Convexity properties of Tr[(a a) n ]

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1 Linear Algebra and its Applications 315 (2) Convexity properties of Tr[(a a) n ] Luis E. Mata-Lorenzo, Lázaro Recht Departamento de Matemáticas Puras y Aplicadas, Universidad Simón Bolívar, Caracas, Venezuela Received 13 March 1999; accepted 18 January 2 Submitted by C. Davis Abstract In this paper, for any even positive integer p = 2n, we consider the function E p (a) = Tr[(a a) n ] definedonac -algebra A with a faithful trace Tr : A C. We show that the quadratic form Q(E p ) a associated to the Hessian of E p at a A is non-negative and for p> 2 we give a simple characterization of the null-space. Finally, we show that g(t) = E p (a + tb)is at least as convex as t p uniformly with respect to b. 2 Elsevier Science Inc. All rights reserved. 1. Introduction In this paper, we consider, for any even positive integer p = 2n, the function E p (a) = Tr[(a a) n ] definedonac -algebra A with a faithful trace Tr : A C. Note that E p (a) = ( a p ) p,where a p is the usual p-norm of Shatten on A (see [7, p. 113] or [2]). We are interested in convexity properties of E p. In Proposition 3.1 of Section 3, we show that the quadratic form Q(E p ) a associated to the Hessian H(E p ) a of E p at a point a is non-negative. Furthermore, in Theorem 4.3, we obtain a simple description of the directions b where it degenerates (for p>2), i.e. Q(E p ) a (b) =. In Section 5, we show that E p : A R is a strictly convex function in spite of the fact that its Hessian may be singular when p>2. In addition, we introduce the notion of convexity of type s>1, which expresses roughly the fact that in every direction b the function is at least as convex as t s (Definition 1). Theorem 5.4 Corresponding author. Present address: Department of Mathematics and Statistics, The University of New Mexico, Albuquerque, NM , USA //$ - see front matter 2 Elsevier Science Inc. All rights reserved. PII:S ()5-1

2 26 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) shows that at least in the finite dimensional case, E p (a) has precisely this type of convexity for s = p. We conjecture that the results above are true in any von Neumann algebra A and for any real p>1 where the function E p (a) is defined in this case by E p (a) = Tr a p. In [1,8], metric properties of homogeneous reductive spaces [5] of the unitary group in a C -algebra are studied. The Finsler metric of these spaces is deduced from the ambient norm in the algebra. In a forthcoming paper with C. Durán, we will examine variational properties of the metric in the Grassmann manifold of a C -algebra where the metric comes from the Finsler structure obtained from the p- norms in the algebra. In this study, the convexity properties of the functions E p are essential. 2. Preliminaries We consider a C -algebra A with a faithful trace Tr: A C, andlett be the real part of Tr. The real trace T is then a bounded linear form T : A R with the following properties (see [7]): 1. a, b A, T(a) = T(a ), T(ab) = T(ba). 2. T is positive, i.e. positive valued on the positive elements of A (strictly positive on non-trivial positive elements of A). We consider the scalar product in A given by a,b =T(b a) a, b A. Some immediate properties of this scalar product are a, b = b, a = a,b = b,a a,b A, and we have the induced norm a 2 = a,a = T(a a) a A. Also, for any positive real s, we shall consider the s-norm in A (see[7]or[2])which is given by a s = s T (a a) s/2. 3. The Hessian of the trace of (a a) n The trace of (a a) n,i.e.tr[(a a) n ], is a real number. Hence we may just look at T[(a a) n ]. In Proposition 3.1, in this section, we show that the Hessian of the mapping E p (a) = T(a a) n = ( a p ) p is non-negative as a quadratic form.

3 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) The first derivative For a,b A, the derivative of E p at a along b, (DE p ) a (b),isgivenby [ ] [ ] d d (DE p ) a (b) = dt E p(a + tb) = t= dt T((a + tb )(a + tb)) n [ ] d = T dt ((a + tb )(a + tb)) n t= = T [ ] d (a a) l dt (a + tb )(a + tb) (a a) m l+m=n 1 = T l+m=n 1 (a a) l (a b + b a)(a a) m = n T(a a) n 1 (a b + b a). t= t= 3.2. The Hessian of E p The Hessian of E p at a, H(E p ) a, is the bilinear form given by H(E p ) a (b, b ) [ ] d = (D 2 E p ) a (b, b ) = DEp dt (a+tb ) (b) t= [ d ( = n T (a + tb ) b + b (a + tb ) )( (a + tb ) (a + tb ) ) ] n 1 dt = n T (b b + b b )(a a) n 1 + n T (a b + b a)(a a) l (b a + a b )(a a) m. We also call the Hessian of E p at a the associated quadratic form Q(E p ) a which in any given direction b is given by Q(E p ) a (b) = H(E p ) a (b, b). We can write explicitly Q(E p ) a (b) = p T (b b)(a a) n 1 + n T (a b + b a)(a a) l (b a + a b)(a a) m. We set the notation Λ = a a, Λ 2 = b b (positive elements in A)andΛ 1 = b a + a b (symmetric in A) and we write Q(E p ) a (b) = p T (Λ 2 )(Λ ) n 1 + n T Λ 1 Λ l Λ 1 Λ m. t=

