Generalized Tractability for Multivariate Problems

Transcription

1 Generalize Tractability for Multivariate Problems Part II: Linear Tensor Prouct Problems, Linear Information, an Unrestricte Tractability Michael Gnewuch Department of Computer Science, University of Kiel, Christian-Albrechts-Platz 4, 4098 Kiel, Germany Henry Woźniaowsi Department of Computer Science, Columbia University New Yor, NY 1007, USA, an Institute of Applie Mathematics, University of Warsaw ul. Banacha, Warszawa, Polan, October 30, 007 Abstract We continue the stuy of generalize tractability initiate in our previous paper Generalize tractability for multivariate problems, Part I: Linear tensor prouct problems an linear information, J. Complexity, 3, 6-95 (007). We stuy linear tensor prouct problems for which we can compute linear information which is given by arbitrary continuous linear functionals. We want to approximate an operator S given as the -fol tensor prouct of a compact linear operator S 1 for = 1,,..., with S 1 = 1 an S 1 has at least two positive singular values. Let n(ε, S ) be the minimal number of information evaluations neee to approximate S to within ε [0, 1]. We stuy generalize tractability by verifying when n(ε, S ) can be boune by a multiple of a power of T (ε 1, ) for all (ε 1, ) Ω [1, ) N. Here, T is a tractability function which is non-ecreasing in both variables an grows slower than exponentially to infinity. We stuy the exponent of tractability which is the smallest power of T (ε 1, ) whose multiple bouns n(ε, S ). We also stuy wea tractability, i.e., when lim ε 1 +,(ε 1,) Ω ln n(ε, S )/(ε 1 + ) = 0. 1

2 In our previous paper, we stuie generalize tractability for proper subsets Ω of [1, ) N, whereas in this paper we tae the unrestricte omain Ω unr = [1, ) N. We consier the three cases for which we have only finitely many positive singular values of S 1, or they ecay exponentially or polynomially fast. Wea tractability hols for these three cases, an for all linear tensor prouct problems for which the singular values of S 1 ecay slightly faster that logarithmically. We provie necessary an sufficient conitions on the function T such that generalize tractability hols. These conitions are obtaine in terms of the singular values of S 1 an mostly limiting properties of T. The tractability conitions tell us how fast T must go to infinity. It is nown that T must go to infinity faster than polynomially. We show that generalize tractability is obtaine for T (x, y) = x 1+ln y. We also stuy tractability functions T of prouct form, T (x, y) = f 1 (x)f (x). Assume that a i = lim inf x (ln ln f i (x))/(ln ln x) is finite for i = 1,. Then generalize tractability taes place iff a i > 1 an (a 1 1)(a 1) 1, an if (a 1 1)(a 1) = 1 then we nee to assume one more conition given in the paper. If (a 1 1)(a 1) > 1 then the exponent of tractability is zero, an if (a 1 1)(a 1) = 1 then the exponent of tractability is finite. It is interesting to a that for T being of the prouct form, the tractability conitions as well as the exponent of tractability epen only on the secon singular eigenvalue of S 1 an they o not epen on the rate of their ecay. Finally, we compare the results obtaine in this paper for the unrestricte omain Ω unr with the results from our previous paper obtaine for the restricte omain Ω res = [1, ) {1,,..., } [1, ε 1 0 ) N with 1 an ε 0 (0, 1). In general, the tractability results are quite ifferent. We may have generalize tractability for the restricte omain an no generalize tractability for the unrestricte omain which is the case, for instance, for polynomial tractability T (x, y) = xy. We may also have generalize tractability for both omains with ifferent or with the same exponents of tractability. 1 Introuction Tractability of multivariate problems has been extensively stuie in information-base complexity an the recent account of the tractability research can be foun in the forthcoming boo [3]. Tractability is the stuy of approximating operators S efine on spaces of functions with variables with proportional to. Problems with huge occur in many applications, see [5]. We approximate S by computing linear information which is given by finitely many, say n, continuous linear functionals, an the error of an algorithm is efine in the worst case setting. Before tractability stuy, the errors of algorithms were stuie as functions of n an the main point was to fin the best possible rate of convergence as n tens to infinity. For large, the errors of algorithms crucially epen also on, an for some problems this epenence is exponential in. Let n(ε, S ) enote the information complexity of S which is the minimal number of continuous linear functionals neee to approximate S to within ε. The main point of

3 tractability is to chec whether n(ε, S ) oes not epen exponentially on ε 1 an. Since there are ifferent ways to measure the lac of exponential behavior, we have ifferent types of tractability. The first type of tractability is polynomial tractability which has been extensively stuie in many papers. In this case we want to verify whether n(ε, S ) can be boune by a multiple of powers of ε 1 an for all (ε 1, ) [1, ) N. There are many positive an negative results for polynomial tractability. Usually, positive results are for problems for which the successive variables or groups of variables of large carinality play a iminishing role, an negative results are for problems for which all variables an groups of variables play the same role. The primary example leaing to negative results is approximation of linear tensor prouct problems. In this case, S is a -fol tensor prouct of a compact linear operator S 1, where S 1 is efine between Hilbert spaces, S 1 = 1 an S 1 has at least two positive singular values. Let { λ j } enote the sequence of the orere singular values of S 1, 0 < λ λ 1 = 1. It is well nown, see [4], that the information complexity of S is n(ε, S ) = { (i 1, i,..., i ) N λ i1 λ i λ i > ε }. Clearly, if λ = 1 then n(ε, ) for all ε < 1, an we have exponential epenence on causing intractability of the problem. That is why we nee to assume that λ < 1. Still, as long as λ is positive, n(ε, ) goes faster to infinity that any power of, see [6], an that is why polynomial tractability oes not hol for linear tensor prouct problems. In [1], we propose to stuy generalize tractability by verifying whether n(ε, S ) can be boune by a multiple of a power of T (ε, ) for all (ε 1, ) Ω [1, ) N. Here T is a tractability function which means that T : [1, ) [1, ) is non-ecreasing in both variables an grows slower than exponentially to infinity, i.e., ln T (x, y) lim x+y x + y = 0. The set Ω is calle tractability omain, an can be a proper subset of [1, ) N but at least one of the parameters ε 1 or is allowe to go to infinity. The exponent of tractability is efine as the smallest (or more precisely as the infimum) power of T (ε 1, ) whose multiple bouns n(ε, S ). There is also the notion of wea tractability when ln n(ε, S ) lim ε 1 + ε 1 + = 0, see [, 3], an it is a necessary conition on the lac of exponential behavior of n(ε, S ). Of course, the hope is that by taing reasonable restricte omains Ω or by allowing tractability functions T that ten to infinity faster than polynomially, we may enlarge the class of tractable problems incluing linear tensor prouct problems. Inee, this is the case. In [1] we showe that polynomial tractability of linear tensor prouct problems hols if we assume that the singular values ten to zero polynomially fast, an we tae the restricte tractability omain Ω = Ω res := [1, ) {1,,..., } [1, ε 1 0 ) N 3

