New Classes of Positive Semi-Definite Hankel Tensors

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1 Miniax Theory and its Applications Volue 017, No., 1 xxx New Classes of Positive Sei-Definite Hankel Tensors Qun Wang Dept. of Applied Matheatics, The Hong Kong Polytechnic University, Hung Ho, Kowloon, Hong Kong wangqun876@gail.co Guoyin Li Dept. of Applied Matheatics, University of New South, Wales, Sydney 05, Australia g.li@unsw.edu.au Liqun Qi Dept. of Applied Matheatics, The Hong Kong Polytechnic University, Hung Ho, Kowloon, Hong Kong aqilq@polyu.edu.hk Yi Xu Departent of Matheatics, Southeast University, Nanjing 10096, P. R. China yi.xu1983@gail.co Received: June 3, 016 Accepted: Septeber 1, 016 A Hankel tensor is called a strong Hankel tensor if the Hankel atrix generated by its generating vector is positive sei-definite. It is known that an even order strong Hankel tensor is a suof-squares tensor, and thus a positive sei-definite tensor. The SOS decoposition of strong Hankel tensors has been well-studied by Ding, Qi and Wei [11]. On the other hand, very little is known for positive sei-definite Hankel tensors which are not strong Hankel tensors. In this paper, we study soe classes of positive sei-definite Hankel tensors which are not strong Hankel tensors. These include truncated Hankel tensors and quasi-truncated Hankel tensors. Then we show that a strong Hankel tensor generated by an absoluate integrable function is always copletely decoposable, and give a class of SOS Hankel tensors which are not copletely decoposable. Keywords: Hankel tensors, generating vectors, positive sei-definiteness, strong Hankel tensors. 010 Matheatics Subject Classification: 15A18, 15A69. This author s work was partially supported by the Australian Research Council. This author s work was partially supported by the Hong Kong Research Grant Council Grant No. PolyU 5011, , and ISSN printed, ISSN electronic / $.50 c Helderann Verlag

2 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors 1. Introduction Hankel tensors have iportant applications in signal processing [, 3, 7, 10], autoatic control [8], and geophysics [0, 30]. For exaple, the positive seidefiniteness of Hankel tensor can be a condition for the ultidiensional oent proble is solvable or not [4, 16, 4]. In [3], two classes of positive sei-definite PSD Hankel tensors were identified: even order strong Hankel tensors and even order coplete Hankel tensors. It was proved in [18] that coplete Hankel tensors are strong Hankel tensors, and even order strong Hankel tensors are SOSsu-of-squares tensors. In[17], generalized anti-circulant tensors were studied, which are one special class of Hankel tensors. The necessary and sufficient conditions for positive sei-definiteness of even order generalized anti-circulant tensors in soe cases were given, and the tensors are strong Hankel tensors and SOS tensors in these cases. Inheritance property was given in [11], which eans that if a lower-order Hankel tensor is positive seidefinite or positive definite, or negative sei-definite, or negative definite, or SOS, then its associated higher-order Hankel tensor with the sae generating vector, where the higher order is a ultiple of the lower order, is also positive sei-definite or positive definite, or negative sei-definite, or negative definite, or SOS, respectively. The SOS decoposition of strong Hankel tensors have also been given in [11]. The inheritance property established in [3] about strong Hankel tensor can be regarded as a special case of this inheritance property. There are other results about PSD Hankel tensors, SOS Hankel tensors and PNS non-sos short for PNS as in [8] Hankel tensors and soe regions which do not exist PNS Hankel tensors were given [6]. A recent detailed study on general SOS tensors can be found in [5]. Denote [n] := {1,,n}. The tensor A is said to be a syetric tensor if its entries a i1 i is invariant under any index perutation. Denote the set of all the real syetric tensors of order and diension n by S,n. Let v = v 0,,v n 1. Define A = a i1 i S,n by a i1 i = v i1 + +i, 1 for i 1,,i [n]. Then A is a Hankel tensor [18, 1, 3] and v is called the generating vector of A. An order diensional n Hilbert tensor H is a Hankel tensor with v = 1, 1, 1,, 1 3 n, and even order Hilbert tensors are positive definite [9]. Let x R n. Then x is a rank-one syetric tensor with entries x i1 x i. For A S,n and x R n, we have a hoogeneous polynoial fx of n variables and degree, fx = Ax a i1 i x i1 x i. i 1,,i [n] Note that there is a one to one relation between hoogeneous polynoials and syetric tensors. If fx 0 for all x R n, then hoogeneous polynoial

