Variable-Coefficient Toeplitz Matrices with Symbols Beyond the Wiener

Operator Theory: Advances and Applications, Vol. 199, 191–202 c 2009 Birkh¨auser Verlag Basel/Switzerland

Variable-coeﬃcient Toeplitz Matrices with Symbols beyond the Wiener Algebra

Albrecht B¨ottcher and Sergei Grudsky

In Memory of Georg Heinig

Abstract. Sequences of so-called variable-coefficient Toeplitz matrices arise in many problems, including the discretization of ordinary differential equations with variable coefficients. Such sequences are known to be bounded if the generating function satisfies a condition of the Wiener type, which is far away from the minimal requirement in the case of constant coefficients. The purpose of this paper is to uncover some phenomena beyond the Wiener condition. We provide counterexamples on the one hand and prove easy-to-check sufficient conditions for boundedness on the other. Mathematics Subject Classification (2000). Primary 47B35; Secondary 15A60, 65F35. Keywords. Toeplitz matrix, variable coefficients, matrix norm.

1. Introduction Let a be a complex-valued continuous function on [0, 1] × [0, 1] × T,whereT is the complex unit circle, a :[0, 1] × [0, 1] × T → C, (x, y, t) → a(x, y, t). For n ∈ Z,we −n puta ˆn(x, y)= T a(x, y, t) t |dt|/(2π) and so have the Fourier series ∞ n a(x, y, t)= aˆn(x, y)t , (1) n=−∞ 2 where equality holds at least in the L sense. Let AN (a)bethematrix N j k AN (a)= aˆj−k , . (2) N N j,k=0

Occasionally we allow us to write a(x, y, t)andAN (a(x, y, t)) for a and AN (a).

This work was partially supported by CONACYT project U46936-F, Mexico. 192 A. B¨ottcher and S. Grudsky

We refer to AN (a) as a Toeplitz matrix with variable coefficients. Clearly, if a does not depend on x and y,thenAN (a) is a pure Toeplitz matrix. It is easily seen that every (N +1)× (N + 1) matrix may be written as AN (a)withsomea.Thus, the notion of a Toeplitz matrix with variable coefficients is rather an asymptotic ∞ notion which makes sense for the entire sequences {AN (a)}N=0 but not for a single (N +1)× (N +1)matrix. Variable-coefficient Toeplitz matrices and their modifications and generaliza- tions are currently emerging in many applications (see, for example, [1], [4], [5], [6], [7], [8], [11], [12]) and go under various names, such as generalized Toeplitz [9], locally Toeplitz [11] or generalized locally Toeplitz matrices [8], Berezin-Toeplitz matrices [1], twisted Toeplitz matrices [12], or generalized discrete convolutions [10], [13]. In our opinion, variable-coefficient Toeplitz matrices [3] is the perhaps best name, at least when considering matrices of the form (2). The problem of primary interest is the understanding of the spectral and pseudospectral properties of AN (a)asN →∞. Accordingly, the works cited above and also [3], [9], [14] deal with extensions of the Szegö and Avram-Parter theorems, that is, with asymptotic formulas for tr f(AN (a)), and with asymptotic inverses and pseudospectra of AN (a).

This paper addresses the problem of the uniform boundedness of AN (a), that is, the question whether AN (a)∞ remains bounded as N →∞.Hereandinwhat follows, A∞ is the spectral norm of A. ∞ n If a is independent of x and y, a(t)= n=−∞ aˆnt ,wedenoteAN (a)by N TN (a). Thus, TN (a)=(ˆaj−k)j,k=0.ItiswellknownthatTN (a)∞ ≤TN+1(a)∞ for all N and that

lim TN (a)∞ = M∞(a):=sup|a(t)|. (3) N→∞ t∈T If a is of the form a(x, y, t)=b(x, y)tn,then N n j + n j AN (a)=TN (t )diag b , N N j=0 (where j + n is taken modulo N + 1) and hence n AN (b(x, y)t )∞ ≤ M∞,∞(b):= sup sup |b(x, y)|. x∈[0,1] y∈[0,1] Consequently, for a given by (1) we have ∞ AN (a)∞ ≤ M∞,∞(ân). (4) n=−∞ Thus, if ∞ M∞,∞(ân) < ∞, (5) n=−∞ Toeplitz Matrices with Variable Coefficients 193