4 with 1 an ε 0 (0, 1). In this paper, we stuy the secon option an we tae the unrestricte omain Ω = Ω unr = [1, ) N, but we allow tractability functions T which go to infinity faster than polynomially. We stuy linear tensor prouct problems for three cases epening on the behavior of the singular values of S 1. In the first case we assume that only finitely many of the singular values are positive, in the secon case we assume that they ecay exponentially fast, an in the thir case that they ecay polynomially fast. For each of these three cases, we have wea tractability. In fact, wea tractability hols if the singular values behave as o((ln(j) ln(ln(j))) 1 ) an it is also almost a necessary conition. We provie necessary an sufficient conitions on T such that generalize tractability hols. These conitions are satisfie if T goes sufficiently fast to infinity. We also provie the formulas for the corresponing exponents of tractability. We illustrate these conitions an formulas for specific tractability functions. For example, tae T (x, y) = x 1+ln y. Then we have tractability for the three cases of singular values. For finitely many positive singular values an for exponentially ecaying singular values, the exponent of tractability is / ln(λ 1 ). Hence it only epens on the secon singular value an is inepenent of how many of them are positive. For polynomially ecaying singular values, λ j = Θ(j β ) for β > 0, the exponent of tractability is max{/β, / ln(λ 1 )}. We also illustrate our results for tractability functions of prouct form, that is when T (x, y) = f 1 (x)f (y) with finite a i = lim inf x (ln ln f i (x))/(ln ln x), i = 1,. Then generalize tractability hols iff a i > 1 an (a 1 1)(a 1) 1, an if (a 1 1)(a 1) = 1 then we nee to assume aitionally conition (1) for = which epens only on the secon singular value. For (a 1 1)(a 1) > 1, the exponent of tractability is zero, whereas for (a 1 1)(a 1) = 1, the exponent of tractability is positive. In fact, in the last case, epening on specific functions f i for which a i are fixe, the exponent of tractability can be arbitrary. Note that a i only epens on the limiting behavior of f i, an is inepenent on the behavior of the singular values. Hence, for (a 1 1)(a 1) > 1, we have the zero exponent of tractability inepenently of the behavior of the singular values, whereas for (a 1 1)(a 1) = 1, the exponent of tractability epens only on the secon singular value an is inepenent of the rest of them. In the final section, we compare the results obtaine in this paper for the unrestricte omain Ω unr with the results from our previous paper obtaine for the restricte omain Ω res. The tractability results for the unrestricte an restricte omains may be quite ifferent. We may have generalize tractability for the restricte omain an no generalize tractability for the unrestricte omain which is the case, as we alreay mentione, for polynomial tractability T (x, y) = xy. We may also have generalize tractability for both omains, however, the exponents of tractability may epen on the omain an can be much larger for the unrestricte omain than for the restricte omain. 4

5 Preliminaries.1 Multivariate Problems For m, N, let F be a norme linear space of functions f : D R m R an let G be a norme linear space. We consier in this paper sequences S = {S } of linear operators S : F G. We call S a multivariate problem. By linear information Λ all = F we mean the class of all continuous linear functionals efine on F. Let Λ F be a class of amissible continuous linear functionals. Without loss of generality, see e.g., [4], we consier linear algorithms that use finitely many amissible information evaluations. An algorithm A n, has the form A n, (f) = n g i L i (f) (1) for some L i Λ an some g i G. In this paper we restrict ourselves to the worst case setting. The worst case error of the algorithm A n, is efine as The initial error is e wor (A n, ) = where A 0, = 0 is the zero algorithm. Let i=1 sup S (f) A n, (f) G. () f F, f F 1 e init (S ) = S = e wor (A 0,), n(ε, S, Λ ) = min{ n A n, : e wor (A n, ) ε e init (S ) } (3) enote the minimal number of amissible information evaluations from Λ neee to reuce the initial error by a factor ε [0, 1]. The number n(ε, S, Λ ) is calle the information complexity of the problem S.. Generalize Tractability A tractability omain Ω is a subset of [1, ) N satisfying [1, ) {1,..., } [1, ε 1 0 ) N Ω (4) for some N {0} an some ε 0 (0, 1] such that + (1 ε 0 ) > 0. In this paper we focus on the unrestricte tractability omain Ω unr := [1, ) N. A function T : [1, ) [1, ) [1, ) is a tractability function if T is non-ecreasing in x an y an ln T (x, y) lim = 0. (5) (x,y) Ω, x+y x + y 5

6 Let now Ω be a tractability omain an T a tractability function. The multivariate problem S = {S } is (T, Ω)-tractable in the class Λ = {Λ } if there exist non-negative numbers C an t such that n(ε, S, Λ ) C T (ε 1, ) t for all (ε 1, ) Ω. (6) The exponent t tra of (T, Ω)-tractability in the class Λ is efine as the infimum of all non-negative t for which there exists a C = C(t) such that (6) hols. The multivariate problem S is strongly (T, Ω)-tractable in the class Λ = {Λ } if there exist non-negative numbers C an t such that n(ε, S, Λ ) C T (ε 1, 1) t for all (ε 1, ) Ω. (7) The exponent t str of strong (T, Ω)-tractability in the class Λ is the infimum of all nonnegative t for which there exists a C = C(t) such that (7) hols. An extensive motivation of the notion of generalize tractability an many examples of tractability omains an functions can be foun in [1]. We say that a multivariate problem S is wealy tractable if ln n(ε, S, Λ ) lim = 0. +ε 1 + ε 1 Obviously, if S is (T, Ω unr )-tractable then S is also wealy tractable. If S is wealy tractable an n(ε, S, Λ ) is at least one an non-ecreasing in ε 1 an, then S is also (T, Ω unr )-tractable for any non-ecreasing extension T : [1, ) [1, ) [1, ) of n(ε, S, Λ )..3 Linear Tensor Prouct Problems We escribe the setting we want to stuy in this paper in more etails. Let F 1 be a separable Hilbert space of real value functions efine on D 1 R m, an let G 1 be an arbitrary separable Hilbert space. Let S 1 : F 1 G 1 be a compact linear operator. Then the non-negative self-ajoint operator W 1 := S 1S 1 : F 1 F 1 is also compact. Let {λ i } enote the sequence of non-increasing eigenvalues of W 1, or equivalently let { λ i } be the sequence of the singular values of S 1. If = im(f 1 ) is finite, then W 1 has just finitely many eigenvalues λ 1, λ,..., λ. Then we formally put λ j = 0 for j >. In any case, the eigenvalues λ j converge to zero. Without loss of generality, we assume that S 1 is not the zero operator, an normalize the problem by assuming that λ 1 = 1. Hence, 1 = λ 1 λ 0. This implies that S 1 = 1 an the initial error is also one. 6

7 For, let F = F 1 F 1 be the complete -fol tensor prouct Hilbert space of F 1 of real value functions efine on D = D 1 D 1 R m. Similarly, let G = G 1 G 1, times. The linear operator S is efine as the tensor prouct operator S = S 1 S 1 : F G. We have S = S 1 = 1, so that the initial error is one for all. We call the multivariate problem S = {S } a linear tensor prouct problem. In this paper we analyze the problem S only for the class of linear information Λ all = {Λ all }. For convenience we write n(ε, S ) instea of n(ε, S, Λ all ). It is nown, see e.g., [4], that n(ε, S ) = {(i 1,..., i ) N λ i1... λ i > ε }, (8) with the convention that the carinality of the empty set is zero. Thus the linear tensor prouct problem S is trivial if λ = 0, since n(ε, S ) = 1 for all ε [0, 1). On the other han, n(ε, S ) grows exponentially in if λ = 1, since n(ε, S ) for all ε [0, 1). Therefore we assume λ (0, 1). We consier here the unrestricte case, i.e., Ω = Ω unr = [1, ) N. We now from [1, Lemma 3.1] that for this tractability omain the linear tensor prouct problem S is not strongly (T, Ω unr )-tractable, regarless of the tractability function T. For ε (0, 1] we efine α(ε) = ln(1/ε)/ ln(1/λ ) 1. (9) Notice that α(ε) is the largest integer n satisfying λ n > ε. We stress that α(ε) epens on λ. It tens to infinity as λ approaches 1, an is zero iff λ ε. From (8) it follows that n(ε, S ) = 1 if α(ε) = 0. For ε (0, 1) an λ (0, 1), let a := min{α(ε), }. Then it is easy to show, see also [1, Lemma 3.], that ( ) n(ε, S ) a 3 Finitely Many Eigenvalues ( ) n(ε, S 1 ) a. (10) a In this section we consier the case when W 1 = S1S 1 has only finitely many positive eigenvalues λ i. First we consier the case where W 1 has eigenvalues ifferent from zero an 1 of them are equal. We now prove an auxiliary lemma which will be helpful in the course of the proof of our first theorem. 7