3 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors 3 fx and syetric tensor A are called positive sei-definitepsd. If fx > 0 for all x R n, x 0, then fx and A are called positive definite PD. The concepts of positive sei-definite and positive definite syetric tensors were introduced in []. The proble for deterining a given even order syetric tensor is positive sei-definite or not has iportant applications in engineering and science [10, 15, 7]. If A is a Hankel tensor, then fx is called a Hankel polynoial. Let A = a ij be an n 1+ n 1+ atrix with a ij v i+j, where is an additional nuber when n 1 is odd. Then A is a Hankel v n 1 atrix, associated with the Hankel tensor A. When n 1 is even, such an associated Hankel atrix is unique. Recall fro [3] that A is called a strong Hankel tensor if there exists an associated Hankel atrix A which is positive sei-definite. Let gy = y Ay, where y = y 1,,yn 1+ and A is an associated Hankel atrix of A. Then, A is a strong Hankel tensor if and only if g is PSD for at least one associated Hankel atrix A of A. Another class of Hankel tensors given by Qi in [3] are coplete Hankel tensors. A vector u = 1,γ,γ,,γ n 1 for soe γ R is called a Vanderonde vector [3]. If tensor A has the for A = i [r]α i u i, where u i for i = 1,,r, are all Vanderonde vectors, then we say that A has a Vanderonde decoposition. It was shown in [3] that a syetric tensor is a Hankel tensor if and only if it has a Vanderonde decoposition. If the coefficients α i for i = 1,,r, are all nonnegative, then A is called a coplete Hankel tensor [3]. It was proved in [3] that even order strong or coplete Hankel tensors are positive sei-definite. Let = k. If fx can be decoposed into a su of squares of polynoials of degree k, then fx is called a su-of-squares SOS polynoial, which eans that there exist fors g 1 x,,g k x of degree k such that fx = k g i x, 3 i=1 and the corresponding syetric tensor A is called an SOS tensor [14] for a recent study see also [5]. Clearly, an SOS polynoial tensor is a PSD polynoial tensor, but not vice versa. In 1888, young Hilbert [13] proved that for hoogeneous polynoials, only in the following three cases, a positive sei-definite for definitely is a su-ofsquares polynoial: 1 = ; n = ; 3 = 4 and n = 3, where is the degree of the polynoial and n is the nuber of variables. Hilbert proved that in