4 28 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) If we set r and q as positive square roots of Λ and Λ 2, respectively (see Chapter 4 in [3] or [4]), i.e. r 2 = Λ, q 2 = Λ 2, we can write Q(E p ) a (b) = p T (Λ 2 )(Λ ) n 1 + n T = p Tqqr n 1 r n 1 + n = p Tr n 1 qqr n 1 + n = p qr n n The computation above shows that: Λ 1 Λ l Λ 1 Λ m TΛ 1 r l r l Λ 1 r m r m Tr m Λ 1 r l r l Λ 1 r m r l Λ 1 r m 2 2. (1) Proposition 3.1. The Hessian of E p in any given direction b is a sum of squares of norms, and hence it is non-negative. 4. The null-space of the Hessian Here we study conditions that characterize the null-space N of the Hessian of E p (p>2) at a. By definition N is the subspace of A given by (see [6]) N ={b A H(E p ) a (b, b ) = foreveryb A}. Due to the fact that Q(E p ) a is a non-negative quadratic form (Proposition 3.1), the Cauchy Schwarz inequality shows that N is just the subspace given by Q(E p ) a (b) = H(E p ) a (b, b) =. If b N, we say that the Hessian is degenerate in the direction b. Next we show that N ={b A a b = = ba }. First we present two lemmas that discuss the equations b(a a) k = andba (aa ) k = which are related to the null-spaces of the Hessian of E p for the cases of n odd or even. Lemma 4.1. Let k N, and a,b A. Suppose b(a a) k =, and b (aa ) k =, then a b = ba = b a = ab =. Proof. Consider the C -algebra A as a subalgebra of bounded operators on a Hilbert space H.LetH 1 = ker(a) H and H 2 = ker(a ) H. Consider the operator a : H H written in block form with respect to the orthogonal decompositions of H induced by H 1 in the domain and by H 2 in the codomain, i.e. a : H1 H 1 H 2 is given by the operator matrix H 2

5 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) a11 a a = 12, where a a 21 a 11 : H1 H 2, etc. 22 Observe that = a 12 : H 1 H2 and = a 22 : H 1 H 2. Observing that H 2 = ker a = ( img a), it follows that = a 21 : H1 H 2. Hence a is given by the operator matrix α a =, where α : H 1 H 2 is invertible. Similarly, the adjoint operator a : H2 H 2 H1 H 1 is given by the operator matrix a α =, where α : H2 H 1 is the adjoint of α. With this notation, we have ( a α a = ) ( α and (a a) k (α = α) k ), where α α : H1 H 1 and (α α) k : H1 H 1 are injective and both have dense images in H1 for any k N. We write b : H H in matrix form with respect to the orthogonal decompositions H = H1 H 1 in the domain and H = H2 H 2 codomain, i.e. b11 b b = 12, where b b 21 b 11 : H1 H 2, etc. 22 From the hypothesis in the lemma, we have b(a a) k =, hence we can write ( b11 b = 12 (α α) k ) ( b11 (α = α) k ) b 21 b 22 b 21 (α α) k. Then we have b 11 (α α) k = andb 21 (α α) k =, but (α α) k has a dense image in H1. Hence by the continuity of the operator b we get b 11 = andb 21 =. Similarly, by hypothesis, b (aa ) k =. Then ( (αα = ) k ) ( ) b12 b22 = b12 (αα ) k. So we have b12 (αα ) k =. Also (αα ) k has a dense image (in H2 ). By continuity, we have that b12 = and hence b 12 =. So we have that b =. b 22 With the computations above we conclude a b = andab = just by looking at these products in operator matrix form. Finally, considering the s of these identities, we get b a = andba = to complete the proof of the lemma.