8 Lemma 3.1. Let, N an let α be an integer satisfying 0 α 1 ( + 1). Then max 0 ν α ( ) ( 1) ν = ν Proof. For 0 ν α the inequality ( ) ( 1) ν 1 ν 1 ( ) ( 1) α. (11) α ( ) ( 1) ν ν hols iff ν ( ν + 1)( 1), an the last inequality hols iff ν 1 ( + 1). This shows that the function ( ) ν ( 1) ν ν is non-ecreasing on [0, α] N. Theorem 3.. Let T be a tractability function. Let λ 1 = 1, 0 < λ =... = λ < 1, an λ l = 0 for l >. Then the linear tensor prouct problem S = {S } is (T, Ω unr )-tractable in the class of linear information iff B := lim inf inf 1 α(ε) 1 ln T (ε 1, ) m (ε, ) (0, ], (1) where m (ε, ) := α(ε) ln( ( 1)) + ( α(ε)) ln( ). α(ε) α(ε) If B > 0, then the exponent t tra of tractability is given by t tra = B 1. (13) Proof. For the eigenvalues specifie in Theorem 3., it is easy to chec that (8) yiels n(ε, S ) = min{α(ε),} ν=0 ( ) ( 1) ν. (14) ν Let us first assume that S is (T, Ω unr )-tractable, i.e., that there exist C, t > 0 such that n(ε, S ) CT (ε 1, ) t. Let 1 α(ε) 1. From (11) an (14) we get the estimate ( ) ( ) ( 1) α(ε) n(ε, S ) (α(ε) + 1) ( 1) α(ε). (15) α(ε) α(ε) 8

9 Using Stirling s formula for factorials m! = m m+1/ e m π (1 + o(1)), we obtain (( ) ln )( 1) α(ε) α(ε) Thus = ln(!) ln(α(ε)!) ln(( α(ε))!) + α(ε) ln( 1) ( = + 1 ) ( ln() α(ε) + 1 ) ( ln(α(ε)) α(ε) + 1 ) ln( α(ε)) ln( π) + ln(o(1)) + α(ε) ln( 1) = m (ε, ) + 1 ( ) ln + O(1). α(ε)( α(ε)) ln T (ε 1, ) m (ε, ) ( 1 ln t + α(ε)( α(ε)) t m (ε, ) ) ln(c) t m (ε, ) + O(1) t m (ε, ). (16) Let {(ε 1 ν, ν )} be a sequence in Ω unr such that 1 α(ε ν ) ( 1)/, an lim ν (α(ε ν )/ ν ) exists (an obviously is at most ( 1)/) with lim ν ν =. If lim ν (α(ε ν )/ ν ) > 0 then m (ε ν, ν ) = Θ( ν ) an the right han sie of (16) tens to 1/t for ν. If lim ν (α(ε ν )/ ν ) = 0 then ( ( )) ν m (ε ν, ν ) = Θ α(ε ν ) ln ( 1), α(ε ν ) since Furthermore, ( ) ν ( ν α(ε ν )) ln ν α(ε ν ) = Θ(α(ε ν )). ( ) ln ν = Θ(ln(α(ε ν ))). α(ε ν )( ν α(ε ν )) Hence, again, the right han sie of (16) tens to 1/t. Since an arbitrary sequence {(ε 1 ν, ν )} with lim ν ν = an 1 α(ε ν ) 1 has a sub-sequence {(ε 1 µ, µ )} for which {α(ε µ )/ µ } converges, we conclue that B 1 t > 0, an ttra B 1. (17) Assume now B > 0. We want to show that for all t > B 1 there exists a C = C(t) > 0 such that n(ε, S ) CT (ε 1, ) t for all N, ε (0, 1). From (14) we see that this inequality is trivial if α(ε) = 0, an, since T (ε 1, ) is non-ecreasing in ε 1, that the case α(ε) > is settle if we have the inequality for α(ε) =. Thus it remains to consier the following two cases: 9

10 Case 1: 1 α(ε) 1. We now show that for all t > B 1 there exists a C = C(t) > 0 such that for all N, ( ) ln T (ε 1, ) 1 ln(1 + α(ε)) ln α(ε)( α(ε)) + + ln(c) m (ε, ) t t m (ε, ) t m (ε, ) t m (ε, ) + O(1) t m (ε, ). (18) ( ( Due to (15) an the formula for ln ) α(ε) ( 1) α(ε)), we conclue that n(ε, S ) CT (ε 1, ) t for all N, 1 α(ε) ( 1. ) To prove (18), observe that ln ln(). Obviously, α(ε)( α(ε)) m (ε, ) α(ε) ln( ( 1)). α(ε) For a given l N let x l R be so large that for all x x l we have ln(1 + x) x ln() 1 l. Let now l (1 + x l ) l+1. Then we get for l an 1 α(ε) 1, ln(1 + α(ε)) α(ε) ln( α(ε) ( 1)) 1 l. Furthermore, let { B l } be a sequence in (0, B ) that converges to B. For each l we fin a l such that for all 1 l an all 1 α(ε) Choose t l := (1 + 1 l ln T (ε 1, ) m (ε, ) 1 ) B l. For all max{ l, l ln T (ε 1, ) m (ε, ) 1 t l ( 1 + B l. } an all 1 α(ε) 1 ln(1 + α(ε)) m (ε, ) ). we have It is now easy to see that (18) hols for t = t l an all N an all 1 α(ε) 1 if we just choose C = C(t l ) suitably large. Observe that t l converges to B 1 as l tens to infinity. 1 Case : < α(ε). Let δ (0, B 1 ) an t (B δ) 1. There exists a δ such that for all δ an all ε with 1 α(ε ) 1, we have ln T (ε 1, ) m (ε, ) B δ. For δ an α(ε) 1, choose ε [ε, 1) such that α(ε ) = 1 =. Then ( m (ε, ) ln() ln 1 + ), 10

11 an t ln T (ε 1, ) (B δ) 1 ln T (ε 1, ) m (ε, ). We fin a number C not epening on such that ln(c) ln(1 + ). From (14) we now that n(ε, S ), an this yiels t ln T (ε 1, ) m (ε, ) ln n(ε, S ) ln(c), implying CT (ε 1, ) t n(ε, S ). Choosing C = C(t) sufficiently large the last inequality extens to all an all ε with 1 α(ε). The statement of the theorem follows from Cases 1 an. We illustrate Theorem 3. by two tractability functions. Let T (x, y) = xy which correspons to polynomial tractability. Then it is easy to chec that B = 0 for all. This means that we o not have polynomial tractability for any linear tensor prouct problem with at least two positive eigenvalues for = 1. This result has been nown before. Let T (x, y) = x 1+ln y. Then it can be chece that B = 1 ln(λ 1 ) for all, an t tra = ln(λ 1 ). Hence, the exponent of tractability only epens on the secon largest eigenvalue an is inepenent of its multiplicity. Note that the exponent of tractability goes to infinity as λ approaches one. We now consier the general case of finitely many positive eigenvalues. Corollary 3.3. Let T be a tractability function. Let an λ 1 = 1, λ (0, 1), an λ l = 0 for l >. Then the linear tensor prouct problem S = {S } is (T, Ω unr )-tractable in the class of linear information iff for some (an thus for all) j {, 3,..., } B j = lim inf inf 1 α(ε) j 1 j ln T (ε 1, ) m j (ε, ) (0, ], (19) where m j (ε, ) = α(ε) ln( of tractability satisfies α(ε) (j 1)) + ( α(ε)) ln( α(ε) ). In this case the exponent ttra B 1 t tra B 1. (0) Proof. Obviously we have B B 3 B. We nee to show that B > 0 implies that B > 0. We first show that B > 0 implies lim inf ln T inf (ε 1, ) 1 α(ε) 1 m (ε, ) > 0. (1) 11