4 4 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors all the other possible cobinations of n and even, there are PNS hoogeneous polynoials. The first PNS hoogeneous polynoial is the Motzkin function [19] with = 6 and n = 3. Other exaples of PNS hoogeneous polynoials was found in [1, 8, 9, 6]. Let A S,n. If there are positive integer r N and vectors x j R n for j [r] such that A = x j, 4 j [r] then we say that A is a copletely decoposable tensor. If A adits a decoposition 4 with x j R n + for all j [r], then A is called a copletely positive tensor [5]. Clearly, a coplete Hankel tensor is a copletely decoposable tensor. By [3], a necessary condition for A to be PSD is that v i 1 0 for i [n]. 5 We know any properties of strong Hankel tensors. However, we know little about PSD Hankel tensors but not strong Hankel tensors. In this paper, we present soe classes of Hankel tensors which are PSD but not strong Hankel tensors, including truncated Hankel tensors and quasi-truncated Hankel tensors. Then we show that strong Hankel tensors are always copletely decoposable, and give a class of SOS Hankel tensors which are not copletely decoposable. The reainder of this paper is organized as follows. In the next section, we introduce truncated Hankel tensors which are of odd diension, i.e., n is odd, and whose generating vector v has only three nonzero entries: v 0, vn 1 and v n 1. Since we are only concerned about PSD Hankel tensors, by 5, we assue that these three entries are all nonnegative. Under this assuption, we show that a truncated Hankel tensor A is not a strong Hankel tensor as long as vn 1 ispositive. Thenweshowthatwheniseven, n = 3and v 0 = v, there are two nubers d 1 = d 1 and d = d with 0 < d d 1, such that if v 0 d 1 v, A is an SOS tensor; and if v 0 d, A is an PSD tensor. Then, for = 6 and n = 3, we show that for a truncated Hankel tensor A, the following three stateents are equivalent: 1 A is PSD; A is SOS; 3 v 0 v 1 v6 d; and we give an explicit value of d. In Section 3, we introduce quasi-truncated Hankel tensors which are of odd diension, i.e., n is odd, and whose generating vector v has only five nonzero entries: v 0, v 1, vn 1, v n 1 1 and v n 1. Again, since we are only concerned about PSD Hankel tensors, by 5, we assue that v 0, vn 1 and v n 1 are nonnegative. Under this assuption, we show that a quasi-truncated Hankel tensor is not a strong Hankel tensor as long as vn 1 is positive. Then we give a necessary condition for a sixth order quasi-truncated Hankel tensor A to be PSD, a sufficient condition for A to be SOS, respectively.

5 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors 5 In Section 4, we define copletely decoposable tensors, show that strong Hankel tensors generated by absoluate integrable functions are always copletely decoposable, and give a class of SOS Hankel tensors which are not copletely decoposable.. Truncated Hankel Tensors In this section, we consider the cases that the Hankel tensors A are generated by v = v 0,0,,0,vn 1,0,,0,v n 1 where n is odd. We call such Hankel tensors truncated Hankel tensors. If v = v 0,0,,0,vn 1,0,,0,v n 1 where n is odd, and gy have the siple for fx = v 0 x 1 +v n 1x n and +vn 1 { t1 t1 t1 t t n t t n 1 x t 1 1 x t x t 1 t t n 1 n : n 1t 1 +n t + +t n 1 = n 1 }. 6 gy = v 0 y1 +v n 1y n 1+ +vn 1 y n i j { y i y j : i+j = n 1 +}. 7 Since we are only concerned about PSD Hankel tensors, we ay assue that 5 holds. Fro 6 and 7, we have the following proposition. Proposition.1. Suppose that 5 holds. If vn 1 = 0, then the truncated Hankel tensor A is a strong Hankel tensor, and furtherore an SOS Hankel tensor if is even. If vn 1 > 0, then A is not a strong Hankel tensor. Proof. When vn 1 = 0, fro 6 and 7, we see that the truncated Hankel tensor A is a strong Hankel tensor, and furtherore an SOS Hankel tensor if is even. If vn 1 > 0, consider ȳ = e i e j where i+j = n 1 +,i j and i 1 or n 1+. We see that gȳ = vn 1 Hankel tensor in this case. < 0. Hence A is not a strong We consider the case that 6 with is even and n = 3. Assue that v 0 = v. We have the following theore.

