A note on Legendre-Fenchel conjugate of the product of two positive-definite quadratic forms

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1 Noname manuscript No. (will be inserted by the editor) A note on Legendre-Fenchel conjugate of the product of two positive-definite quadratic forms Yong Xia Received: date / Accepted: date Abstract The Legendre-Fenchel conjugate of the product of two positivedefinite quadratic forms was posted as an open question in the field of nonlinear analysis and optimization by Hiriart-Urruty [ Question 11 in SIAM Review 49, , (2007)]. Under a conve assumption on the function, it was answered by Zhao [SIAM J. Matri Analysis & Applications, 31(4), (2010)]. In this note, we answer the open question without making the conveity assumption. Keywords Legendre-Fenchel conjugate quadratic form PACS 15A48 65F15 65K05 90C25 1 Introduction The Legendre-Fenchel conjugate of the function h(y) is defined as h () = sup y R n T y h(y). For eample, if h(y) = 1 2 yt Ay, where A is positive definite, then h () = 1 2 T A 1. In the field of nonlinear analysis and optimization, J.B. Hiriart- Urruty [1] raised an open question what is the epression or formula of the conjugate of the product function f(y) = 1 4 (yt Ay)(y T By), This research was supported by National Natural Science Foundation of China under grants and /A011702, and by the fund of State Key Laboratory of Software Development Environment under grant SKLSDE-2011ZX-15. Y. Xia State Key Laboratory of Software Development Environment, LMIB of the Ministry of Education, School of Mathematics and System Sciences, Beihang University, Beijing , P. R. China dearyia@gmail.com

2 2 Yong Xia where A,B are two n n positive definite matrices. Recently, Zhao [3] answered the open question under the assumption that f(y) is conve. The main result is as follows. Theorem 1 ([3]) Let the function f(y) be conve. At any point R n, the value of the conjugate f () is finite, and f () = 0 if = 0, otherwise if 0, ( f () = p(α) := 3α 1/3 T (A+αB) 1 ) 2/3, 4 where α is any real root of the univariate equation g(α) := α T (A+αB) 1 A(A+αB) 1 T (A+αB) 1 B(A+αB) 1 = 0. (1) The following gives a sufficient condition under which the real root to the equation (1) is unique. Theorem 2 ([3]) If ma{κ(a), κ(b)} 2.5, where κ( ) denotes the condition number, then f(y) is conve and for any 0, there eists a unique real root to the equation (1). In this note, we answer the open question without the conveity assumption. Our proof is much shorter than that of Theorem 1 given in [3]. As a corollary, the sufficient condition (Theorem 2) is improved. 2 Main Result In this section, we study the conjugate f () = sup y R n T y f(y) without assuming that f(y) is conve. Lemma 1 The conjugate f () is finite. f () = 0 if = 0, otherwise if 0, f () > 0. Proof. The finiteness of f () follows from lim y T y (yt Ay)(y T By) lim y T y λ min(a)λ min (B) y =, where λ min (A) > 0 and λ min (B) > 0 are the minimum eigenvalues of A and B, respectively. If = 0, it is trivial to verify f () = 0. Now we assume 0. Since ma ǫ { T (ǫ) f(ǫ)} = ma{ T ǫ ( T A)( T B)ǫ 2 } = ǫ ( T ) 2 ( T A)( T B) > 0, we have f () > 0. The proof is complete.

3 Legendre-Fenchel conjugate of two quadratic forms 3 Lemma 2 Suppose 0. y is a stationary point of T y f(y) if and only if T y f(y) = p(α), where α is a real root of the univariate equation (1). Proof. Suppose y is a stationary point of T y f(y), we have ( 1 = f(y) = 2 (yt Ay)B + 1 ) 2 (yt By)A y. The assumption 0 implies that y 0. Therefore, y = 2(βB +γa) 1, (2) where β = y T Ay > 0 and γ = y T By > 0. Substituting (2) into y T Ay and y T By, we have Dividing (3) by (4) yields β = 4 T (βb +γa) 1 A(βB +γa) 1, (3) γ = 4 T (βb +γa) 1 B(βB +γa) 1. (4) β γ T (A+ β γ B) 1 A(A+ β γ B) 1 T (A+ β γ B) 1 B(A+ β γ B) 1 = 0. That is, α := β γ is a real root of (1). According to (3) and (4), we have T (A+αB) 1 = T (A+αB) 1 A(A+αB) 1 +α T (A+αB) 1 B(A+αB) 1 = 1 4 βγ αγ3 = 1 2 αγ3, which implies that ( 2 T (A+αB) 1 ) 1/3 γ =. (5) α Therefore, it follows from (2), (5) and the definitions of β, γ and α that T y f(y) = 2 γ T (A+αB) 1 1 ( T 4 βγ = (A+αB) 1 ) 2/3 3α1/3. 4 On the other hand, let α be a real root of (1). Define γ as in (5). Let β = αγ. Define y as in (2). Then y is a stationary point. The proof is complete. Lemma 3 Suppose 0. T y f(y) has at most 2n 1 stationary points, which can be solved to any given precision in polynomial time.