12 Let ε satisfy α(ε ) =. We have m (ε, ) ln() ln(1 + ). Thus for 1 < α(ε) we get m (ε, ) ln() + ln() Cm (ε, ) for an C sufficiently large. Since T is non-ecreasing with respect to the first variable, it is easy to see that B > 0 implies (1). Now we prove that m (ε, ) m (ε, ) = 1 + α(ε) ln( 1) α(ε) ln( ) + ( α(ε)) ln( ) () α(ε) α(ε) is boune uniformly for all N an all ε with 1 α(ε) 1. This follows easily from ( ) ( ) m (ε, ) α(ε) ln α(ε) ln. α(ε) 1 Thus B > 0 implies ( B lim inf ) ( ln T inf (ε 1, ) 1 α(ε) 1 m (ε, ) inf N ; 1 α(ε) 1 m (ε, ) m (ε, ) Since the linear tensor prouct problem S having only the two non-zero eigenvalues λ 1 = λ 1 an λ = λ is at most as ifficult as S an the problem S having eigenvalues λ 1 = λ 1, λ =... = λ = λ an λ l = 0 for l > is at least as as ifficult as S, the corollary follows from Theorem 3.. Remar 3.4. Theorem 3. shows that in the case λ 3 =... = λ = 0 we have t tra = B 1, while in the case λ = λ 3 =... = λ we have t tra = B 1. If we consier a fixe tractability function T, a sequence {S (n) } of tensor prouct problems whose eigenvalues {λ (n) i } satisfy λ (n) 1 = λ 1 = 1, λ (n) = λ (0, 1), λ (n) 3,..., λ (n) > 0, an lim n λ (n) 3 = 0, then we o not necessarily have that the corresponing exponents of tractability t tra n converge to B 1 as the following counterexample shows. Let T (ε 1, ) = min{α(ε),} ν=0 ( ). ν Then it is not har to see that T is inee a tractability function an that B = 1 (we showe that implicitly in the proof of Theorem 3.). Accoring to Corollary 3.3 each problem S (n) is (T, Ω unr )-tractable. For N we obviously have sup ε (0,1) T (ε 1, ) =. If we choose ε = ε (n) = 1 (λ(n) )/, we get n(ε, S (n) for all n. This shows that the sequence {t tra n ) > 0. ) =. This implies t tra n ln()/ ln() } oes not converge to B 1 = 1. 1

13 Example 3.5. Let the conitions of Corollary 3.3 hol. We consier the special tractability function T (x, y) = exp(f 1 (x)f (y)), where f i : [1, ) (0, ), i = 1,, are nonecreasing functions. Let a i := lim inf x f i (x) ln x for i = 1,. Let us assume that S is (T, Ω unr )-tractable. Accoring to Corollary 3.3 we have B > 0, an from m (ε, ) α(ε)(ln() ln(α(ε))) for all ε satisfying 1 α(ε) / we get Thus a > 0, an f 1 (ε 1 )f () 0 < B lim inf α(ε)(ln() ln(α(ε))) f ( ) ( ) 1(ε 1 ) f () ln() lim inf lim sup α(ε) ln() ln() ln(α(ε)) = f 1(ε 1 ) α(ε) a. 0 < B a lim inf ε 0 f 1 (ε 1 ) α(ε) = lim inf ε 0 ( ln(ε 1 ) α(ε) f 1 (ε 1 ) ln(ε 1 ) ) = ln(λ 1 ) a 1. Hence, a 1 > 0 an a > 0 are necessary conitions for the problem S to be (T, Ω unr )- tractable, an the exponent of tractability is boune from below by t tra B 1 a 1 a ln(λ 1 ). In Corollary 5. we will show in particular that the conitions a 1 > 0, a > 0 are also sufficient for (T, Ω unr )-tractability. Remar 3.6. Uner the conitions of Corollary 3.3 we can state a slightly simpler criterion to characterize (T, Ω unr )-tractability. The linear tensor prouct problem S = {S } is (T, Ω unr )-tractable in the class of linear information iff B := lim inf inf 1 α(ε) / ln T (ε 1, ) α(ε) ln(/α(ε)) (0, ]. (3) The necessity an sufficiency of B > 0 follows from (19) an the (easy to chec) inequalities ( ) 1 m (ε, ) α(ε) ln m (ε, ) α(ε) for all ε satisfying 1 α(ε) an large. A rawbac of (3) is that the quantity B is not relate to the exact exponent of tractability as B in Theorem 3.. Example 3.7. The tractability criteria (19) an (3) epen on the secon largest eigenvalue λ via α(ε). In fact, for a given tractability function T, a linear tensor prouct problem S = {S } with only two positive eigenvalues for S 1S 1 may be (T, Ω unr )-tractable, 13

14 but if we increase the value of λ this may not necessarily be the case any more. Choose, e.g., { 1 if x [1, λ 1/ ], T (x, y) := e ln(x)(1+ln(y)) otherwise. From criterion (3) it follows easily that S is (T, Ω unr )-tractable. But if we consier the problem S where we only increase the secon eigenvalue to λ > λ, we see that for λ 1/ < ε 1 λ 1/ we have n(ε, S ) min{ α(ε),} ν=0 ( ) ν ( ) =, where α(ε) := 1 ln(ε 1 ) 1 1. ln( λ 1 ) Thus the problem S is obviously not (T, Ω unr )-tractable since CT (ε 1, ) t = C cannot be larger than for > C. The counterexample above motivates us to state a sufficient conition on T ensuring (T, Ω unr )-tractability of all linear tensor prouct problems S with finitely many eigenvalues regarless of the specific value of λ. Corollary 3.8. Let T be a tractability function. If B := lim inf ln T (ε 1, ) inf 1<ε 1 e ln(ε 1 ) (1 + ln (/ ln(ε 1 ))) (0, ] (4) then arbitrary linear tensor prouct problem S with finitely many eigenvalues is (T, Ω unr )- tractable. However, the exponent of tractability goes to infinity as λ approaches one. Proof. The proof of the corollary is easy. For values of ε [e, 1) satisfying α(ε) [1, /] one can simply show that α(ε) ln(/α(ε)) C ln(ε 1 )(1 + ln(/ ln(ε 1 ))), where the constant C epens only on λ. If we substitute the upper boun on α(ε) in the efinition of B in (3) by the minimum of / an / ln(λ 1 ) 1 we therefore see that this moifie quantity is strictly positive. From that we can euce similarly as in Case in the proof of Theorem 3. that B > 0, an ue to Remar 3.6, the problem S is (T, Ω unr )-tractable. Obviously, n(ε, S ) for ε < λ. Hence, the exponent of tractability must go to infinity as λ goes to one. Remar 3.9. Conition (4) in the corollary above is sufficient for (T, Ω unr )-tractability for all linear tensor prouct problems S with finitely many eigenvalues, but not necessary as the example T (ε 1, ) = exp(ln(ε 1 )(1 + ln())) shows, see Corollary Exponential Decay of Eigenvalues We begin to stuy linear tensor problems with infinitely many positive eigenvalues. As we shall see, tractability results epen on the behavior of the eigenvalues for = 1. In this section we assume that they are exponentially ecaying whereas in the next section that they are polynomially ecaying. 14