6 6 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors Theore.. Suppose that A = a i1 i where 6, is even and n = 3 is a truncated Hankel tensor and 5 holds. Assue that v 0 = v. Let v = v 0,0,,0,v,0,,0,v 0 and d 1 = inf{d > 0 : A is an SOS tensor for all v such that v 0 dv }, d = inf{d > 0 : A is a PSD tensor for all v such that v 0 dv }. Then 1 d = d 1 = 0 if v = 0; 0 d d 1 < + if v > 0. Proof. If v = 0, it is clear that fx = v 0 x 1 +v 0 x 3 is SOS and in particular PSD because v 0,v 0 and is even. Thus, d 1 = d = 0 in this case. Now, let us consider the case where v > 0. We rewrite 6 as fx = f 1 x+f x+f 3 x, where f 1 x = v p p p x p x p 1 +x p 3, f x = v 0 x 1 + v x v p p p x p x p 1 and f 3 x = v 0 x 3 + v x v p p p x p x p 3. Clearly, f 1 x is PSD and SOS. We now consider the ters in f x. For each p = 1,...,, choose a positive constant δp such that 1 p δp > 0. For each p = 1,...,, let p be another positive constant such that p p δp p = p p p. Then, by the arithetic-geoetric inequality, for each p = 1,...,, p p p x p x p 1 = δpx px 1 p p p δpx + p px 1.

7 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors 7 This shows that, for each p = 1,...,, p δpx + p px 1 p p p x p x p 1 is a PSD diagonal inus tail for, and hence SOS [1]. Note that f can be written as f x = v 1 Therefore, if + v +v 0 x 1 v p δp x p δpx p p px 1. v 0 p p p pv, x p x p 1 + p px 1 then f is PSD and SOS. Siilarly, we ay show that under the sae condition, f 3 is also PSD and SOS. This, in particular, shows that 0 d d 1 p p < +. We consider the siple case that the tensors A are sixth order three diensional truncated Hankel tensors. Here we allow v 0 v 1. We give a necessary and sufficient condition that the sixth order three diensional truncated Hankel tensors to be PSD, and show that such tensors are PSD if and only if they are SOS. The sixth order three diensional truncated Hankel tensor A is generated by v = v 0,0,0,0,0,0,v 6,0,0,0,0,0,v 1. Now, 6 and 7 have the siple for and fx = v 0 x 6 1 +v 6x 6 +30x 1x 4 x 3 +90x 1 x x 3 +0x3 1 x3 3 +v 1x gy = v 0 y 1 +v 6y 4 +y 1y 7 +y y 6 +y 3 y 5 +v 1 y 7. 9 Theore.3. Suppose that A is a sixth order three diensional truncated Hankel tensor, the following stateents are equivalent: i The truncated Hankel tensor A is a PSD Hankel tensor; ii The truncated Hankel tensor A is an SOS Hankel tensor; iii The relation 5 holds and v0 v v 6. 10

8 8 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors Furtherore, the truncated Hankel tensor A is positive definite if and only if v 0,v 6,v 1 > 0 and strict inequality holds in 10. Proof. [i iii] Suppose that A is PSD, then clearly v 0,v 6,v 1 0. To see iii, we only need to show 10 holds. Let t 0 and let x = x 1, x, x 3, where x 1 = v 1 6 1, x = tv 0 v 1 1 1, x3 = v Substitute the to 8. If A is PSD, then f x 0. It follows fro 8 that Fro this, we have v 0 v 1 +v 6 t 3 30t +90t 0 v 0 v 1 +v 0 v 1 0. v0 v 1 t3 +30t 90t+0 v 6. Substituting t = to it, we have 10. [iii ii] We now assue that 5 and 10 hold. We will show that A is SOS. If v 6 = 0, then by Proposition.1, A is an SOS Hankel tensor. Assue that v 6 > 0. By 10, v 0 > 0 and v 1 > 0. We now have 1 v0 4 fx = 10v 6 x 3 v v1 +v 6 v x x x 1 x x 3 +f 1 x, where f 1 x = + v 0 10v 6 v0 v 1 v 1 10v 6 v1 v x v 6 x 6 1 x v 6 x 1 x x We see that f 1 x is a diagonal inus tail for [1]. By the arithetic-geoetric inequality, we have v 0 10v 6 v0 v x v 6 x 6 + v1 v 1 10v 6 v 0 v 6 v 0 v 1 10v x 1x x 3. 1 x 6 3

