Various Proofs for the Decrease Monotonicity of the Schatten s Power Norm, Various Families of R n Norms and Some Open Problems

Transcription

1 Int. J. Oen Problems Comt. Math., Vol. 3, No. 2, June 2010 ISSN ; Coyright c ICSRS Publication, Various Proofs for the Decrease Monotonicity of the Schatten s Power Norm, Various Families of R n Norms and Some Oen Problems Mustaha Raïssouli and Iqbal H. Jebril Alied Functional Analysis Team, AFACSI Laboratory, My Ismaïl University, Faculty of Sciences, P.O.Box 11201, Menès, Morocco raissouli 10@hotmail.com Deartment of Mathematics, King Faisal University, Saudi Arabia ijebril@fu.edu.sa Abstract Let 1 be a extended) real number, x = x 1, x 2,..., x n ) R n and the ma x := x i. In the resent aer, some different roofs for the decrease monotonicity of x are given. Afterwards, we construct an iterative algorithm that converges oint-wisely to a arameterized norm. At the end, the interconnection between. and, together with extensions for 2 norms, are utted as oen roblems. Keywords: Norms of R n, monotonicity, normalized norm, algorithm, oint-wise convergence. MSC 2000): 47A63. 1 Introduction Let R n be the real finite dimensional sace. In the literature, three secial norms of R n are well nown: for x = x 1, x 2,..., x n ) R n, ) 1/2 x 1 = x i, x 2 = x i 2, x = max x i, 1 i n which satisfy the following inequalities x x 2 x 1 n x.

2 Various Proofs for the Decrease Monotonicity of The second norm. 2, so-called euclidian norm, derives from the standard inner roduct defined by x = x 1, x 2,..., x n ) R n, y = y 1, y 2,..., y n ) R n, x, y = x i y i. There are many other norms of R n, so-called Schatten ower norm, extending the above three ones: for 1 extended) real number, we set x = x i. Such arameterized norm immediately gives. 1 and. 2 for = 1 and = 2 resectively, and extends. in the following sense x R n x = lim x. If we denote by the conjugate of defined by 1/+1/ = 1 i.e = / 1) with the convention = 1, = and =, = 1, the following inequality, so-called the Hölder inequality in R n, is well nown: for all x, y R n there holds x, y x y. For = = 2, the above Hölder inequality is exactly the classical Cauchy- Schwartz one. Further, it is not difficult to rove the following inequality: x R n x 1 x x. 1.1) Indeed, writing x i = x i 1 x i ) 1 max x i 1 i n x i, we deduce that x i ) 1)/ max x i 1 i n x i, which, with the Young inequality, yields the desired result 1.1). Finally, we may state the following theorem that is also nown in the literature. Theorem 1.1. Let 1 be a extended) real number. For fixed vector x := x 1, x 2,..., x n ) R n, the maing x := x i, is monotone decreasing, that is, > q 1 = x x q.

3 166 Mustaha Raïssouli and Iqbal H. Jebril In the following section, we will resent some different roofs of the above theorem. For this, we state the next lemma which will be needed in the sequel. Lemma 1.2. Let c 1, c 2,..., c N be N ositive real numbers and 0 < m < 1. Then there holds N ) m N c i c m i. Proof. It is a simle exercise for the reader. 2 Various Proofs of Theorem 1.1 In what follows, we will resent seven different roofs of Theorem 1.1. Proof 1. The short roof, based on Lemma 1.2, is first as follows. If > q 1 then 0 < q/ < 1 and Lemma 1.2 with m = q/, c i = x i gives The desired result follows. ) q/ x i x i q. Proof 2. The second roof, simle and elementary, is based on the homogeneity rincile of a norm. We can assume without loss the generality that x q = 1. Then, x i 1 and so x i x i q, for all i = 1, 2,..., n. It follows that x i x i q = 1, and so which is the desired result. x i 1 = x q, Proof 3. The third roof is an exlicit and manual statement of the above one. Setting x i = a i 0, we can assume that a i > 0. We write successively n n a i )1/ = a i n aq i )1/q aq i )/q = )/q aq i )/q = a q i aq i ) /q.

4 Various Proofs for the Decrease Monotonicity of Since 0 aq i 1 and /q 1, we deduce that a q i aq i ) /q Summing over i from 1 to n we infer that. aq i a q i aq i ) /q aq i = 1, from which we deduce a q i aq i ) /q 1. The desired result follows. Proof 4. The fourth roof is also simle and elementary, with exlicit comutations. We can assume that a i := x i > 0 for all i = 1, 2,..., n. Let us set )) 1 Ψ) = ex ln a i, for ]1, + [. We wish to establish that the ma Ψ) is monotone decreasing. A simle comutation of Ψ ), after a reduction, yields a i ln a i 1ln The fact that Ψ ) = 1 for all i = 1; 2,..., n imlies that a i a i a i, ln a i 1 ln a i, for all i = 1, 2,..., n, and thus a i ln a i 1 ln a i ) a i ) a i ) a i. Ψ).