∞ then {AN (a)∞}N=0 is a bounded sequence. Condition (5) is a condition of the Wiener type. In the case where a does not depend on x and y, it amounts to saying that TN (a)∞ remains bounded as N →∞if a belongs to the Wiener algebra, ∞ that is, if n=−∞ |aˆn| < ∞. This is clearly far away from (3) and is therefore a source of motivation for looking whether the uniform boundedness of AN (a)∞ can be guaranteed under weaker assumptions. The ﬁrst question one might ask is whether the sole continuity of the generating function a on [0, 1]2 ×T := [0, 1]×[0, 1]×T ensures the uniform boundedness of AN (a)∞. We show that, surprisingly, the answer to this question is no.

2 Theorem 1.1. There exist a ∈ C([0, 1] × T) such that sup AN (a)∞ = ∞.

On the other hand, we will prove that AN (a)∞ remains bounded if a(x, y, t) has certain smoothness in x and y. (Notice that (5) is a requirement on the smoothness in t.) Our results will imply the following.

Theorem 1.2. If a ∈ C4,0([0, 1]2 × T), which means that the function a(x, y, t) has continuous partial derivatives with respect to x and y up to the order 4,then sup AN (a)∞ < ∞.

For a(x, y, t) independent of x and y, this is equivalent to the statement that TN (a) is uniformly bounded if a ∈ C(T). In fact, we can sharpen Theorem 1.2 as follows.

Theorem 1.3. If a ∈ L∞(T,C4([0, 1]2)), that is, if a is an L∞ function on T with values in the Banach space of all functions on [0, 1]2 that have continuous partial derivatives up to the order 4,thensup AN (a)∞ < ∞.

In the case of constant coeﬃcients, this theorem is best possible: it says that ∞ sup TN (a)∞ < ∞ if a ∈ L (T). The next question is whether the exponent 4 is close to a kind of a minimum. Let 0 <α≤ 1. For a continuous function a(x, y, t)on[0, 1]2 × T we deﬁne

|a(x2,y,t) − a(x1,y,t)| Mα,∞,∞(a)=sup sup sup α , t∈T y∈[0,1] x1,x2 |x2 − x1|

|a(x, y2,t) − a(x, y1,t)| M∞,α,∞(a)=sup sup sup α , t∈T x∈[0,1] y1,y2 |y2 − y1| and |∆2a(x1,x2,y1,y2,t)| Mα,α,∞(a)=sup sup sup α α t∈T x1,x2 y1,y2 |x2 − x1| |y2 − y1| where ∆2a(x1,x2,y1,y2,t) is the second diﬀerence

∆2a(x1,x2,y1,y2,t)=a(x2,y2,t) − a(x2,y1,t) − (a(x1,y2,t) − a(x1,y1,t)). ∈ Here supz1,z2 means the supremum over all z1,z2 [0, 1] such that z1 = z2.We say that the function a(x, y, t) belongs to Hα,α,∞ if the three numbers Mα,α,∞(a), 194 A. B¨ottcher and S. Grudsky

Mα,∞,∞(a), M∞,α,∞(a) are ﬁnite and we denote by H1+α,1+α,∞ the set of all functions a(x, y, t) that have continuous partial derivatives up to the order 2 in x and y and for which the three numbers

M1+α,∞,∞(a):=Mα,∞,∞(∂xa),M∞,1+α,∞(a):=M∞,α,∞(∂ya), M1+α,1+α,∞(a):=Mα,α,∞(∂x∂ya) 4,0 2 are ﬁnite. Notice that C ([0, 1] × T) ⊂ H2,2,∞.Wehereprovethefollowing.