6 3 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) Lemma 4.2. Let k N, and a,b A. Suppose ba (aa ) k =, and b a(a a) k =, then a b = ba = b a = ab =. Proof. Similar to the proof of Lemma 4.1. We leave the details for the reader. Theorem 4.3. The Hessian of the mapping E p, for p>2, is degenerate at a along the direction b if and only if a b = ba = (= b a = ab ). Proof. Suppose b N. Recall the expression of the Hessian of E p,asgivenby formula (1). H(E p ) a (b, b) = p T (Λ 2 )(Λ ) n 1 + n T Λ 1 Λ l m Λ 1 Λ = p qr n n r l Λ 1 r m 2 2 with T (Λ 2 )(Λ ) n 1 = qr n and T Λ 1Λ l Λ 1 Λ m = r l Λ 1 r m 2 2. All these terms must be for the Hessian to vanish. In particular, the first term T (Λ 2 )(Λ ) n 1 must be, i.e. T b b(a a) n 1 =. Now observe that E p (a) = E p (a ) for all a A. Hence, Q(E p ) a (b) = H(E p ) a (b, b) = H(E p ) a (b,b ) = Q(E p ) a (b ) and we have T bb (aa ) n 1 =. To conclude we separate the cases n odd from n even: n odd:ifnisodd, say n=2k+1andh(e p ) a (b, b)=, we have T b b(a a) 2k =. Hence b(a a) k 2 2 =, and T bb (aa ) 2k =. Hence b (aa ) k 2 2 =. Then b(a a) k = andb (aa ) k =. By Lemma 4.1, we have that a b = ba = b a = ab =. n even: Ifn is even, say n = 2k + 2andH(E p ) a (b, b) =, we have T b b (a a) 2k+1 =. Hence ba (aa ) k 2 2 =, and T bb (aa ) 2k+1 =. Hence b a (a a) k 2 2 =. Then ba (aa ) k = andb a(a a) k =. By Lemma 4.2, we have that a b = ba = b a = ab =. 5. A convexity property of Tr[(a a) n ] For any positive integer n and p = 2n, the function E p : A R is convex, for E p is the composition of the p-norm (which is a convex function) with the monomial x p (p 2) which is convex and non-decreasing for x>. In fact, E p : A R is strictly convex as the following argument shows. Suppose that for a 1 /= a 2 in A, there exists <t < 1 such that E p (t a 1 + (1 t )a 2 ) = t E p (a 1 ) + (1 t )E p (a 2 ). We consider the polynomial h(t) = E p (ta 1 + (1 t)a 2 ) te p (a 1 ) (1 t)e p (a 2 ) which has degree p 2, is non-positive in [, 1], is convex and has

7 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) a local maximum at t (h(t ) = ). If the local maximum at t is strict, then we have a contradiction with the fact that h(t) is convex. On the other hand, if the local maximum at t is not strict, then we have an infinite number of zeros of the polynomial h(t) which contradicts the fact that h has degree larger than 1. We will show that E p has a stronger convexity property: namely that it is convex of type p according to Definition 1, at least in the case when A is a sub-algebra of M d d (C), the algebra of complex d d matrices and Tr is the usual trace. Recall our previous notation Λ = a a, Λ 1 = a b + b a and Λ 2 = b b used earlier in this work. The mapping h(t) : R R, givenbyh(t) = E p (a + tb), can be written as E p (a + tb)=t ( (a + tb )(a + tb) ) n =T [(a a + (a b + b a)t + (b b) t 2 ) n] [ =T (Λ + Λ 1 t + Λ 2 t 2 ) n]. The expression (Λ + Λ 1 t + Λ 2 t 2 ) n expands to a sum of 3 n terms, namely all the possible words of size n in the three symbols : Λ, Λ 1 t and Λ 2 t 2. The expression (Λ + Λ 1 t + Λ 2 t 2 ) n can also be written as a polynomial in t of degree p with coefficients in A. We write (Λ + Λ 1 t + Λ 2 t 2 ) n as (Λ + Λ 1 t + Λ 2 t 2 ) n = p G a,b (i) t i, i= where G a,b (i) is the sum of all the words in the three symbols Λ, Λ 1 and Λ 2 of order i relative to the variable t, i.e. those words that have say, k instances of Λ 1 and l instances of Λ 2 with k + 2l = i. The summands of G a,b (i) shall be called i-words. Finally, we define C a,b (i) = TG a,b (i) and we have E p (a + tb) = T(Λ + Λ 1 t + Λ 2 t 2 ) n = p C a,b (i) t i. Proposition 5.1. If the Hessian of the mapping E p is at a along the direction b, then G a,b (i) = (and C a,b (i) = ) for i {1, 2,...,p 1}. Proof. It is a straightforward exercise to see that every i-word W a,b (i), fori {1, 2,...,p 1} contains at least one factor from the set {Λ 1, Λ 2 Λ, Λ Λ 2 }.From Theorem 4.3, we have that a b = ba = = b a = ab. Hence Λ 1 = b a + a b =, Λ 2 Λ = b ba a = andλ Λ 2 = a ab b =. Then for i {1, 2,...,p 1}, any i-word is, and hence G a,b (i) = andc a,b (i) = T G a,b (i) =. i=

