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volume 6, issue 3, article 87, 2005.

Received 16 July, 2004;

accepted 29 June, 2005.

Communicated by:F. Hansen

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Journal of Inequalities in Pure and Applied Mathematics

A NUMERICAL METHOD IN TERMS OF THE THIRD DERIVATIVE FOR A DELAY INTEGRAL EQUATION FROM BIOMATHEMATICS

MARKUS SIGG

Department of Mathematics and Statistics University of Konstanz

Germany.

EMail:mail@MarkusSigg.de

c

2000Victoria University ISSN (electronic): 1443-5756 135-04

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A Minkowski-Type Inequality for the Schatten Norm

Markus Sigg

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J. Ineq. Pure and Appl. Math. 6(3) Art. 87, 2005

http://jipam.vu.edu.au

Abstract

LetF be a Schattenp-operator andR, Spositive operators. We show that the inequality|F(R+S)¹^c|_p^c≤ |F R¹^c|_p^c+|F S¹^c|_p^cfor the Schattenp-norm| · |_pis true forp≥c= 1and forp≥c= 2, conjecture it to be true forp≥c∈[1,2], give counterexamples for the other cases, and present a numerical study for2×2 matrices. Furthermore, we have a look at a generalisation of the inequality which involves an additional factorσ(c, p).

2000 Mathematics Subject Classification:47A30, 47B10

Key words: Schatten class, Schatten norm, Norm inequality, Minkowski inequality, Triangle inequality, Powers of operators, Schatten-Minkowski constant.

This work was supported by the Dr. Helmut Manfred Riedl Foundation.

1. Introduction

Let H and K be complex Hilbert spaces and 0 < p ≤ ∞. Following [1], we denote bycp(H, K)the space of Schattenp-operatorsT :H −→K, equipped with the Schattenp-norm or quasi-norm| · |_p. Note that [1] deals only with the spacesc_p(H) :=c_p(H, H). The generalisationsc_p(H, K)can be found in text- books like [2] and [3] (there written asB_p(H, K)andS_p(H, K)respectively).

By L(H) we denote the space of bounded linear operators on H, and by L(H)₊ the subset of positive operators. With |T| := (T^∗T)^1/2 ∈ L(H)₊ for T ∈L(H, K)we have forp <∞

|T|^p_p = tr|T|^p forT ∈c_p(H, K), and consequently

|T|^p_p = tr T^p forT ∈c_p(H)₊:=c_p(H)∩L(H)₊. Applying|T|_p =|T^∗|_p forT ∈c_p(H, K), this shows in case ofp < ∞

F U¹²

2 p =

U¹²F^∗

2 p =

tr (F U F^∗)^p² _p²

=|F U F^∗|^p

2

forF ∈c_p(H, K)andU ∈L(H)₊. Because| · |_∞is the usual operator norm,

F U¹²

2 p

=|F U F^∗|^p

2

is also true forp=∞, with the common convention ^∞₂ :=∞.

Our question, which arose while studying the integration of Schatten operator valued functions in [4], is: For what values ofp∈ (0,∞]andc∈(0,∞)is

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the Minkowski-like inequality (MS)

F (R+S)¹^c

c p

≤ F R¹^c

c p

+ F S¹^c

c p

true for allF ∈c_p(H, K)andR, S ∈L(H)₊?

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2. The Conjecture

LetH, K, p, c, F, R, Sbe as above.

Theorem 2.1. Inequality (MS) is true forp≥c= 1and forp≥c= 2.

Proof. Forp≥c= 1, the triangle inequality for| · |_p shows

|F(R+S)|_p =|F R+F S|_p ≤ |F R|_p+|F S|_p. Forp≥c= 2, the triangle inequality for| · |^p

2 shows

F(R+S)¹²

2

p =|F(R+S)F^∗|^p

2 ≤ |F RF^∗|^p

2+|F SF^∗|^p

2 =

F R¹²

2 p+

F S¹²

2 p.

Theorem2.1suggests the following conjecture.

Conjecture 1. Inequality (MS) is true forp≥c∈[1,2].

For c ∈ (1,2)we have at the present time no proof of this conjecture for other than trivial situations, not even for the special case of 2× 2 matrices.

However, some justification will be given in Section4.

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3. The Case p < c and the Case c 6∈ [1, 2]

In this section we will demonstrate, by providing counterexamples, that inequality (MS) is not necessarily true for other values of (c, p) than those stated in Conjecture 1. We will offer one example for 0 < p < c < ∞, and one for arbitrary p when c < 1 or c > 2, both examples using 2×2 matrices. The power U^t for t > 0of a non-negative matrix U can be calculated easily with help of the spectral decomposition ofU.

Example 3.1. Inequality (MS) is violated for0< p < c < ∞by F :=

1 0 0 1

, R :=

1 0 0 0

, S :=

0 0 0 1

. Proof. FromU^t =U forU ∈ {R, S, R+S}andt∈(0,∞)we get

F U¹^c

p =|U|_p = (trU^p)¹^p = (trU)¹^p, yielding

F R¹^c

p

= 1, F S¹^c

p

= 1,

F(R+S)¹^c p

= 2^p¹, and usingp < c,

F R¹^c

c p+

F S¹^c

c

p = 2 <2^c^p =

F(R+S)¹^c

c p.

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The second example makes use of an inequality which is interesting in its own right. Seeming simple, it is surprisingly fiddly to prove:

Lemma 3.1. Forx∈(0,1)∪(2,∞)we have

1 + 1

√5

3 +√ 5 2

!x

+

1− 1

√5

3−√ 5 2

!x

<1 + 3^x. Proof. Settingr := √

5,α₁ := 1 + ¹_r,α₂ := 1− ¹_r, andω := ^3+r₂ , we have to show

α₁ω^x+α₂ω^−x <1 + 3^x.

