dimensional weighted ℓ2 couple n n n ℓ2 (λ) := (ℓ2 ,ℓ2 (λ)) , n where we write λ for the sequence (λi)1 . n C n We will henceforth identify a vector x = xiei with the point x = (xi)1 in ; n ∗ C accordingly, the space (ℓ2 ) is identified with the C -algebra Mn( ) of complex n n matrices. L P × n It will be convenient to reparametrize the K-functional for the couple ℓ2 (λ) and write n (2.4) kλ(t, x) := K 1/t,x; ℓ2 (λ) . By Lemma 1.1 we have n λ (2.5) k (t, x)= i x 2, x C n. λ t + λ | i| ∈ i=1 i X n 2.4. Basic reductions. To prove that the couple ℓ2 (λ) has the K-property, we introduce an auxiliary parameter ρ> 1. The exact value of ρ will change meaning during the course of the argument, the main point being that it can be chosen arbitrarily close to 1. Initially, we pick any ρ > 1 such that ρλi < λi+1 for all i; we assume also that we are given two elements x0,y0 C n such that ∈ 1 (2.6) k t,y0 < k t, x0 , t 0. λ ρ λ ≥ We must construct a matrix T M (C) such that ∈ n (2.7) T x0 = y0 and k (t,Tx) k (t, x) , x C n,t> 0. λ ≤ λ ∈ 8 YACIN AMEUR
Define x˜0 = ( x0 )n and y˜0 = ( y0 )n and suppose that | i | 1 | i | 1 1 k (t, y˜0) < k (t, x˜0), t 0. λ ρ λ ≥ 0 0 Suppose that we can find an operator T0 Mn(C) such that T0x˜ =y ˜ and C n ∈ 0 iθk 0 0 iϕk 0 kλ (t,T0x) < kλ(t, x) for all x and t> 0. Writing xk = e x˜k and yk = e y˜k where θ , ϕ R, we then have∈ T x0 = y0 and k (t,Tx) < k (t, x) where k k ∈ λ λ iϕk −iθk T = diag(e )T0 diag(e ). 0 0 0 0 0 0 Replacing x ,y by x˜ , y˜ we can thus assume that the coordinates xi and yi are non-negative; replacing them by small perturbations if necessary, we can assume that they are strictly positive, at the expense of slightly diminishing the number ρ.
Now put βi = λi and αi = ρλi. Our assumption on ρ means that 0 <β < α < <β < α . 1 1 ··· n n Using the explicit expression for the K-functional, it is plain to check that k (t, x) k (t, x) ρk (t, x), x Cn, t 0. β ≤ α ≤ β ∈ ≥ Our assumption (2.6) therefore implies that (2.8) k (t,y0) < k (t, x0), t 0. α β ≥ We shall verify the existence of a matrix T = Tρ = Tρ,x0,y0 such that (2.9) T x0 = y0 and k (t,Tx) k (t, x) , x Cn,t> 0. α ≤ β ∈ It is clear by compactness that, as ρ 1, the corresponding matrices Tρ will cluster at some point T satisfying T x0 = y0↓and T 1. (See Remark 1.4.) k k L( H ) ≤ In conclusion, the proof of Theorem 2.4 will be complete when we can construct a matrix T satisfying (2.9) with ρ arbitrarily close to 1.
2.5. Construction of T . Let k denote the linear space of complex polynomials of degree at most k. We shall useP the polynomials n n Lα(t)= (t + αi) ,Lβ(t)= (t + βi) , 1 1 Y Y and the product L = LαLβ. Notice that (2.10) L′( α ) < 0 ,L′( β ) > 0. − i − i Recalling the formula (2.5), it is clear that we can define a real polynomial P by ∈ P2n−1 P (t) (2.11) = k t, x0 k t,y0 . L(t) β − α Clearly P (t) > 0 when t 0. Moreover, a consideration of the residues at the poles of the right-hand member≥ shows that P is uniquely defined by the values (2.12) P ( β ) = (x0) 2β L′ ( β ) , P ( α )= (y0) 2α L′ ( α ) . − i i i − i − i − i i − i Combining with (2.10), we conclude that (2.13) P ( α ) > 0 and P ( β ) > 0. − i − i INTERPOLATION BETWEEN HILBERT SPACES 9