Theorem 1.4. If β<1/2, there exist functions a in the space Hβ,β,∞ such that sup AN (a)∞ = ∞.Ifβ>1,thensup AN (a)∞ < ∞ for every function a in Hβ,β,∞.

The theorem leaves a gap. We can actually remove this gap and prove that sup AN (a)∞ < ∞ whenever a ∈ Hβ,β,∞ and β>1/2, which is guaranteed if a ∈ C2,0([0, 1]2 × T). However, the proof of this result is very sophisticated. We see the purpose of this paper in revealing the delicacy of the problem of the uniform boundedness of AN (a)∞ and in providing results that might be suﬃcient for applications. Drawing down things to β>1/2 is a matter of mathematical ambition and will be the topic of our subsequent paper [2].

2. H¨older continuity

We already deﬁned the space Hβ,β,∞ and the quantities Mβ,β,∞(a), Mβ,∞,∞(a), 2 M∞,β,∞(a) for a continuous function a(x, y, t)on[0, 1] × T and for 0 <β≤ 2. In addition, we put

M∞,∞,∞(a)=sup sup sup |a(x, y, t)|. t∈T x∈[0,1] y∈[0,1] Note that if a(x, y, t)=xγ + yγ with 0 <γ<1, then

Mβ,∞,∞(a)=M∞,β,∞(a)=∞,Mβ,β,∞(a)=0 for γ<β<1, which shows that the assumption Mβ,β,∞(a) < ∞ does not imply that Mβ,∞,∞(a)andM∞,β,∞(a) are finite. In the introduction we also mentioned 2,0 2 that C ([0, 1] × T) is contained in H1,1,0 and thus in Hβ,β,0 for 0 <β<1. This follows from the representations x2 a(x2,y,t) − a(x1,y,t)= ∂xa(ξ,y,t)dξ, x1 y2 a(x, y2,t) − a(x, y1,t)= ∂ya(x, η, t)dη, y1 x2 y2 ∆2a(x1,x2,y1,y2,t)= ∂x∂ya(ξ,η,t)dηdξ, x1 y1 which show that in fact the continuity of the partial derivatives ∂xa, ∂ya, ∂x∂ya 2,0 2 4,0 2 would suffice. Since C ([0, 1] × T) ⊂ H1,1,0, it follows that C ([0, 1] × T)isa subset of H2,2,0. Toeplitz Matrices with Variable Coefficients 195

In what follows we work with functions a(x, y, t) that are independent of one of the variables x and y and therefore need the following modifications. Let 0 <α≤ 1. We say that a continuous functions a(x, t)on[0, 1] × T is in Hα,∞ if | − | a(x2,t) a(x1,t) ∞ Mα,∞(a):=sup sup α < , t∈T x1,x2 |x2 − x1| and we denote by M∞,∞(a)themaximumof|a(x, t)| on [0, 1] × T. The function a(x, t)issaidtobeinH1+α,∞ if it is continuously differentiable in x and ∂xa is in Hα,∞.InthatcaseM1+α,∞(a) is defined as Mα,∞(∂xa). Analogously, we say that a continuous functions a(y,t)on[0, 1] × T is in Hα,∞ if | − | a(y2,t) a(y1,t) ∞ Mα,∞(a):=sup sup α < , t∈T y1,y2 |y2 − y1| we denote by M∞,∞(a)themaximumof|a(y,t)| on [0, 1] × T, and say that a(y,t) belongs to H1+α,∞ if ∂ya ∈ Hα,∞,inwhichcaseM1+α,∞(a) is defined as Mα,∞(∂ya). Finally, Hα is the set of all continuous functions f(x)on[0, 1] for which | − | f(x2) f(x1) ∞ Mα(f):=sup α < , x1,x2 |x2 − x1| and H1+α is the space of all continuously differentiable functions f(x)on[0, 1] with M1+α(f):=Mα(f ) < ∞; we put M∞(f)=supx∈[0,1] |f(x)|.