8 32 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) Corollary 5.2. If C a,b (2) =, then G a,b (i) = (and C a,b (i) = ) for i {1, 2,...,p 1}. Proof. Observe that Q(E p ) a (b) = 2C a,b (2) =. By Proposition 5.1, it follows that G a,b (i) = (andc a,b (i) = ) for i {1, 2,...,p 1}. Corollary 5.3. If the Hessian of the mapping E p is at a along the direction b, then E p (a + tb) = T(Λ ) n + T(Λ 2 ) n t p = a p p + b p p t p. Proof. This follows immediately from the following calculation and Proposition 5.1 (C a,b (i) = fori {1, 2,...,p 1}): [ E p (a + tb) = T (Λ + Λ 1 t + Λ 2 t 2 ) n] p = C a,b (i) t i i= = T(Λ ) n + T(Λ 2 ) n t p = T(a a) n + T(b b) n t p. Corollaries 5.2 and 5.3 show that the term C a,b (2)t 2 in the expression of E p (a + tb) dominates the terms C a,b (i)t i for i {1,...,p 1} in the sense that C a,b (2) = implies C a,b (i) = fori {1,...,p 1}. This suggests that either for t small the function E p (a + tb) has quadratic convexity as a function of t (if C a,b (2) /= ), or else C a,b (2) = ande p (a + tb) is just ( a p ) p + ( b p ) p t p. The non-linear part of E p (a + tb) is the excess of E p (a + tb) over its linear approximation. The remarks above reveal that this excess (difference) is at least of order t p. This leads us to the definition of convexity of type s below. Definition 1. Let V be a normed vector space. We say that a differentiable function g : V R is convex of type s>1ata V if there exists a constant L>and some ε>suchthatifb V with b =1and t <ε,then d g(a + tb) g(a) t g(a + tb) L t s. dt It is easy to see that a convex function of type s>1 is strictly convex. Theorem 5.4. Let A M d d (C) with the p-norm. For every positive even integer p the function E p : A R is convex of type p at every a A. Moreover, the constant L may be set equal to 1. Proof. The idea of the proof is to show that the second degree term dominates uniformly (in b) the higher degree terms up to degree p 1 (Proposition 5.5). We start recalling the explicit form for E p (a + tb) with the notation introduced earlier t=

9 Hence L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) [ E p (a + tb) = T (Λ + Λ 1 t + Λ 2 t 2 ) n] = E p (a + tb) E p (a) t d dt E p(a + tb) p C a,b (i) t i. = C a,b (i) t i + C a,b (p) t p = C a,b (i) t i + T((Λ 2 ) n )t p i= t= = C a,b (i) t i + t p (since T((Λ 2 ) n ) = b p p = 1). The equalities above show that the theorem follows immediately from Proposition 5.5. Proposition 5.5. t <ε, In the conditions of Theorem 5.4, there exists ε> such that for C a,b (i) t i b A, with b p = 1. The proof of this proposition boils down to the analysis of the coefficients C a,b (i). We shall need some preliminary results and constructions presented in the following section. Our proof gives an estimate of ε which depends on a, d and p and it will be completed in Section Preliminary constructions and propositions Considering a fixed element a A as a linear mapping a : H H we look at two orthogonal decompositions of H associated to a and a, i.e. we consider the orthogonal subspaces K a = ker(a), andka = ker (a) of H and the orthogonal decomposition H = H a = Ka K a. Similarly, the orthogonal subspaces K a = ker(a ),andka = ker (a ) give the orthogonal decomposition H = H a = Ka K a. We look at any endomorphism m of H, as a linear mapping m : H a H a but look at its dual m,asm : H a H a. Choosing some fixed basis for each of the subspaces in the two decompositions, we give a matrix representation to any m A. In order to look closely at the estimates of ε, we shall choose an orthonormal basis of H a = Ka K a and another orthonormal basis of H a = Ka K a such that a a : H a H a and aa : H a H a are represented by diagonal matrices, both with non-negative entries (a a and aa are positive matrices).