4 4 Yong Xia Proof. According to Lemma 2, the number of stationary points of T y f(y) is equal to the number of the real roots of Equation (1). Since A,B are positive definite, there eists a nonsingular matri P such that both P T AP and P T BP are diagonal matrices. Assume P T AP = Diag(a 1,...,a n ) and P T AP = Diag(b 1,...,b n ). Then for i = 1,...,n, a i > 0 and b i > 0. Let z = P T. Since P is nonsingular and 0, we have z 0. It is trivial to verify that each root of Equation (1) satisfies ( n n (a i +αb i ) )(α 2 b i zi 2 n (a i +αb i ) 2 a i z 2 i (a i +αb i ) 2 ) = 0, (6) which is a univariate polynomial equation of degree (2n 1). Therefore, E- quation (1) has at most 2n 1 real roots, which can be solved to any given precision in polynomial time [2]. The proof is complete. Now we present our main result. Theorem 3 At any point R n, the value of the conjugate f () is finite, and f () = 0 if = 0, otherwise if 0, f () = p(α ), where α is the maimizer of p(α), solved in polynomial time. Proof. Since the maimizer of T y f(y) must be a stationary point, it follows fromlemmas1and2thatf () = p(α ),whereα = argma g(α)=0 p(α). Itis not difficult to verify that the equations dp(α) dα = 0 and g(α) = 0 areequivalent. Therefore, α = argmap(α). Moreover, α can be solved by enumerating all the roots of g(α) = 0, which is done in polynomial time according to the proof of Lemma 3. As a corollary, we show that Theorem 2 is improved. Theorem 4 ([3]) If f(y) is conve, then for any 0, there eists a unique real root to the equation (1). Proof. If f(y) is conve, a stationary point of T y f(y) is also a maimizer. It follows from Lemma 2 and Theorem 3 that any stationary point of p(α) is also a global maimizer. According to Lemma 3 and the equivalence between dp(α) dα = 0 and g(α) = 0, p(α) has k( 2n 1) stationary points, denoted by α 1 <... < α k. It is sufficient to show k = 1. Suppose this is not true, i.e., k 2. Since p(α 1 ) = p(α 2 ) are the maimum value of p(α), for any α (α 1,α 2 ),wehavep(α) < p(α 1 ). Thenthereis alocalminimizerin (α 1,α 2 ), which is a stationary point and can not be a maimizer. Now, we obtain a contradiction. The proof is complete.

5 Legendre-Fenchel conjugate of two quadratic forms 5 3 Conclusion The Legendre-Fenchel conjugate of the product of two positive-definite quadratic forms was posted as an open question by Hiriart-Urruty [1]. Under a conve assumption on the function, it was answered by Zhao [3]. In this note, we give an answer to the open question without making the conveity assumption. Our proof is much shorter than that of [3]. Finally, we raise an open question what is the epression of the conjugate of f(y) = 1 4 (yt Ay)(y T By)+ 1 2 yt Cy, where A, B, C are positive definite matrices. References 1. J.B. Hiriart-Urruty, Potpourri of conjectures and open questions in nonlinear analysis and optimization, SIAM Review, 49, , (2007) 2. M. Sagraloff, When Newton meets Descartes: A Simple and Fast Algorithm to Isolate the Real Roots of a Polynomial, in Proceedings of the 37th International Symposium on Symbolic and Algebraic Computation, , ACM New York, NY, USA, Y.B. Zhao, The Legendre-Fenchel conjugate of the product of two positive-definite quadratic forms, SIAM J. Matri Analysis & Applications, 31(4), (2010)