3. Counterexamples In this section we prove Theorem 1.1 and the ﬁrst statement of Theorem 1.4. We show that counterexamples can even be found within the functions a(x, y, t)that are independent of one of the variables x and y. Theorem 3.1. There exist functions a(x, t) in C([0, 1] × T) such that

sup AN (a)∞ = ∞. N≥0

Proof. Assume the contrary, that is, sup AN (a)∞ < ∞ for every function a in ∞ C([0, 1]×T). Let S denote the Banach space of all sequences {BN }N=0 of matrices (N+1)×(N+1) BN ∈ C such that ∞ {BN }N=0 := sup BN ∞ < ∞. N≥0 By our assumption, the map ∞ T : C([0, 1] × T) →S,a→{AN (a)}N=0 is a linear operator defined on all of C([0, 1] × T). To show that T is bounded, we employ the closed graph theorem. Thus, let an,a ∈ C([0, 1] × T) and suppose ∞ an → a in C([0, 1] × T)andTan → b = {BN }N=0 in S. Then, for fixed N ≥ 0, AN (an) − BN ∞ → 0asn →∞and hence the jk entry of AN (an)convergesto 196 A. Böttcher and S. Grudsky the jk entry of BN as n →∞,thatis,[AN (an)]jk → [BN ]jk for 0 ≤ j, k ≤ N.On the other hand, j j | [AN (an)]jk − [AN (a)]jk| = (ân)j−k − aˆj−k N N j j −(j−k) |dt| = an ,t − a ,t t ≤ M∞,∞(an − a)=o(1), T N N 2π which yields the equality [BN ]jk =[AN (a)]jk.Consequently,Ta = b. The closed graph theorem therefore implies that T is bounded. We have shown that there is a constant C<∞ such that

AN (a)∞ ≤ CM∞,∞(a)(6) for all a ∈ C([0, 1] × T). Fix N ≥ 2andforj =1,...,N − 1, denote by Ij the segment j 1 j 1 Ij = − , + . N 2N N 2N Let aj be the function that is identically zero on [0, 1] \ Ij, increases linearly from 0 to 1 on the left half of Ij, and decreases linearly from 1 to 0 on the right half of Ij.Put 1 2 N−1 a(x, t)=a1(x)t + a2(x)t + ···+ aN−1(x)t . As the spectral norm of a matrix is greater than or equal to the 2 norm of its ﬁrst column and asa ˆj(x)=aj(x)for1≤ j ≤ N − 1, it follows that N−1 2 N−1 2 j 2 AN (a)∞ ≥ aj = 1 = N − 1. j=1 N j=1 j Since a(x, t)=0forx/∈∪Ij and |a(x, t)| = |aj(x)t |≤1forx ∈ Ij,weobtain 2 2 that M∞,∞(a) = 1. Consequently, (6) gives N − 1 ≤ C · 1 for all N ≥ 2, which is impossible.

Theorem 3.2. If 0 <α<1/2, there exist functions a(x, t) in Hα,∞ such that

sup AN (a)∞ = ∞. N≥0

Proof. Assume that sup AN (a)∞ < ∞ for every a ∈ Hα,∞.ThespaceHα,∞ is a Banach space under the norm a := M∞,∞(a)+Mα,∞(a) and hence the same argument as in the proof of Theorem 3.1 gives

AN (a)∞ ≤ C(M∞,∞(a)+Mα,∞(a)) (7) for all a ∈ Hα,√∞.Leta(x, t) be exactly as the proof of Theorem 3.1. We then have α AN (a)∞ ≥ N −√1, M∞,∞(a) = 1, and it is easily seen that Mα,∞(a)=O(N ). Thus, (7) delivers N − 1=O(N α), which is impossible for α<1/2.