9 By 10, 3 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors v 6 v 0 v 1 10v x 1x x v 6 x 1x x 3. 1 Thus, f 1 is a PSD diagonal inus tail for. By [1], f 1 is an SOS polynoial. Hence, f is also an SOS polynoial if 5 and 10 hold. [ii i] This iplication follows direct fro the definition. We now prove the last conclusion of this theore. First, we assue that A is positive definite. Then, v 6 = fe > 0 as e 0. Siilarly, v 0 = fe 1 > 0 and v 1 = fe 3 > 0. Note that in the above [i iii] part, f x > 0 as x 0. Then strict inequality holds for the last two inequalities in the above [i iii] part. This iplies that strict inequality holds in 10. On the other hand, assue that v 0,v 6,v 1 > 0 and strict inequality holds in 10. Let x = x 1,x,x 3 0. If x 1 0,x 0 and x 3 0, then strict inequality holds in 1 as v 6 > 0 and strict inequality holds in 10. Then f 1 x > 0. If x 0 but x 1 x 3 = 0, then fro 11, we still have f 1 x > 0. If x = 0 and one of x 1 and x 3 are nonzero, then we still have f 1 x > 0 by 11. Thus, we always have f 1 x > 0 as long as x 0. This iplies fx > 0 as long as x 0. Hence, A is positive definite. 3. Quasi-Truncated Hankel Tensors In this section, we consider the case that the Hankel tensor A is generated by v = v 0,v 1,0,,0,vn 1,0,,0,v n 1 1,v n 1 where nisodd. Adding v 1 and v n 1 1 to the case in the last section, we get this case. We call such a Hankel tensor a quasi-truncated Hankel tensor. Hence, truncated Hankel tensors are quasi-truncated Hankel tensors. Since we are only concerned about PSD Hankel tensors, we ay assue that 5 holds. Now, and gy have the siple for fx = v 0 x 1 +v 1x 1 1 x +v n 1 1 x n 1 x 1 n +v n 1 x n and +vn 1 { t1 t1 t1 t t n t t n 1 x t 1 1 x t x t 1 t t n 1 n : n 1t 1 +n t + +t n 1 = n 1 gy = v 0 y1 +v 1y 1 y +v n 1 1 yn 1+yn 1 +vn 1 y n i j }. 13 +v n 1 y n 1+ { y i y j : i+j = n 1 +}. 14 We first show that a result that Proposition.1 continues to hold in this case.

10 10 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors Proposition 3.1. Suppose that 5 holds and is even. If vn 1 = 0, then the quasi-truncated Hankel tensor A is PSD if and only if v 1 = v n 1 1 = 0. In this case, A is a strong Hankel tensor and an SOS Hankel tensor. If vn 1 > 0, then A is not a strong Hankel tensor. Proof. Suppose that vn 1 = 0. Assue that v 1 0. If v 0 = 0, consider ˆx = 1, v 1,0,,0. Then fˆx = v1 < 0. If v 0 > 0, consider x = 1, v 0 v 1,0,,0. Then f x = 1 v 0 < 0. Thus, A is not PSD in these two cases. Siilar discussion holds for the case that v n 1 1 = 0. Assue now that v 1 = v n 1 1 = 0. By Proposition.1, we see that the quasi-truncated Hankel tensor A is a strong Hankel tensor and an SOS Hankel tensor in this case. This proves the first part of this proposition. Suppose that vn 1 > 0. Consider ȳ = e i e j where i+j = n 1 +,i j and i 1 or n 1+. We see that gȳ = vn 1 strong Hankel tensor in this case. < 0. Hence A is not a We consider the siple case that the tensors A are sixth order three diensional quasi-truncated Hankel tensors. We give a necessary condition that the sixth order three diensional quasi-truncated Hankel tensors to be PSD, and a sufficient condition that the sixth order three diensional quasi-truncated Hankel tensors to be SOS. The sixth order three diensional quasi-truncated Hankel tensor A is generated byv = v 0,v 1,0,0,0,0,v 6,0,0,0,0,v 11,v 1 R and14havethesiple for and fx = v 0 x v 1x 5 1 x +v 6 x 6 +30x 1x 4 x 3 +90x 1 x x 3 +0x3 1 x3 3 +6v 11 x x 5 3 +v 1 x 6 3, 15 gy = v 0 y 1 +v 1y 1 y +v 6 y 4 +y 1y 7 +y y 6 +y 3 y 5 +v 11 y 6 y 7 +v 1 y To present a necessary condition for a sixth order three diensional quasi-truncated Hankel tensor to be PSD, we first prove the following lea. Lea 3.. Consider ˆfx1,x = v 0 x v 1x 5 1 x +v 6 x 6. Then ˆf is PSD if and only if v 0 0, v 6 0 and v 1 5 v0 6 v