Perturbing the problem slightly, it is clear that we can assume that P has exact degree 2n 1, and that all zeros of P have multiplicity 1. (We here diminish the value of ρ>− 1 somewhat, if necessary.) Now, P has 2n 1 simple zeros, which we split according to − P −1 ( 0 )= r 2m−1 c , c¯ n−m , { } {− i}i=1 ∪ {− i − i}i=1 where the ri are positive and the ci are non-real, and chosen to have positive imaginary parts. The following is the key observation. Lemma 2.5. We have that (2.14) L′ ( β ) P ( β ) > 0 ,L′ ( α ) P ( α ) < 0 − i − i − i − i and there is a splitting r 2m−1 = δ m γ m−1 such that { i}i=1 { i}i=1 ∪{ i}i=1 (2.15) L ( δ ) P ′ ( δ ) > 0 ,L ( γ ) P ′ ( γ ) < 0. − j − j − k − k Proof. The inequalities (2.14) follow immediately from (2.13) and (2.10). It remains to prove (2.15). Let h denote the leftmost real zero of the polynomial LP (of degree 4n 1). We claim− that P ( h)=0. If this were not the case, we would have h = α . Since− − n the degree of P is odd, P ( t) is negative for large values of t, and so P ( αn) < 0 contradicting (2.13). We have− shown that P ( h)=0. Since all zeros of−LP have multiplicity 1, we have (LP )′( h) =0, whence− − 6 L( h)P ′( h) = (LP )′( h) > 0. − − − We write δm = h and put P∗(t) = P (t)/(t + δm). Since t + δm > 0 for t α , β n, we have by (2.13) that for all i ∈ {− i − i}1 P ( α ) > 0 and P ( β ) > 0. ∗ − i ∗ − i Denote by r ∗ 2m−2 the real zeros of P . Since the degree of LP is even and {− j }j=1 ∗ ∗ the polynomial (LP )′ has alternating signs in the set α , β n r ∗ 2m−2, ∗ {− i − i}i=1 ∪{− i }i=1 we can split the zeros of P as δ , γ m−1, where ∗ {− i − i}i=1 (2.16) L( δ )P ′ ( δ ) > 0 ,L( γ )P ′ ( γ ) < 0. − i ∗ − i − i ∗ − i ′ ∗ ∗ ′ ∗ ∗ ′ ∗ Since P ( rj ) = (δm rj )P∗( rj ) and δm > rj , the signs of P ( rj ) and P ′( r ∗) −are equal, proving− (2.15).− − ∗ − j n−m Recall that ci 1 denote the zeros of P such that Im ci > 0. We put (with the convention{− that} an empty product equals 1)
m m−1 n−m Lδ(t)= (t + δi) ,Lγ(t)= (t + γi) ,Lc(t)= (t + ci). i=1 i=1 i=1 Y Y Y We define a linear map F : Cn+m Cn+m−1 in the following way. First define a subspace U by → ⊂P2n−1 U = L q ; q . { c ∈Pn+m−1 } Notice that U has dimension n + m 1 and that P U; in fact P = aL L ∗L L − ∈ c c δ γ where a is the leading coefficient and the -operation is defined by L ∗(z)= L(¯z). ∗ 10 YACIN AMEUR
For a polynomial Q U we have ∈ n n Q(t) 2 β δ | | = x 2 i + x′ 2 i L(t)P (t) | i| t + β | i| t + δ i=1 i i=1 i X X (2.17) n m−1 α γ y 2 i y′ 2 i , − | i| t + α − | i| t + γ i=1 i i=1 i X X where, for definiteness,
Q( βi) ′ Q( δj ) (2.18) xi = − ; x = − ′ j ′ β L ( β )P ( β ) δj L ( δj)P ( δj ) i − i − i − − Q( αi) ′ Q( γj) (2.19) yi = p − ; y = p − . ′ j ′ α L ( α )P ( α ) γj L ( γj )P ( γj ) − i − i − i − − − The identitiesp in (2.18) give rise to a linear mapp (2.20) M : C n C m U ;[x; x′] Q. ⊕ → 7→ We can similarly regard (2.19) as a linear map
(2.21) N : U C n C m−1 ; Q [y; y′] . → ⊕ 7→ Our desired map F is defined as the composite F = NM : C n C m C n C m−1 ;[x; x′] [y; y′]. ⊕ → ⊕ 7→ Notice that if Q = M [x; x′] and [y; y′]= F [x; x′] then (2.17) means that
Q(t) 2 k (t, [x; x′]) k (t, F [x; x′]) = | | 0, t 0. β⊕δ − α⊕γ L(t)P (t) ≥ ≥
n+m n+m−1 This implies that F is a contraction from ℓ2 (β δ) to ℓ2 (α γ). We now define T as a "compression" of F . Namely,⊕ let E : C n ⊕C m−1 C n be the projection onto the first n coordinates, and define an operator⊕ T on C→n by T x = EF [x; 0] , x C n. ∈ Taking Q = P in (2.17) we see that T x0 = y0. Moreover,
n n β α k (t, x) k (t,Tx)= x 2 i y 2 i β − α | i| t + β − | i| t + α i=1 i i=1 i X X n n m−1 β α γ x 2 i y 2 i y′ 2 i ≥ | i| t + β − | i| t + α − | i| t + γ i=1 i i=1 i j=1 i X X X Q(t) 2 = k (t, [x; 0]) k (t, F [x; 0]) = | | . β⊕δ − α⊕γ L(t)P (t) Since the right-hand side is non-negative, we have shown that k (t,Tx) k (t, x), t> 0, x Cn, α ≤ β ∈ as desired. The proof of Theorem 2.4 is finished. q.e.d. INTERPOLATION BETWEEN HILBERT SPACES 11