5 168 Mustaha Raïssouli and Iqbal H. Jebril Combining the revious, we then have roved which yields the desired result. ]1, [ Ψ ) 0, Proof 5. Here, we use a mathematical induction on n. For n = 1 the desired result is trivial. Assume that, for > q, We can write +1 ) 1/q a i. 1/ a i = a i +1) + a, which, according to the induction hyothesis, becomes +1 ) /q a i + a +1 1/, or again a i +1 ) /q + a +1 q/ 1/q. Now, Lemma 1.2 with yields the following N = 2, m = q/ ]0, 1[, c 1 = +1 1/q a i n+1) + aq. Summarizing what revious, the desired result follows. ) /q, c 2 = a +1 Proof 6. This roof is based on a duality rincile using the Hölder inequality. It is easy to deduce that x = max { x, y, y 1} = max{ x, y, y i 1}.

6 Various Proofs for the Decrease Monotonicity of If q then q and so there holds {y R n, y i 1} {y R n, y i q 1}. Then we can write x = max{ x, y, y i 1} max{ x, y, y i q 1} = x q, which is the desired result. Proof 7. Here we only resent the ey idea of this roof and we omit the details to the reader. First, we easily show the theorem when and q are both integers. Secondly, if and q are rational the roof of the desired result can be reduced, with a simle maniulation, to the above case. Finally, for and q real numbers, it is sufficient to remar that the ma x, for fixed x R n, is continuous and the desired result follows by density of Q in R. 3 Another Family of R n Norms As already ointed, the aim of this section turns out of to construct another family of norms in R n indexed by [0, + ]. We need additional basic notions which we will state below. Let α be a norm in R n. The dual of α is the norm α defined by x R n α x, y x) = max{ x, y, αy) 1} = max{ αy), y 0}. It is easy to see that α := α ) = α and, if α and β are two norms of R n such that α β i.e αx) βx) for every x R n ) then α β. The norm α will be called a normalized norm if αe i ) = 1 for all i = 1, 2,..., n, where e 1, e 2,..., e n ) refers to the canonical basis of R n. It is clear that the set of all normalized norms of R n is not stable for the oeration addition but it is a convex set. It is easy to see that if α is a normalized norm then so is α. The above norm. is a normalized norm for all 1, with the relationshi. ) =.. In articular,. 1 =.,. =. 1,. 2 =. 2, and it is not hard to verify that. 2 is the unique self-dual norm of R n. A sequence α ) of norms in R n will be called oint-wise) converging if the two following conditions are both satisfied i) For all x R n, α x) converges in R, ii) Setting αx) := lim α x), then α defines a norm in R n. It is clear that i) doesn t ensure ii). In fact, if α ) satisfies i) with lim α x) =

7 170 Mustaha Raïssouli and Iqbal H. Jebril αx) then α is only a semi-norm i.e the searation axiom of a norm is not generally satisfied by α). For examle, α x) = x / + 1), for a given norm. of R n, satisfies i) but not ii). However, the following result holds. Lemma 3.1. Let α ) be a sequence of normalized norms of R n with lim α x) = αx) for every x R n. Then, α is a normalized norm of R n. If moreover α ) is monotone decreasing then α ) ) converges to α. Proof. It is a simle exercise for the reader. The following lemma will be needed in the sequel. Lemma 3.2. Let α and β be two norms of R n. Then, for all 1, there holds 1 α + 1 ) β 1 α + 1 β. 3.1) If α β then the ma 1 α + 1 β is oint-wise) monotone decreasing. Proof. By virtu of the definition of the dual norm, it is sufficient to rove that 1 a + 1 ) 1 b 1 a b 1, for all real numbers a > 0, b > 0. Since the real maing x 1/x is convex on ]0, + [, the desired inequality follows. The second art of the lemma is immediate so comletes the roof. Now, for 1, we define a sequence ) by the following iterate rocess +1 = ), 0; 0 = ) It is easy to see that is a normalized norm for all and 0, with 1 = It is clear that 1 =. 1 and =. for all 0. Below, we can then assume that 1 < <. Proosition 3.3. For all 1 < <, the following assertions are met i) For every 0, ) ii) The sequence ) is monotone decreasing.