11 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors 11 Proof. Suppose that v 0 0, v 6 0 and 17 holds. Then, by the aritheticgeoetric inequality, one has v 0 x 6 1 +v 6 x 6 = 1 5 v 0x v 0x v 0x v 0x v 0x 6 1 +v 6 x 6 v0 1 5x v 6x v1 x 5 1 x. This iplies that ˆfx 1,x 0 for any x 1,x R, i.e., ˆfx1,x is PSD. Suppose that ˆfx 1,x is PSD. It is easy to see that v 0 0 and v 6 0. Assue now that 17 does not hold, i.e., v 1 > 5 v0 6 v If v 0 = v 6 = 0, let x 1 = 1 and x = v 1. Then ˆfx 1,x < 0. We get a contradiction. If v 0 = 0 and v 6 0, let x 1 = v and x = v Again, ˆfx 1,x < 0. We get a contradiction. Siilarly, if v 0 0 and v 6 = 0, we ay get a contradiction. If v 0 0 and v 6 0, let x 1 = 5v and x = v 1 v 1 v Then by 18, ˆfx 1,x = 6v 0 v 6 6 v 1 5v v < 0. We still get a contradiction. This copletes the proof. We now present a necessary condition for a sixth order three diensional quasitruncated Hankel tensor to be PSD. Proposition 3.3. Suppose that 5 holds. If A is a sixth order three diensional PSD quasi-truncated Hankel tensor, then 17 and the following inequalities v 1 v v and hold. If furtherore v0 v 1 10v 6 0 v 1 v = v 11 v 5 6 0, 1 then 10 also holds. Proof. Suppose that A is PSD. In 15, let x 3 = 0. By Lea 3., 17 holds. In 15, let x 1 = 0. By an arguent siilar to Lea 3., 19 holds. In 15, let x = 0. Since A is PSD, we ay easily get 0.

12 1 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors Suppose further that 1 holds. As in the part [i iii] of the proof of Theore.3, we let t 0 and let x = x 1, x, x 3, where x 1 = v 1 6 1, x = tv0 v 1 1 1, x 3 = v It follows fro 1 that This, together with 15 iplies that 6v 1 x 5 1 x +6v 11 x x 5 3 = 0. f x = v 0 v 1 +v 6 t 3 30t +90t 0 v 0 v 1 +v 0 v 1 0. Proceed as in the part [i iii] of the proof of Theore.3: we see that 10 holds in this case. This copletes the proof. We can also present a sufficient condition for a sixth order three diensional quasi-truncated Hankel tensor to be SOS. Proposition 3.4. Let A be a sixth order three diensional quasi-truncated Hankel tensor. Suppose that v 0,v 6,v 1 > 0. If there exist t 1,t > 0 such that v 1 1 t 1 10v 6 t 1 v0 v 1 3 and v 0 10v 6 v0 v 1 v v 6 t t v0 v v 1 + v 11 v 6 t 1 v 0 t v v1 t 1 v 0 v 1 10v 6 v1 v 0 1 v11 t v v 6 v 1 v 11 t 1 v 0 t v v hold, then A is SOS. Proof. We write fx = 5 i=1 f ix, where 1 v0 f 1 x = v 0 10v 6 v1 t 1 v v 1 v 1 10v 6 v1 v 0 x v 6 v 1 v 11 x 6 t 1 v 0 t v 1 1 v11 t v 1 x 6 3 v x 1x x 3,