Since AN (a(y,t)) is the transpose of AN (a(x, 1/t)), the above two theorems also deliver counterexamples with functions of the form a(y,t). Toeplitz Matrices with Variable Coeﬃcients 197

4. Suﬃcient conditions In this section we prove the second half of Theorem 1.4 and thus also Theorem 1.2. The following result is well known; see, for example, [15, Chap. 2, Sec. 4]. We cite it with a full proof for the reader’s convenience.

Lemma 4.1. If f(x) is a function in H1+α and f(0) = f(1),then ∞ 2πinx f(x)= fne n=−∞ with Mα(f ) |fn|≤ for |n|≥1. 22+απ|n|1+α Proof. Let |n|≥1. Then 1 1 −2πinx −2πinx e fn = f(x)e dx = f(x)d 0 0 −2πin −2πinx 1 1 1 e 1 −2πinx 1 −2πinx = f(x) + f (x)e dx = f (x)e dx, −2πin 0 2πin 0 2πin 0 the last equality resulting from the requirement that f(0) = f(1). The substitution x → x − 1/(2n) yields 1 1 1 f (x)e−2πinxdx = − f x − e−2πinxdx, 0 0 2n whence 1 1 1 −2πinx fn = f (x) − f x − e dx. 4πin 0 2n α Taking into account that |f (x) − f (x − 1/(2n))|≤Mα(f )/(2n) , we arrive at the asserted inequality.

We ﬁrst consider functions a(x, y, t) that are independent of either x or y.

Theorem 4.2. Let α>0. There exists a constant C(α) depending only on α such that AN (a)∞ ≤ C(α)(M∞,∞(a)+M1+α,∞(a)) for all functions a(x, t) in H1+α,∞.

Proof. We write a = a0 + a1 with

a1(x, t)=(a(1,t) − a(0,t))x + a(0,t),a0(x, t)=a(x, t) − a1(x, t).

Then AN (a)=AN (a0)+AN (a1). Obviously, N j AN (b(x)c(t)) = b cˆj−k = DN (b)TN (c), (8) N j,k=0 198 A. B¨ottcher and S. Grudsky

N N where DN (b)=diag(b(j/N))j=0 and TN (c)=(ˆcj−k)j,k=0.Takingintoaccount that DN (b)∞ ≤ M∞(b)andTN (c)∞ ≤ M∞(c), we obtain that

AN (a1)∞ ≤ M∞(x)M∞(a(1,t) − a(0,t)) + M∞(a(0,t)) ≤ 3M∞,∞(a).

As a0(0,t)=a0(1,t) (= 0), Lemma 4.1 gives ∞ 0 2πinx a0(x, t)= an(t)e n=−∞ with 0 Mα(∂xa0(x, t)) |an(t)|≤ (9) 22+απ|n|1+α for |n|≥1. From (8) we infer that 0 2πinx 2πinx 0 0 AN (an(t)e )∞ ≤ M∞(e )M∞(an(t)) = M∞(an). Thus, by (9), 0 0 AN (a0)∞ ≤ M∞(a0)+ M∞(an) |n|≥1 0 1 Mα,∞(∂xa0(x, t)) ≤ M∞(a )+ . 0 2+απ |n|1+α 2 |n|≥1

Since a0(x, t)=a(x, t) − a1(x, t)and∂xa1(x, t) is independent of x,weget

Mα,∞(∂xa0(x, t)) = Mα,∞(∂xa(x, t)) = M1+α,∞(a). Furthermore, 1 0 M∞(a0)=sup a0(x, t)dx ≤ M∞,∞(a0)=M∞,∞(a − a1) t∈T 0 ≤ M∞,∞(a)+M∞,∞(a1) ≤ 4 M∞,∞(a). In summary,    1 1  AN (a)∞ ≤ 7 M∞,∞(a)+ M1+α,∞(a), 2+απ |n|1+α 2 |n|≥1 which implies the assertion at once. Theorem 4.3. Let α>0. There exists a constant C(α) that depends only on α such that AN (a)∞ ≤ C(α)(M∞,∞(a)+M1+α,∞(a)) for all functions a(y,t) in H1+α,∞.