15 Theorem 4.1. Let T be a tractability function. Let S be a linear tensor prouct problem with exponentially ecaying eigenvalues λ j, exp( β 1 (j 1)) λ j exp( β (j 1)) for all j N, for some positive numbers β 1, β. For i = 1,, efine B (i) e ln T := lim inf (ε 1, ) ε 1 + ε<σ i m (i) e (ε, ), where σ 1 = e β1/, σ = λ, an ( m (i) e (ε, ) := z i ln 1 + ) ( + ln 1 + z ) i, z i with Then z i = z i (ε) := β i ln(ε 1 ) 1. S is (T, Ω unr )-tractable iff B () e (0, ]. Furthermore, B e () > 0 is equivalent to B e (1) (0, ] an B (0, ] with B given by (1) for =. If S is (T, Ω unr )-tractable then the exponent t tra of tractability satisfies ( min{b, B e (1) } ) 1 t tra ( ) B e () 1. If β 1 = β then t tra = (B () e ) 1. Before we prove Theorem 4.1, we state an auxiliary lemma. Lemma 4.. For N an x > 1 let { µ e (x, ) := (i 1,..., i ) N i j < x + + 1}. j=1 Then Proof. For = 1 we have ( ) x + µ e (x, ) =. µ e (x, 1) = {i N i < x + } = x + 1. Assume by inuction that ( ) y + µ e (y, ) = 15

16 for some N an all y > 1. If x > 1 then µ e (x, + 1) = x +1 =1 µ e (x + 1, ) = x +1 =1 x ( ) ( ) ν + x = =. + 1 ν=0 ( ) x Proof of Theorem 4.1. Let µ e (x, ) be efine as in Lemma 4.. Then { } µ e (z 1, ) = (i 1,..., i ) N exp( β 1 (i j 1)) > ε n(ε, S ). j=1 Similarly, we get n(ε, S ) µ e (z, ). Let us first assume that S is (T, Ω unr )-tractable, i.e., that there exist positive t, C such that n(ε, S ) CT (ε 1, ) t for all (ε 1, ) Ω unr. Let us assume that ε < e β1/, which implies that z 1 1. From this inequality we get ue to Lemma 4. ln T (ε 1, ) ln(c 1 ) + ln ( z 1 ) +. m (1) e (ε, ) t m (1) e (ε, ) Similarly as in the proof of Theorem 3. we use Stirling s formula for factorials, an conclue ( ) z1 + ln = m (1) e (ε, ) + 1 ( ) ln z1 + + O(1). (5) z 1 We have min{ln(), ln z 1 } ln ( ) ( z1 + 1 = ln z 1 z ) ln(). So it is easy to chec that we get B e (1) 1/t, implying B e (1) > 0 an t tra (B e (1) ) 1. Furthermore, we get from Corollary 3.3 that B > 0 an t tra B 1. Let us now show that B > 0 an B e (1) > 0 imply B e () > 0. As a careful analysis reveals, we get m (1) e (ε, ) K := lim inf inf ε 1 + ε<e β 1 / m () e (ε, ) > 0, which gives us lim inf ε 1 + ε<e β 1 / ln T (ε 1, ) m () e (ε, ) B (1) e K > 0. 16

17 In the case e β1/ ε < λ both functions α(ε) an z (ε) are boune. Thus we have m (ε, ) = Θ(ln()) = m () e (ε, ), where m is given in Theorem 3.. Hence which yiels L := lim inf lim inf ε 1 + e β 1 / ε λ m (ε, ) inf e β 1 / ε< λ m () e (ε, ) > 0, ln T (ε 1, ) m () e (ε, ) B L > 0. This means that B e () is positive, as claime. Now let us assume that B e () > 0 an let t δ := ((1 δ)b e () ) 1 for a given δ (0, 1). Then there exists an R(δ) such that for any pair (ε, ) with ε 1 + > R(δ) (an ε < λ, but for convenience we will not mention this restriction in the rest of the proof) we get ( ln T (ε 1, ) > 1 δ ) B e (). m () e (ε, ) We want to show that there exists a number C δ such that n(ε, S ) C δ T (ε 1, ) t δ for all (ε 1, ) Ω unr. Since n(ε, S ) µ e (z, ), it is sufficient to verify the inequality ln T (ε 1, ) ln(c 1 δ ) + ln ( z ) +. (6) m () e (ε, ) t δ m () e (ε, ) The left han sie is at least (1 δ/)b e (). Using Stirling s formula (5) for z instea of z 1, we see that the right han sie can be written as (1 δ)b () e + ln( z + z ) 1 δ ) t δ m () e (ε, ) + ln(c t δ m () e (ε, ) + O(1) t δ m () e (ε, ). The limes superior of all the summans, except of (1 δ)b e (), goes to zero as ε 1 + tens to infinity. Hence, there exists an R(δ) such that for all pairs (ε, ) with ε 1 + > R(δ) inequality (6) hols. Choosing C δ sufficiently large, we see therefore that (6) hols for all (ε 1, ) Ω unr. This shows that we have (T, Ω unr )-tractability an, since δ (0, 1) was arbitrary, the exponent of tractability t tra satisfies t tra (B e () ) 1. As we alreay have seen, tractability implies B e (1) > 0 an B > 0. Finally, if β 1 = β then B e (1) = B e (), an therefore (min{b, B e (1) }) 1 t tra (B e () ) 1 implies that B B e (1) an t tra = (B e () ) 1. We illustrate Theorem 4.1 by taing again the tractability function T (x, y) = x 1+ln y. For β 1 = β = β > 0, we have λ = exp( β). It can be chece that B () e ) = β = ln(λ 1. 17

18 Thus the exponent of tractability is t tra = (B () e ) 1 = β = ln(λ 1 ). We can simplify the necessary an sufficient conitions in Theorem 4.1 for (T, Ω unr )- tractability at the expense of getting goo estimates on the exponent of tractability. Corollary 4.3. Let T be a tractability function. Let S be a linear tensor prouct problem with 0 < λ < λ 1 = 1, an with exponentially ecaying eigenvalues λ j, K 1 exp( β 1 j) λ j K exp( β j) for all j N, for some positive numbers β 1, β, K 1 an K. Then S is (T, Ω unr )-tractable iff lim inf ε 1 + ε< λ ln T (ε 1, ) min{, α(ε)} (1 + ln(/α(ε)) ) (0, ]. (7) Proof. Since λ j min{λ, K exp( β j)} for j, we can choose positive β 1 β 1, β β such that exp( β 1(j 1)) λ j exp( β (j 1)) for all j N. Thus we can apply Theorem 4.1. There we showe that B e () > 0 is necessary an sufficient for (T, Ω unr )-tractability. For 1 α(ε) / an large, we have m (ε, )/ α(ε) ln(/α(ε)) m (ε, ). Furthermore, one can also verify that lim inf inf / α(ε) where q { 1, +1}. Thus (7) hols iff B () e ( ) m () q e (ε, ) > 0, min{, α(ε)}(1 + ln(/α(ε)) ) 5 Polynomial Decay of Eigenvalues (0, ], which proves the corollary In this section we stuy tractability for linear tensor prouct problems with polynomially ecaying eigenvalues for = 1. We believe that such behavior of eigenvalues is typical an therefore the results of this section are probably more important than the results of the previous sections. Theorem 5.1. Let T be a tractability function. Let S be a linear tensor prouct problem with 1 = λ 1 > λ > 0 an λ j = O(j β ) for all j N an some positive β. A sufficient conition for (T, Ω unr )-tractability of S is F := lim inf ε 1 + ε< λ ln T (ε 1, ) ln(ε 1 )(1 + ln()) 18 (0, ].