10 34 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) With the basis chosen as above, we have that any m A is written as 2 2 block-matrix. Say m11 m m = 12, m 21 m 22 where m 11 represents a homomorphism m 11 : Ka K a and, so on with the other three blocks. Among the advantages of carefully choosing the orthogonal decomposition above is the ease of representation of the adjoint m as ( m m = 11 m ) 21 m 12 m, 22 where m 21 : K a K a is the adjoint of m 21 : Ka K a and so on. The special fixed elements a A and a are represented as α a = and a α = such that α α and αα are both diagonal square blocks with positive (real) diagonal entries (non-singular matrices). We notice also that the trace of the matrices representing m 1 m 2 : H a H a and m 1 m 2 : H a H a are the usual traces of these elements in A. In the matrix representation above, we refer to the block-coordinates of the elements of A. For example, if m A is given by m11 m m = 12 m 21 m 22 then the real and imaginary parts of the entries of m 11 shall be called the (real) 1 1-coordinates of m, and so on for the other three blocks of m. In what follows, we shall consider the special element a fixed, but we need to take b variable in A. We write y11 y b = 12 x y 21 ( and b y = 11 y21 ) y12 x. So we shall refer to the y ij and y ij variables as well as the x and x variables in expressions containing b and b, meaning the real coordinate variables of b in the corresponding blocks. Proposition 5.6. For 2 i p 1, any i-word W a,b (i) has the 1 2 and 2 1- coordinates of degree at most i 1 with respect to the variables {x,x } and has the 1 1 and 2 2-coordinates of degree at most i 2 with respect to the variables {x,x }. Proof. This is proved by studying all possible cases of i-words W a,b (i) and keeping track of the degrees which can occur in each block of the matrices. 1 1 The details have been checked by the referee, and are available from the authors.

11 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) Proposition 5.6 has the following corollaries. Corollary 5.7. For 2 i p 1, any i-word W a,b (i) has trace TW a,b (i) of degree at most i 2 with respect to the variables {x,x }. Proof. Immediate. Corollary 5.8. If the i-word W a,b (i) has order n in the symbols Λ, Λ 1 and Λ 2, with 3 i p 1, then TW a,b (i) is a sum of monomials of degree 2 with respect to the 1 1, 1 2 and 2 1-coordinates of b. Hence the coefficient C a,b (i) of t i in E p (a + tb) is a sum of monomials with degrees 2 with respect to the 1 1, 1 2 and 2 1-coordinates. Proof. By Proposition 5.6, the expression TW a,b (i) is a sum of monomials of degree i 2 with respect to the 2 2-coordinates. But the word W a,b (i) is homogeneous of order i with respect to b, and hence each entry of W a,b (i) is homogeneous of order i with respect to the coordinates of b. We conclude that TW a,b (i) is a sum of monomials with degrees 2 with respect to the 1 1, 1 2 and 2 1-coordinates of b. It follows immediately that C a,b (i) = TG a,b (i) is a sum of monomials with degrees 2 with respect to the 1 1, 1 2 and 2 1-coordinates of b. Lemma 5.9. The coefficient C a,b (2) = TG a,b (2) satisfies C a,b (2) M [ T(y11 y 11) + T(y12 y 12) + T(y21 y 21) ], where M> is a constant which depends only on d, n, and a. Furthermore, M may be taken as the minimum of the union of the spectra of the invertible operators (α α) n 1 and (αα ) n 1. Proof. Recall that for the block-matrix representation we have chosen, the matrices a a and aa are diagonal with non-negative entries (eigenvalues). Furthermore, the matrices α α and αα are non-singular, positive, diagonal matrices. We look at the summands in C a,b (2), which are the traces of all the 2-words. Consider the 2-word Λ 2 Λ n 1 given by (( y Λ 2 Λ n 1 = 11 y 11 + y21 y ) 21 (α α) n 1 ) ( y 12 y 11 + x ) y 21 (α α) n 1. Then ( (y T(Λ 2 Λ n 1 ) ) = T 11 y 11 (α α) n 1) ( (y ) + T 21 y 21 (α α) n 1) ( M 1 T(y 11 y 11 ) + T(y21 y 21) ), where M 1 = min{λ n 1 λ is eigenvalue of α α} >. Consider also the 2-word Λ 1 2 Λ n 2 given by