13 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors 13 1 v0 4 f x = 10v 6 x 3 v v1 v x 3 3, and 5 f 3 x = v 1 t 1 v 0 x v 1x 5 1 x + v 1 t 1 v 0 5 x 6, 5 5 f 4 x = v 11 t v 1 x v 11x 5 3 x + v 11 x 6 t v 1 f 5 x = v x x 1 x x 3. Clearly, f and f 5 are squares. Fro Lea 1, we deduce that f 3 and f 4 are PSD. Since each of f 3 and f 4 has only two variables, they are SOS. If 3 6 hold, by the arithetic-geoetric inequality, f 1 is PSD. In this case, f 1 is a PSD diagonal inus tail for. By [1], f 1 is SOS. Thus, if 3 6 hold, then f, hence A, is SOS. 4. A Class of SOS Hankel Tensors In this section, we provide further classes for SOS Hanke tensors and exaples for SOS Hankel tensors which are not strong Hankel tensors. We say that A is a strong Hankel tensor generated by an absolutely integrable real valued function h :,+ [0,+ if it is a Hankel tensor and its generating vector v = v 0,v 1,,v n 1 satisfies v k = t k htdt, k = 0,1,,n 1. 7 SucharealvaluedfunctionhiscalledthegeneratingfunctionofthestrongHankel tensor A. Ithasbeenshownin[3]thatanystrongHankel tensor generatedbyan absolutely integrable real valued nonnegative function is a strong Hankel tensor. We now define the copletely decoposable tensor. Definition 4.1. Let A S,n. If there are positive integer r N and vectors x j R n for j [r] such that A = j [r] then we say that A is a copletely decoposable tensor. x j, 8 Then in the following theore, we will show that when the order is even, a strong Hankel tensor generated by an absolutely integrable real valued nonnegative function is indeed a liiting point of coplete Hankel tensors, which is a copletely decoposable tensor.

14 14 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors Theore 4.. Copletely decoposability of strong Hankel tensors Let,n N. Let A be an th-order n-diensional strong Hankel tensor generated by an absolutely integrable real valued nonnegative function. If the order is an even nuber, then A is a copletely decoposable tensor and a liiting point of coplete Hankel tensors. If the order is an odd nuber, then A is a copletely r-decoposable tensor with r = n 1+1. Proof. Let h be the generating function of the strong Hankel tensor A. Then, for any x R n, where fx := Ax = = = = n i 1,i,...,i =1 + n i 1,i,...,i =1 + + f l x = v i1 +i +...+i x i1 x i...x i t i 1+i +...+i htdt n i 1,i,...,i =1 n i=1 l l x i1 x i...x i t i 1+i +...+i x i1 x i...x i htdt t i 1 x i htdt = li l + f lx, 9 n t i 1 x i htdt. By the definition of Rieann integral, for each l 0, we have f l x = li where fl k x is a polynoial defined by kl fl k x := n h i=1 j k li 1 j x i l k. k j=0 Fix any l 0 and k N. Note that kl n h fl k i=1 x : = j k li 1 j x i l k k j=0 kl n j k = li 1 h j l 1 k x i j=0 i=1 i=1 kl = u j,x, j=0 k 1 k f k l x, where u j = h j l 1 k 1, j k 1 k l,...,j ln 1. Here u k j are always well-defined as h takes nonnegative values [3]. Define A k l be a syetric tensor such that