19 If F (0, ], then the exponent of tractability satisfies { } B 1 t tra max β, ln(λ 1 F 1, ) with B given in (1) for =. Proof. Let C 1 be a positive constant satisfying λ j C 1 j β for all j. With C := C 1/β 1 we have n(ε, S 1 ) = max{j λ j > ε } max{j C 1 j β > ε } C ε /β C ε p for all p > /β. From the ientity n(ε, S ) = it now follows by simple inuction that n(ε, S ) C ( j=1 i=1 λ p/ j n (ε/ ) λ i, S 1 ) 1 ε p for all p > /β. (8) Thus for each 0 N an all p > /β there exists a number C( 0, p) such that n(ε, S ) C( 0, p) ε p for all 0 an ε (0, λ ). Let now δ (0, 1) an ε δ < λ such that for all ε (0, ε δ ) an all 0 ln T (ε 1, ) ln(ε 1 )(1 + ln()) (1 δ) F, where F is assume to be positive. Then for t = t(δ, p, 0 ) := p(1 δ) 1 F 1 an C = C( 0, p) we have ln(ct (ε 1, ) t ) ln C + p(1 + ln()) ln(ε 1 ) ln n(ε, S ) for all 0 an ε (0, ε δ ). This implies that for each t > (/β)f 1 there exists a sufficiently large number C = C t such that n(ε, S ) C T (ε 1, ) t for all 0 an ε (0, λ ). (9) We now consier arbitrarily large. Let us estimate the sum on the right han sie of inequality (8). For this purpose we choose N such that λ > C 1 β. Since λ C 1 β, we have obviously >. We have j=1 λ p/ j 1 + λ p/ + + λ p/ + C p/ 1 19 j=+1 j pβ,

20 an j=+1 j pβ x pβ pβ +1 x = (pβ/) 1. Now we choose p = p() such that ( λ p/ + (C ) 1 β ) p/ = 1 (pβ/) 1. From λ p/ 1/ we conclue From λ > C 1 β we get implying ( ) ln + ln p ln(λ 1. ) ( ) λ p/ 1 (pβ/) 1, ( p ln + ln + ln ln(λ 1 ) (pβ/) 1 ) Thus we have p = ln() ln(λ 1 ) (1 + o (1)) as. Let now σ (0, 1) an σ N such that o (1) σ an ln T (ε 1, ) ln(ε 1 )(1 + ln()) (1 + σ) 1 F for all σ an all ε (0, λ ). For these an ε we have ( n(ε, S ) C 1 + ) 1 1 ( ) ln() ε p e C exp ln(λ 1 ) (1 + σ) ln(ε 1 ) ( ) C 3 exp ln(λ 1 ) F 1 (1 + σ) ln T (ε 1, ), where C 3 := e C. Hence for τ = τ(σ, p, δ ) := (ln(λ 1 )) 1 (1 + σ) F 1 we get n(ε, S ) C 3 T (ε 1, ) τ for all σ an ε (0, λ ). (30) The estimates (9) an (30) show that we have (T, Ω unr )-tractability. Choosing 0 = σ in (9) an letting σ ten to zero yiels the claime upper boun for t tra. Since our problem is at least as har as the problem with only two positive eigenvalues 0 < λ < λ 1 = 1 for = 1, the lower boun t tra B 1 follows from Theorem 3. for =. 0.

21 The upper boun on the exponent t tra in Theorem 5.1 is, in general, sharp. Inee, assume that λ j = Θ(j β ) an tae T (x, y) = x 1+ln y. Then n(ε, S 1 ) = Θ(ε /β ) which easily implies that t tra /β. In this case, we have F = 1 an B = 1 ln(λ 1 ). This shows that the upper boun on t tra in Theorem 5.1 is sharp an { } t tra = max β, ln(λ 1. ) Hence, for β ln(λ 1 ) the exponent of tractability is the same as for the problem with only two positive eigenvalues 0 < λ < λ 1 = 1. For this tractability function, the problem S with polynomially ecaying eigenvalues is as har as the problem with only two positive eigenvalues. However, for β < ln λ 1, the exponent of tractability epens on β an the problem S is harer than the problem with only two positive eigenvalues. Corollary 5.. Let 1 = λ 1 > λ > 0 an λ j = O(j β ) for all j N an some fixe β > 0. Let f i : [1, ) (0, ), i = 1,, be non-ecreasing functions such that f 1 (x)f (y) lim x+y x + y = 0. For T (x, y) = exp(f 1 (x)f (y)), we have (T, Ω unr )-tractability iff a i := lim inf x f i (x) ln x (0, ] for i = 1,. If a 1, a (0, ], then the exponent of tractability satisfies { } a 1 a ln(λ 1 ) ttra max β, 1 ln(λ 1 ) min{a 1 b, b 1 a }, where b 1 = f 1 (ε 1 ) inf ε< λ ln(ε 1 ) an f () b = inf N 1 + ln(). Proof. We have alreay seen in Example 3.5 that even for two non-zero eigenvalues λ 1, λ an 0 = λ 3 = λ 4 =... the conition a 1, a > 0 is necessary for S to be (T, Ω unr )-tractable, an that t tra /(a 1 a ln(λ 1 )). Let us now assume that a 1, a (0, ]. It is easy to see that F = lim inf ε 1 + ε< λ ln T (ε 1, ) ln(ε 1 )(1 + ln()) f 1 = lim inf (ε 1 )f () ε 1 + ln(ε 1 )(1 + ln()) = min{a 1b, b 1 a }, ε< λ an that a 1, a > 0 implies b 1, b > 0. Thus F > 0 an ue to Theorem 5.1 we have (T, Ω unr )-tractability an the state upper boun for t tra. 1

22 We illustrate Corollary 5. again for T (x, y) = x 1+ln y = exp ((ln x)(1 + ln y)). We now have a 1 = a = b 1 = b = 1. If we assume that λ j = Θ(j β ) then, as we have alreay chece, t tra = max{/β, / ln λ 1 }. Hence, the upper boun on t tra in Corollary 5. is, in general, sharp. This proves that for tractability functions T of the form T (x, y) = exp(f 1 (x)f (x)), the exponent of tractability may epen on β, i.e., on how fast the eigenvalues ecay to zero for = 1. We now consier ifferent tractability functions of the form T (x, y) = f 1 (x)f (x) = exp(ln f 1 (x) + ln f (x)) an show that for such functions the exponent of tractability oes not epen on β. The following theorem generalizes a result from [7] which correspons to f i (x) = exp(ln 1+α i (1 + x)). Theorem 5.3. Let S be a linear tensor prouct problem with 1 = λ 1 > λ > 0 an λ j = O(j β ) for all j N. For i = 1, let f i : [1, ) [1, ) be a non-ecreasing function with ln ln f i (x) a i := lim inf <. x ln ln x Then the function T efine by T (x, y) = f 1 (x)f (y) is a tractability function. S is (T, Ω unr )-tractable iff a 1 > 1, a > 1, (a 1 1)(a 1) 1, an B (0, ], where B is given by (1) for =. If a 1 > 1, a > 1 an (a 1 1)(a 1) > 1 then B = an the exponent of tractability t tra is zero. If a 1 > 1, a > 1, (a 1 1)(a 1) = 1 an B > 0 then the exponent of tractability is t tra = B 1 = ln f 1 lim inf (ε 1 ) + ln f () ε 1 + α(ε) ln() ε< λ Proof. Since a 1, a <, it is obvious that T is a tractability function. Let first S be (T, Ω)-tractable, i.e., there exist positive constants C, t such that Due to (10) we have which implies n(ε, S ) C f 1 (ε 1 ) t f () t for all (ε 1, ) Ω unr. n(ε, S ) ( ) ( ) α(ε), α(ε) α(ε) ( ) α(ε) ln ln(c) + t ln f 1 (ε 1 ) + t ln f (). (31) α(ε) Keeping ε fixe an letting grow, we see that for any δ > 0 there exists a = (δ, ε) such that for all we have α(ε) ln() (t + δ) ln f (), an therefore 1 + ln α(ε) ln ln() ln ln f () ln(t + δ) + ln ln() ln ln(). 1.