12 36 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) (( Λ 2 1 Λ n 2 (α = y 11 + y11 α)2 + α y 12 y12 α) (α α) n 2 ) y12 α ( α y 11 + y11 α) (α α) n 2. Then T(Λ 2 1 Λ n 2 ) = T((α y 11 + y11 α)2 (α α) n 2 ) + T(α y 12 y12 α(α α) n 2 ). But α y 11 + y11 α is symmetric, and hence (α y 11 + y11 α)2 is a positive (maybe singular) matrix. Then it has non-negative diagonal entries and (α y 11 + y11 α)2 (α α) n 2 has also non-negative diagonal entries and T((α y 11 + y11 α)2 (α α) n 2 ). On the other hand, T(α y 12 y12 α(α α) n 2 ) = T(y 12 y12 (αα ) n 1 ) M 2 T(y 12 y12 ), where M 2 = min{λ n 1 λ is eigenvalue of αα } >. Any other 2-word not considered so far has the form Λ l Λ 2 Λ n l 1 with <l n 1, or has the form Λ l Λ 1 Λ k Λ 1 Λ n l k 1 with l,k n 1and<l+ k n 1. These words have non-negative trace. To see this, one rewrites Λ as the square of a positive matrix r and chooses cyclic permutations of the factors in the words until they can be seen formally to be positive. We conclude that [ C a,b (2) M 1 T(y 11 y 11 ) + T(y21 y 21) ] [ + M 2 T(y 12 y 12 ) ] M [ T(y11 y 11) + T(y12 y 12) + T(y21 y 21) ], where M = min{m 1,M 2 } >. Lemma 5.1. For i {3,...,p 1}, the quotient C a,b (i) / C a,b (2) is bounded for all b A with b p = 1 and <C a,b (2). Proof. ByLemma5.9wehavethat C a,b (2) M ( T(y11 y 11) + T(y12 y 12) + T(y21 y 21) ) ( ) = M y y y Hence C a,b (i) C a,b (2) C a,b (i) M ( y y y 21 2 ). 2 Observe that y y y is the sum of the squares of all the 1 1, 1 2 and 2 1-(real) coordinates of b. By Corollary 5.8, we have that for i {3,..., p 1} the coefficient C a,b (i) is a sum of monomials of degree 2 with respect to the 1 1, 1 2 and 2 1-coordinates of b. Hence, C a,b (i) / C a,b (2) is bounded above in any set of b s, where <C a,b (2) and b p = 1 (this unit sphere is compact for we are in the finite dimensional case).

13 L.E. Mata-Lorenzo, L. Recht / Linear Algebra and its Applications 315 (2) We shall use the following basic result about polynomials. Lemma Let q(t) = t p + d t + d p 2 t p 2 + +d be any monic real polynomial of even degree p. If t is real and t > max{1, i= d i }, then q(t )>. Proof. This is a straightforward exercise on polynomials. Notice that q(t)is a monic real polynomial and has even degree The proof of Proposition 5.5 We shall use the change of variables t = 1/u to get 1 i 1 C a,b (i) t i = C a,b (i) = u u p C a,b (i) u p i. Recalling that C a,b (2) we shall consider two cases: <C a,b (2) and C a,b (2)=. 1. If <C a,b (2), we consider ĵ(u) = C a,b (i) C a,b (2) up i = u p i + i=3 C a,b (i) C a,b (2) up i, which has the same sign as C a,b(i) t i. By Lemma 5.1, for i {3,...,p 1} the fraction C a,b (i) / C a,b (2) is bounded for all b A with b p = 1and<C a,b (2). Hence there exists a constant δ 1 > such that C a,b (i) C a,b (2) <δ 1 i=3 for all b A with b p = 1and<C a,b (2). By Lemma 5.11, we have that ĵ(u) > if u >δ 2 = max{1,δ 1 }. In this case, we can take ε = 1/δ 2 >. 2. If C a,b (2) =, then, by the condition for degeneracy of the Hessian as described in Corollary 5.2, we have that C a,b (i) = fori {3,...,p 1}. Hence, C a,b (i) t i = t R. In this case, we can take ε = 1 (or any positive number). The proof of Proposition 5.5 is complete. References [1] E. Andruchow, A. Larotonda, L. Recht, D. Stojanoff, Infinite dimensional homogeneous reductive spaces and finite index conditional expectations, Illinois J. Math. 41 (1997)

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