15 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors 15 fl kx = Ak l x. Then, it is easy to see that each A l k is a coplete Hankel tensor and thus a copletely decoposable tensor. Note that the copletely decoposable tensor cone CD,n is a closed convex cone when is even. It then follows that A = li k li l A l k is a copletely decoposable tensor and a liiting point of coplete Hankel tensors. To see the assertion in the odd order case, we use a siilar arguent as in [3]. Pick real nubers γ 1,...,γ r with r = n 1+1 andγ i γ j for i j. Consider the following linear equation in α = α 1,...,α r with v k = r α i γi k, k = 0,...,n 1. i=1 Note that this linear equation always has a solution say ᾱ = ᾱ 1,...,ᾱ r because the atrix in the above linear equation is a nonsingular Vanderonde atrix. Then, we see that A i1,...,i = v i i = r i=1 ᾱ i γ i i i = r ᾱ i ui i 1,...,i, i=1 where u i R n is given by u i = 1,γ i,...,γi n 1 T. This shows that A = i [r]ᾱiu i. Now, as is an odd nuber, we have A = r. i=1 ᾱ 1 i u i Therefore, A is a copletely decoposable tensor and the last conclusion follows. In [3], the author has provided an exaple of positive sei-definite Hankel tensors with order = 4, which is not a strong Hankel tensor generated by an absolutely integrable real valued nonnegative function. We now extend this exaple to the general case where = k for any integer k. We will also further show that such tensors are indeed SOS tensors but not copletely decoposable and so, are not strong Hankel tensors generated by an absolutely integrable real valued nonnegative function by Theore 4.. Let = k, n =, k is an integer and k. Let v 0 = v = 1, v l = v l = 1 l, l = 1,...,k 1, and v j = 0 for other j. Let A = a i1 i be defined by a i1 i = v i1 + +i, for i 1,,i = 1,. Then A is an even order Hankel tensor. For any x R, we have k 1 Ax = x 1 j=1 k x j 1 x j +x = j=0. x k j 1 x j x k j 1 x j+ Thus, A is an SOS-Hankel tensor, hence a positive sei-definite Hankel tensor. On the other hand, A is not a copletely decoposable tensor. Assue that A is a copletely decoposable tensor. Then there are vectors u j = a j,b j for

16 16 Q. Wang, G. Li, L. Qi, Y. Xi / Positive Sei-Definite Hankel Tensors j [r] such that A = r j=1 u j. Then for any x R, Ax = On the other hand, r a p x 1 +b p x = k 1 Ax = x 1 j=1 r j=0 j x j 1 x j +x. a j p b j p x j 1 x j. Coparing the coefficients of x 1 x in the above two expressions of Ax, we have r a p b p = 1. This is ipossible. Thus, A is not copletely decoposable and so, is not a strong Hankel tensor generated by an absolutely integrable real valued nonnegative function. We now have two further questions: 1. Is a copletely decoposable Hankel tensor always a strong Hankel tensor?. Is a truncated Hankel tensor copletely decoposable? References [1] A. A. Ahadi, P. A. Parrilo: A convex polynoial that is not sos-convex, Math. Prograing [] R. Badeau, R. Boyer: Fast ultilinear singular value decoposition for structured tensors, SIAM J. Matrix Analysis Appl [3] R. Boyer, L. De Lathauwer, K. Abed-Merai: Higher order tensor-based ethod for delayed exponential fitting, IEEE Transactions Signal Processing [4] C. Berg: The ultidiensional oent proble and seigroups, oents in Matheatics, Proc. Syposia Applied Matheatics [5] H. Chen, G. Li, L. Qi: SOS tensor decoposition: theory and applications, to appear in Co. Math. Sciences 016. [6] Y. Chen, L. Qi, Q. Wang: Positive sei-definiteness and su-of-squares property of fourth order four diensional Hankel tensors, J. Coputational Appl. Math [7] Y. Chen, L. Qi, Q. Wang: Coputing extree eigenvalues of large scale Hankel tensors, J. Scientific Coputing [8] G. Chesi: On the gap between positive polynoials and SOS of polynoials, IEEE Transactions Auto. Control

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