23 Thus a 1. Let now ε vary an tae = α(ε). Since ln f () = o() = o(α(ε)), we get from (31) for arbitrary δ > 0, for ε = ε (δ) sufficiently small, an for all ε ε that α(ε) ln() (t + δ) ln f 1 (ε 1 ). Since ln α(ε) = ln() + ln ln(ε 1 ) ln ln(λ 1 ) + O(1) as ε tens to zero, the estimate a 1 1 easily follows. Let now η > a 1 1. Define Then (31) yiels = (ε) = α(ε) α(ε)η. (α(ε) η+1 α(ε)) ln(α(ε)) ln(c) + t ln f 1 (ε 1 ) + t ln f (). Due to the choice of η an the fact that α(ε) = ln(ε 1 )/ ln(λ 1 ) + O(1), the function ln f 1 (ε 1 ) is of orer o(α(ε) η+1 ). We thus have for arbitrary δ, for ε(δ) sufficiently small, an for all ε ε(δ), α(ε) η+1 ln(α(ε)) (t + δ) ln f (), leaing to This implies η + 1 η ( lim inf ln ln(α(ε)) ln(α(ε)) ) ( ln ln f () ln ln() ln(t + δ) ln(α(ε)) + ln ln f () ln ln() ln ln() ln(α(ε)). lim ε 1 ) η ln(α(ε)) + ln ln(α(ε)) = a η. ln(α(ε)) Thus η(a 1) 1. Letting η ten to a 1 1 we get (a 1 1)(a 1) 1. This proves that a 1 > 1 an a > 1. Furthermore, ue to Theorem 3., B has to be positive or infinite for any tractable problems with two positive eigenvalues 0 < λ < λ 1 = 1. Assume now that a 1 > 1, a > 1, (a 1 1)(a 1) 1, an B > 0. Due to Theorem 5.1, to prove (T, Ω unr )-tractability it is enough to verify that ln f 1 F = lim inf (ε 1 ) + ln f () ε 1 + ln(ε 1 )(1 + ln()) ε< λ (0, ]. Assume we have an arbitrary sequence {(ε 1 m, m )} such that {ε 1 m + m } tens to infinity, ε m < λ, an the sequence {F m }, where F m := ln f 1(ε 1 m ) + ln f ( m ) ln(ε 1 m )(1 + ln( m )), converges to F. Then we fin a sub-sequence {(ε 1 n, n )} for which {ln ln( n )/ ln ln(ε 1 n )} converges to an element x [0, ]. For this sub-sequence we show that F > B/ ln(λ 1 If the sequence {ε 1 n ). } or { n } is boune, then {F n } tens to infinity, since a 1 an a are } as well as { n } ten to infinity. both strictly larger than 1. So we can assume that {ε 1 n 3

24 First, let us assume that x [0, (a 1 1)). Then ln( n ) ln(ε 1 n ) a1 1 δ for δ sufficiently small an sufficiently large n n(δ). Thus F lim inf n ln f 1 (ε 1 n ) ln(ε 1 n ) =. a 1 δ If x ((a 1) 1, ], we just change the roles of the parameters ε 1 an to get F lim inf n ln f ( n ) ln( n ) a δ =. If (a 1 1)(a 1) > 1, then we have consiere all possible values of x in [0, ] since then [0, (a 1 1)) ((a 1) 1, ] = [0, ], an we have shown that F =. Theorem 5.1 implies then that the exponent of tractability is t tra = 0 an B =. If (a 1 1)(a 1) = 1, we still have to consier the case x = a 1 1. Then ln(α(ε n )) = ln ln(ε 1 n ) + ln() ln ln(λ 1 ) + O(1) [(a 1) δ, (a 1) + δ] ln ln( n ) for arbitrary δ an sufficiently large n n(δ). Then α(ε n ) (ln n ) a 1+δ = o( n ). Hence we have F = lim inf n ln f 1 (ε 1 n ) + ln f ( n ) α(ε n )(1 + ln( n /α(ε n ))) α(ε n )(1 + ln( n /α(ε n ))) ln(ε 1 n )(1 + ln( n )) = B ln(λ 1 ) > 0. To obtain the formula for the exponent t tra we can use the boun on t tra from Theorem 5.1. For β ln λ 1 we get t tra = B 1. To obtain the same result for β < ln λ 1 we procee as follows. In the proof of Theorem 5.1 we showe that for small positive δ there is a positive number C β,δ epening only on β an δ such that ( { } ) + δ (1 + δ) ln() n(ε, S ) C β,δ exp max, β ln(λ 1 ln(ε 1 ) for all (ε 1, ) Ω unr. To show that the last right sie function is at most C (f 1 (ε 1 )f ()) t it is enough to chec that (1 + δ) ln(λ 1 ) ln(ε 1 ) ln() t ( ln(f 1 (ε 1 )) + ln(f ()) ) for large ε 1 an. Or equivalently that t (1 + δ) ln(f 1 lim inf (ε 1 )) + ln(f ()) ε 1 + α(ε) ln() ε< λ The last limit inferior is achieve if α(ε) is a power of ln(), an therefore it is the same as B. Since δ can be arbitrarily small we conclue that t tra B 1. The lower boun on t tra from Theorem 5.1 then implies t tra = B 1, as claime. This completes the proof of Theorem ) 1.

25 Remar 5.4. Let the conitions of Theorem 5.3 hol an assume that a 1 > 1, a > 1 an (a 1 1)(a 1) = 1. Then conition B (0, ] oes not necessarily hol as the following example shows. Let δ : [1, ) [0, ) be a ecreasing function with lim x δ(x) = 0. Define f i (x) = exp ( ln(x) δ(x)) for i = 1,. Then we have obviously a 1 = = a an (a 1 1)(a 1) = 1. But ( ) ln λ 1 1 ln(ε 1 ) δ(ε 1) + ln() δ() B lim inf ε 1 + ln(ε 1 ) ln() ε 1 = = lim inf exp ( δ() ln ln()). If we choose, e.g., δ(x) = (ln ln ln(x)) 1, then we see that B = 0. = lim inf ln() δ() We stress again that the exponent of tractability in Theorem 5.3 oes not epen on β an it is B 1 for all polynomial ecaying eigenvalues with the same two largest eigenvalues 0 < λ < λ 1 = 1. However, B epens on particular functions f i satisfying the conitions of Theorem 5.3. We now show that B can tae any positive value or even be infinite. Inee, tae f i (x) = exp ( c i [ln x] ) (1+α i) for positive c i an α i. Then a i = 1 + α i. For α 1 α = 1 it can be chece that ( ) 1/(1+α1 c1 α ) 1 ln(λ 1 ) B = c (1 + α ). (3) c Taing, c = c 1 = c an varying c for fixe α i, we see that B can be any positive number with the same limits a i. On the other han, for f i (x) = exp (ln(e + ln x) [ln x] 1+α i ), an α 1 α = 1 we get a i = 1 + α i as before, but B =. We also stress that in Theorem 5.3 we assume that the eigenvalues ecay at least polynomially. This assumption hols, in particular, for finitely many positive or exponentially ecaying eigenvalues. We summarize this iscussion in the following remar. Remar 5.5. As long as a tractability function T is of prouct form, T (x, y) = f 1 (x)f (x), then (T, Ω unr )-tractability of S as well as the exponent of tractability epen only on the functions f 1, f an the secon eigenvalue λ as long as the eigenvalues λ j ecay at least polynomially. Hence, if we have two problems, one with only two positive eigenvalues 0 < λ < λ 1 = 1, an the secon with the same two eigenvalues an the rest of them are non-negative an ecaying polynomially, then these two problems lea to the same tractability conitions an to the same exponents of tractability. We stress that this property oes not hol for more general tractability functions. For instance, if we consier T (x, y) = exp(g 1 (x)g (y)), i.e, when ln T is of prouct form, then the exponent of tractability may epen on the rate of ecay of eigenvalues. This hols, for instance, for T (x, y) = exp(ln(x) (1 + ln())) as shown after Corollary 5.. 5