Proof. This follows from Theorem 4.2 by taking transposed matrices. We now turn to the case where a(x, y, t) depends on all of the three variables. Toeplitz Matrices with Variable Coeﬃcients 199

Lemma 4.4. Let f(x, y, t) be a function in H1+α,1+α,∞ (α>0) and suppose f(0,y,t)=f(1,y,t) for all y and t.Then ∞ 2πinx f(x, y, t)= fn(y,t)e n=−∞ with 1 M∞,∞(fn) ≤ M1+α,∞,∞(f), (10) 22+απ|n|1+α 1 M1+α,∞(fn) ≤ M1+α,1+α,∞(f) (11) 22+απ|n|1+α for |n|≥1.

Proof. Estimate (10) is immediate from Lemma 4.1 and the deﬁnition of the num- ber M1+α,∞,∞(f). To prove (11), we ﬁrst note that 1 −2πinx ∂yfn(y2,t) − ∂yfn(y1,t)= (∂yf(x, y2,t) − ∂yf(x, y1,t))e dx (12) 0 and ∂yf(0,y,t)=∂y(1,y,t). Integrating by parts we therefore see that (12) equals 1 1 −2πinx (∂x∂yf(x, y2,t) − ∂x∂yf(x, y1,t))e dx, 2πin 0 and the substitution x → x − 1/(2n) shows that this is 1 1 1 −2πinx ∆2∂x∂yf x + ,x,y1,y2,t e dx. 4πin 0 2n Consequently, 1 | − |≤ 1 1 | − |α ∂yfn(y2,t) ∂yfn(y1,t) M1+α,1+α,∞(f) α y2 y1 dx, 4π|n| 0 |2n| which gives (11).

Theorem 4.5. Let α>0. Then there exists a constant D(α) < ∞ depending only on α such that

AN (a)∞ ≤ D(α)(M∞,∞,∞(a)+M1+α,∞,∞(a)+M∞,1+α,∞(a)+M1+α,1+α,∞(a)) for all N ≥ 0 and all functions a(x, y, t) in H1+α,1+α,∞.

Proof. We write a = a0 + a1 and accordingly AN (a)=AN (a0)+AN (a1)with

a1(x, y, t)=(a(1,y,t) − a(0,y,t))x + a(0,y,t), a0(x, y, t)=a(x, y, t) − a1(x, y, t). For every function c(y,t)wehave N N k j j AN (c(y,t)x)= cˆj−k =diag AN (c(y,t)) N N j,k=0 N j=0 200 A. B¨ottcher and S. Grudsky and the spectral norm of the diagonal matrix is 1. Hence

AN (a1)∞ ≤AN (a(1,y,t) − a(0,y,t))∞ + AN (a(0,y,t))∞ ≤AN (a(1,y,t))∞ +2AN (a(0,y,t))∞. Theorem 4.3 gives

AN (a(1,y,t))∞ ≤ C(α)(M∞,∞(a(1,y,t)) + M1+α,∞(a(1,y,t))) ≤ C(α)(M∞,∞,∞(a)+M∞,1+α,∞(a)), AN (a(0,y,t))∞ ≤ C(α)(M∞,∞(a(0,y,t)) + M1+α,∞(a(0,y,t))) ≤ C(α)(M∞,∞,∞(a)+M∞,1+α,∞(a)). Thus, AN (a1)∞ ≤ 3 C(α)(M∞,∞,∞(a)+M∞,1+α,∞(a)).