2.6. Real scalars. Theorem 2.4 holds also in the case of Euclidean spaces over the real scalar field. To see this, assume without loss of generality that the vectors x0,y0 Cn have real entries (still satisfying k t,y0 k t, x0 for all t> 0). ∈ λ ≤ λ By Theorem 2.4 we can find a (complex) contraction T of ℓn(λ) such that T x0 = 2 y0. It is clear that the operator T ∗ defined by T ∗x = T (¯x) satisfies those same 1 ∗ R conditions. Replacing T by 2 (T + T ) we obtain a real matrix T Mn( ), which n 0 0 ∈ is a contraction of ℓ2 (λ) and maps x to y . 2.7. Explicit representations. We here deduce an explicit representation for the operator T constructed above. Let x0 and y0 be two non-negative vectors such that k t,y0 k t, x0 , t> 0. λ ≤ λ For small ρ > 0 we perturb x0, y0 slightly to vectors x˜ 0, y˜0 which satisfy the conditions imposed the previous subsections. We can then construct a matrix T = Tρ such that (2.22) T x˜ 0 =y ˜ 0 and k (t,Tx) k (t, x) , t> 0, x C n, α ≤ β ∈ where β = λ and α = ρλ. As ρ, x˜0, y˜0 approaches 1, x0, resp. y0, it is clear that any cluster point T of the set of contractions Tρ will satisfy T x0 = y0 and k (t,Tx) k (t, x) , t> 0, x C n. λ ≤ λ ∈ n Theorem 2.6. The matrix T = T̺ = (τik)i,k=1 where 1 x˜0 β L ( α )L ( α )L ( β ) (2.23) τ = Re k k δ − i c − i α − k ik α β y˜0 α L ( β )L ( β )L′ ( α ) i − k i i δ − k c − k α − i satisfies (2.22). Proof. The range of the map C n U, x M [x; 0] (see 2.20) is precisely the n-dimensional subspace → 7→ (2.24) V := L L = L L R; R U. δ c ·Pn−1 { δ c ∈Pn−1}⊂ n We introduce a basis (Qk)k=1 for V by L (t)L (t)L (t) β L′( β )P ( β ) Q (t)= δ c β k − k − k . k t + β L ( β )L ( β )L′ ( β ) k δp− k c − k β − k Then Q ( β ) 1 i = k, k − i = ′ βiL ( βi)P ( βi) (0 i = k. − − 6 C n Denoting by (ei) the canonicalp basis in and using (2.18), (2.19) we get
Qk( αi) τik = (Tek)i = − α L′( α )P ( α ) i − i − i ′ 1/2 1 Lpδ( αi)Lc( αi)Lβ( αi) βkL ( βk)P ( βk) = − − − − − . β α L ( β )L ( β )L′ ( β ) α L′( α )P ( α ) k − i δ − k c − k β − k − i − i − i Inserting the expressions (2.12) for P ( αi) and P ( βk) and taking real parts (see the remarks in §2.6), we obtain the formula− (2.23).− 12 YACIN AMEUR
Remark 2.7. It is easy to see that, if we pick all matrix-elements real, some elements 0 0 τik of the matrix T in (2.23) will be negative, even while the numbers xi and yk are positive. It was proved in [2], Theorem 2.3, that this is necessarily so. Indeed, 5 one there constructs an example of a five-dimensional couple ℓ2 (λ) and two vectors x0,y0 R5 having non-negative entries such that no contraction T = (τ )5 on ∈ ik i,k=1 ℓ 5(λ) having all matrix entries τ 0 can satisfy T x0 = y0. On the other hand, if 2 ik ≥ one settles for using a matrix with T √2, then it is possible to find one with only non-negative matrix entries. Indeed,k k≤ such a matrix was used by Sedaev [35], see also [39].
2.8. On sharpness of the norm-bounds. We shall show that if m Q(1)(t) 2 k (t, x(1)) k (t,y(1))= | | , t> 0. β − α L(t)P (t) (1) 2 (1) 2 Choosing t =0 we conclude that x n T x n =0, whence T 1, k kℓ2 −k kℓ2 k k L(H0) ≥ proving our claim. Similarly, the condition m Q(2)(t) 2 k (t, x(2)) k (t,y(2))= | | , t> 0. β − α L(t)P (t) (2) 2 Multiplying this relation by t and then sending t , we find that x n → ∞ k kℓ2 (β) − (2) 2 −1/2 T x n =0, which implies T L H ρ . k kℓ2 (α) k k ( 1) ≥ 2.9. A remark on weighted ℓ -couples. As far as we are aware, if 1