26 6 Wea Tractability So far we iscusse (T, Ω unr )-tractability of linear tensor prouct problems with exponentially an polynomially ecaying eigenvalues. We now verify what we have to assume about the ecay of eigenvalues to obtain wea tractability. As we shall see, in particular, exponential or polynomial ecay of eigenvalues implies wea tractability. Let us consier a logarithmic ecay of the eigenvalues, i.e., λ j = Θ((1 + ln j) β ) for all j an some fixe β > 0. In [1] we prove that ln n(ε, S 1 ) = Θ(ε /β (1 + o(1))). Thus for β not even the one-imensional problem S 1 is tractable. For β >, we characterize (T, Ω res )-tractability for Ω res = [1, ) [ ] [1, ε 1 0 ) N, with + (1 ε 0 ) > 0, see [1]. Here we consier the unrestricte tractability omain an prove, in particular, wea tractability for β >. Theorem 6.1. Let λ 1 = 1, λ (0, 1) an λ j = o ( ( ln j ) ( ln(ln j) ) ) as j. (33) Then the linear tensor prouct problem S is wealy tractable. If S is wealy tractable then λ < 1 an λ j = o ( ( ln j ) ) as j. Proof. To prove the first point, we may assume without loss of generality that λ j > 0 for all j N. Then there exists a function f : N (0, ) with lim j f(j) = 0 an λ j = f(j)(1 + ln j) (1 + ln(1 + ln j)) for all j N. We now show that ln n(ε, 1) = o(ε 1 (ln(ε 1 )) 1 ). Accoring to (8) we have n(ε, 1) = max{j g(j)(1 + ln j)(1 + ln(1 + ln j)) < ε 1 }, (34) where g(j) := f(j) 1/. Now let j = j(ε) = exp( cε 1 (ln(ε 1 )) 1 1) for some c > 0. Then εg(j)(1 + ln j)(1 + ln(1 + ln j)) g(j(ε))(c + (1 + ln c)c(ln(ε 1 )) 1 c ln ln(ε 1 )(ln(ε 1 )) 1 ), which tens to infinity as ε approaches zero. From this calculation an from (34) we conclue that ln n(ε, 1) = o(ε 1 (ln(ε 1 )) 1 ). With a := min{α(ε), } we get from (10) ln n(ε, ) + ε 1 ln ( ) a + a ln n(ε, 1) + ε 1 a(ln(/a) ln n(ε, 1)) + ε 1. 6

27 Case 1 : α(ε). Then ln n(ε, ) + ε 1 α(ε)(ln(/α(ε)) + 1) + ε 1 + o(α(ε)ε 1 (ln(ε 1 )) 1 ) + ε 1. (35) Since α(ε) ln(ε 1 ), the secon term on the right han sie of (35) goes to zero as + ε 1 tens to infinity. If α(ε) = Θ(), the first term goes obviously also to zero if + ε 1 tens to infinity. So let us consier the case α(ε) = o(). If α(ε) = Ω(/(ln )), then ε 1 = exp(ω(/(ln ))), an α(ε)(ln )ε 0 as + ε 1. If α(ε) = o(/(ln )), then α(ε)(ln )/ 0 as + ε 1. Case : α(ε) >. Then ln n(ε, ) + ε 1 + ε + (o(ε 1 (ln(ε 1 ) 1 ))) = o(ε 1 ) ε 1 ε 1 as + ε 1. Altogether we prove lim +ε 1 ln n(ε, )/( + ε 1 ) = 0. We switch to the secon point an assume that S is wealy tractable. Then λ < 1 since otherwise n(ε, S 1 ) for all ε (0, 1). For = 1 we have n(ε, S 1 ) = min{ j λ j+1 ε } = exp ( o ( ε 1)). This can happen only if λ j = o((ln j) ), as claime. This completes the proof. 7 Comparison We briefly compare tractability results of this paper for the unrestricte omain Ω unr = [1, ) N with tractability results of [1] for the restricte omain Ω res = [1, ) {1,,..., } [1, ε 1 0 ) N for 1 an ε 0 (0, 1). We consier linear tensor prouct problems S with ε 0 < λ < λ 1 = 1. Strong (T, Ω unr )-tractability of S as well as strong (T, Ω res )-tractability of S oes not hol regarless of the tractability function T, see [1, Lemma 3.1]. Consier finitely many, say, positive eigenvalues as in Section. This case has not been formally stuie in [1] for Ω res. However, it is easy to see from (14) that for (ε, ) with, the information complexity n(ε, S ) is uniformly boune in ε 1. Therefore the more interesting case is when (ε 1, ) [1, ε 1 0 ) N. Then n(ε, S ) = Θ( α(ε) ) with the factors in the Theta-notation only epenent on ε 0, 7

28 λ an. So we have a polynomial epenence on which obviously implies wea tractability. It follows from [1, Theorem 4.1] that (T, Ω res )-tractability of S hols iff B res := lim inf ln T (ε 1, ) inf 1 α(ε) α(ε 0 ) α(ε) ln() (0, ], (36) an the exponent of tractability is 1/B res. In particular, we have polynomial tractability, i.e., when T (x, y) = xy, with the exponent ln(ε 1 0 ) α(ε 0 ) = ln(λ 1 1. This exponent can be arbitrarily large if ε 0 is small or λ close to one. On the other han, it is interesting that the exponent oes not epen on the total number of positive eigenvalues. As we alreay sai, for the unrestricte omain Ω unr we o not have polynomial tractability of S. This agrees with the fact that the exponent of polynomial tractability for the restricte omain goes to infinity as ε 0 approaches zero, an for the unrestricte omain formally ε 0 = 0. Consier exponentially ecaying eigenvalues λ j = exp( β(j 1)) for a positive β. Then [1, Theorem 4.8] states that (T, Ω res )-tractability of S hols iff ) A e,res := lim inf x ln T (x, 1) ln ln(x) (0, ] an B res (0, ], where B res is given by (36). Furthermore, if A e,res = then the exponent of tractability is Bres 1. Hence, we again have polynomial tractability, an inee since A e,res = an λ = exp( β), the exponent of polynomial tractability is ln ε 1 0 α(ε 0 ) = 1. β As we now, for the unrestricte omain Ω unr we o not have polynomial tractability. Tae now T (x, y) = x 1+ln y. Then A e,res = an B res = β/. Furthermore, as we alreay now, B e () = β/. So we have (T, Ω res )-tractability as well as (T, Ω unr )- tractability with the same exponents /β. Hence, there is no much ifference between the restricte an unrestricte omains for this particular tractability function. Note also the ifference in the exponents for the last two tractability functions an for the restricte omain. For polynomial tractability, the exponent epens on ε 0 an goes to infinity as ε 0 approaches zero. For the secon tractability function, the exponent oes not epen on ε 0. 8