On the other hand, since a0(0,y,t)=a0(1,y,t), Lemma 4.4 implies that ∞ 0 2πinx a0(x, y, t)= an(y,t)e n=−∞ 0 where the functions an(y,t) satisfy estimates like (10) and (11). This time N 0 2πinx 0 k 2πinj/N AN (an(y,t)e )= (ˆan)j−k e N j,k=0 N 2πijn/N 0 =diage AN (an(y,t)) j=0 and the spectral norm of the diagonal matrix is again 1. It follows from Theorem 4.3 and Lemma 4.4 that 0 0 AN (a0)∞ ≤AN (a0)∞ + AN (an)∞ |n|≥1 0 1 ≤AN (a )∞ + (M1+α,∞,∞(a0)+M1+α,1+α,∞(a0)) 0 2+απ|n|1+α |n|≥1 2 0 =: AN (a0)∞ + D0(α)(M1+α,∞,∞(a0)+M1+α,1+α,∞(a0))

We have M∗(a0)=M∗(a − a1) ≤ M∗(a)+M∗(a1). Since a1(x, y, t) depends on x linearly, we get M1+α,∞,∞(a1)=M1+α,1+α,∞(a1)=0.Thus,

M1+α,∞,∞(a0)+M1+α,1+α,∞(a0) ≤ M1+α,∞,∞(a)+M1+α,1+α,∞(a) Finally, Theorem 4.3 shows that 0 0 0 AN (a0)∞ ≤ C(α)(M∞,∞(a0)+M1+α,∞(a0)). 0 1 As a0(t)= 0 a0(x, y, t)dx, it results that 0 M∞,∞(a0) ≤ M∞,∞,∞(a0)=M∞,∞,∞(a − a1) ≤ M∞,∞,∞(a)+M∞,∞,∞(a1) ≤ 4 M∞,∞,∞(a0) Toeplitz Matrices with Variable Coeﬃcients 201 and 1 | − | 0 ≤ ∂ya0(x, y2,t) ∂ya0(x, y1,t) M1+α,∞(a0) sup sup α dx t∈T y1,y2 0 |y2 − y1| ≤ M∞,1+α,∞(a0) ≤ M∞,1+α,∞(a)+M∞,1+α,∞(a1) = M∞,1+α,∞(a)+3M∞,1+α,∞(a).

Putting things together we arrive at the theorem with D(α)=7C(α)+D0(α).

Clearly, Theorem 4.5 implies the second half of Theorem 1.4 and thus also Theorem 1.2.

5. Discontinuous generating functions Sofarwehaveassumedthata ∈ C([0, 1]2 × T). Inequality (4) implies that sup AN (a)∞ <∞ if a is given by (1) anda ˆn are any bounded functions on 2 ∞ [0, 1] such that n=−∞ M∞,∞(ân) < ∞. Thus, sufficient smoothness in t allows us to admit arbitrary bounded coefficienta ˆn(x, y). For 0 <α≤ 1, we denote by H1+α,1+α the Banach space of all continuous functions f :[0, 1]2 → C which have continuous partial derivatives up to the order 2 and for which

f1+α := M∞,∞(f)+Mα,∞(∂xf)+M∞,α(∂yf)+Mα,α(∂x∂yf) < ∞, 2 where M∞,∞(f) is the maximum modulus of f(x, y)on[0, 1] and

|g(x2,y) − g(x1,y)| Mα,∞(g)= sup sup α , y∈[0,1] x1,x2 |x2 − x1|

|g(x, y2) − g(x, y1)| M∞,α(g)= sup sup α , x∈[0,1] y1,y2 |y2 − y1|

|∆2g(x1,x2,y1,y2)| Mα,α(g)= sup sup α α . x1,x2 y1,y2 |x2 − x1| |y2 − y1| ∞ Let L (T,H1+α,1+α) be the set of all measurable and essentially bounded functions a : T → H1+α,1+α. A check of the proofs shows that these work literally ∞ ∞ also for functions a in L (T,H1+α,1+α). Thus, if a is in L (T,H1+α,1+α)then 4 2 sup AN (a)∞ < ∞.SinceC ([0, 1] ) ⊂ H2,2, we obtain in particular Theorem 1.3.

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Albrecht Böttcher Fakultät für Mathematik TU Chemnitz D-09107 Chemnitz, Germany e-mail: [email protected] Sergei Grudsky Departamento de Matemáticas CINVESTAV del I.P.N. Apartado Postal 14-740 07000 México, D.F., México e-mail: [email protected]