8 Various Proofs for the Decrease Monotonicity of Proof. i) For = 0 it is trivial. By 3.2) with Lemma 3.2, we obtain ) +1 1 ) 1 + = +1, which gives the desired inequality. ii) Substituting the above inequality in 3.2), we immediately deduce the decrease monotonicity of the sequence ). This comletes the roof. Theorem 3.4. The norm sequence ) converges oint-wisely) to a normalized norm of R n, with the following estimation ). 3.3) Proof. By Proosition 3.3, for all 0 we have.... ) 1 ) ) It follows that, for all x R n, the real sequences x) ) is monotone decreasing and lower bounded and so converges in R. Since is a normalized norm, the first art of Lemma 3.1 tells us that ) converges oint-wisely) to a normalized norm denoted by. To rove the estimation 3.3), we first remar that ), which with the fact that +1 = 1 yields the following double inequality 0 +1 ) 1 ) + ), 1 ), for all 0. The desired result follows by a mathematical induction with a simle maniulation. The roof of the theorem is comlete. Corollary 3.5. For all 1 < <, the following roerties hold true i) ii) ) =. In articular, one has 2 =. 2.

9 172 Mustaha Raïssouli and Iqbal H. Jebril Proof. i) By the decrease monotonicity of ) we have 1 = Letting in this later inequality we obtain the desired result. ii) Since ) converges to for every 1 < < then ) converges to. Algorithm 3.2), with the second art of Lemma 3.1, yields when the first relation of ii). In articular, if = 2 then = 2 and so 2 ) = 2. Since. 2 is the unique self-dual norm of R n, the second relation of ii) is so obtained, thus concludes the roof. Proosition 3.6. For 1 < <, the ma is monotone decreasing. Proof. By a mathematical induction on 1, we wish to establish that q for q. For = 1, it is immediate from the second art of Lemma 3.2, since.. 1. Now, assume that for and q such that q we have q. One has +1 = ) 1 q + 1 ). q 3.5) According to Proosition 3.3,i), we have q ) q which, with the second art of Lemma 3.2 and 3.5) yields +1 1 q q + 1 q q ) = +1 q. Summarizing, for all 0 and q we have q. Letting in this latter inequality we deduce the desired result. A Hölder tye inequality satisfied by the norm is recited in the following. Proosition 3.7. Let 1 < < and x, y R n then one has x, y x) y) Proof. Algorithm 3.2), with the definition of the dual of a norm, gives +1 x) 1 x) + 1 ) x) 1 x) + 1 x, y y), for all x, y R n with y 0. By Young inequality we deduce that +1 x) x) x, y 1/ y). This, when and after a simle reduction, yields the desired inequality thus comletes the roof.

10 Various Proofs for the Decrease Monotonicity of Some Oen Problems As we have seen throughout the revious study, the norms. and have similar roerties. In summary, we have obtained the following: 1.. and are both normalized norms of R n, for all 1, 2.. and are both monotone decreasing, 3.. ) =. and ) = for all 1, and , for all 1, 5. x, y x y and x, y x) y), for all 1 and x, y R n, = 1,. 2 = 2 and. =. This allows us to ut the following. Oen Problem 1. What is the exlicit form of? What is the interconnection between. and? Is it ossible to comare algebraically) these two norms? Now, in order to state a conjecture we need some additional notions about tensorial roduct theory. For further details, the reader can consult [1, 2] for examle. Let n 2 be an integer and a chosen decomosition n = rs with r, s integers. Then, R n = R r R s and every x R n can be written as follows x = N y i z i, y i R r, z i R s, N 1. Let 1 < < and. R r and. R s be two fixed norms of R r and R s resectively. For all x R n we define d r,s x) = inf x= P N y i z i N { N }) y i R su 1/ z r i, t, t R s 1, where the inf is taen over all finite decomosition i y i z i of x R n = R r R s. For fixed integers r, s such that n = rs, the above exression of d r,s defines a

11 174 Mustaha Raïssouli and Iqbal H. Jebril norm on R n, called cross norm on R r R s. After these notions, our conjecture is now recited in the following. Conjecture. Let n 2 be a integer and 1 < <. Then there exist r, s integers with n = rs and two norms. R r,. R s such that = d r,s. Finally, we briefly recall the notions of 2 norms in order to state our second oen roblem. For further details concerning the introduction of 2 norms, the reader can consult [3] for examle. Let E be a linear vector sace on a field K = R, C. A binary ma.,. defined from E E into [0, + [ is called 2 norm if the four following conditions are simultaneously satisfied: i) u, v = 0 if and only if u and v are linearly deendent, ii) u, v = v, u for all u, v E, iii) λ.u, v = λ. u, v for all u, v E and λ K, iv) u + v, w u, w + v, w for all u, v, w E. Now, our question arising from the above can be recited as follows: Oen Problem 2. What should be the reasonable analogues of the above family of norms for 2 norms? In this direction, we invite the reader to consult [4]. References [1] E.N.Cheney and W.A.Light, Aroximation Theory in Tensor Product Saces, Lecture Notes in Mathematics, [2] R.Schatten, A Theory of Cross Saces, Princeton University Press, New- Yor, [3] S.Gähler, Lineare 2 Normiete Raume. Math. Nachr ), [4] I.H.Jebril and M.Raïssouli, An Algorithm Converging to a Famify of 2 Norms. In rearation.