The case x ∈ (2,∞): Setf(x) := α₁ω^x, g(x) := α₂ω^−x, h(x) := 1 + 3^x for x ∈ (0,∞). Because α₂ > 0 and ω > 1, g is strictly decreasing, thus f(x) +g(x) < f(x) +g(2)forx > 2. We will showf(x) +g(2) < h(x)for x > 2. Because f(2) +g(2) = h(2), this is done if we prove f⁰(x) < h⁰(x) for x > 2, which is equivalent to α₁(^ω₃)^x lnω < ln 3. This inequality is true forx = 2. All factors of its left side are positive, andω < 3, so the left side is strictly decreasing forx≥2. Hence the inequality is true forx >2as well.

The casex ∈ (0,1): After substitutings := ω^x and settingδ := _ln^{ln 3}_ω, we have to prove the equivalent inequality

s+ 1 s + 1

r

s− 1 s

<1 +s^δ

fors ∈(1, ω), which can be done by building a sandwich with a suitable poly-

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nomial function inside: Set ϕ(s) :=s+ 1

s + 1 r

s−1

s

, p(s) := 2

1 + s−1 ω−1

, q(s) := (s−1)(s−ω)

(3−1)(3−ω)(ϕ(3)−p(3)) fors >0. The claim is

ϕ(s)< p(s) +q(s)<1 +s^δ

fors ∈ (1, ω). The left inequality is verified by the fact thats·(p(s) +q(s)− ϕ(s))defines a polynomial of degree 3with three zeros {1, ω,3}, where 1 <

ω < 3, and with positive leading coefficientλ := ¹₂(ϕ(3)−p(3))/(3−ω). To prove the second inequality, we inspect

ψ(s) := 1 +s^δ−p(s)−q(s)

for s > 0and getψ⁰⁰(s) = δ(δ−1)s^δ−2 −2λ. Because 1 < δ < 2, ψ⁰⁰ has a unique zero

s₀ :=

δ(δ−1) 2λ

_2−δ¹

, 1< s₀ < ω,

withψ⁰⁰(s)>0fors ∈(0, s0)andψ⁰⁰(s)<0fors ∈(s0,∞). Nowψ(1) = 0, ψ⁰(1) >0, andψ⁰⁰(s) >0fors ∈(1, s₀)showψ(s) >0fors ∈(1, s₀], while ψ(s₀)>0,ψ(ω) = 0,ψ⁰(ω)<0, andψ⁰⁰(s)<0fors∈(s₀, ω)showψ(s)>0 fors∈[s0, ω).

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Example 3.2. Inequality (MS) is violated for0< p≤ ∞andc <1as well as c > 2by

F :=

0 0 1 0

, R:=

0 0 0 1

, S :=

2 −1

−1 1

. Proof. Evaluation of the matrix powers fort ∈(0,∞)gives

R^t=R, S^t =

1

2(α1ω^t+α2ω^−t) ¹_r(ω^−t−ω^t)

1

r(ω^−t−ω^t) ¹₂(α₂ω^t+α₁ω^−t)

! ,

(R+S)^t= 1 2

1 + 3^t 1−3^t 1−3^t 1 + 3^t

!

withr:=√

5,α₁ := 1 +¹_r,α₂ := 1−¹_r,ω := ^3+r₂ . ForU ∈ {R, S, R+S}we get in case ofp <∞

|F U^t|_p =

tr (F U^2tF^∗)^p²¹_p

=√ u_t

withu_t being the top left entry ofU^2t. Using|F U^t|²_∞ = |F U^2tF^∗|_∞, the case p=∞yields the same result, thus for allp:

F R¹^c

p

= 0, F S¹^c

p

= r1

2(α₁ω^2/c+α₂ω^−2/c),

F(R+S)¹^c p =

r1

2(1 + 3^2/c).

Substituting ²_c byx, we have to proveα₁ω^x+α₂ω^−x <1 + 3^x forx∈(2,∞) and forx∈(0,1), which is the statement of Lemma3.1.

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4. Some Numerical Evidence

To justify Conjecture 1, we present the results of a numerical study performed with2×2matrices.

From functional calculus it is known: For an operatorT ≥ 0on a complex Hilbert space the powers T^α, T^β for α, β ∈ (0,∞) obey the rule T^αT^β = T^α+β. IfT is invertible, thenT^αcan be defined forα≤0as well, andT^αT^β = T^α+β is true for allα, β ∈R.

Before turning to the matrix case, we note the following general lemma.

Lemma 4.1. LetH, K, F be as above andα∈(0,∞).

(a) LetT ∈L(H)₊. ThenF T^α = 0if and only ifF T = 0.

(b) Let R, S ∈ L(H)₊. ThenF(R+S)^α = 0if and only if F R^α = 0 and F S^α = 0.

Proof. (a) SupposeF T^α = 0. Then|F T^α/2|² =|F T^αF^∗|= 0, henceF T^α/2 = 0. Repeated application yieldsβ∈(0,1)withF T^β = 0, thusF T =F T^βT^1−β

= 0.

Now suppose F T = 0. There is nothing to prove in the case of α = 1, so assume α 6= 1. IfT is invertible, then F T^α = F T T^α−1 = 0. IfT is not invertible, then we have0∈σ(T), the spectrum ofT. Choose polynomialsf_n∈ R[t]forn ∈Nsuch thatf_n(x)→x^αforn → ∞uniformly forx∈σ(T). Then f_n(T)→ T^α andF f_n(T)→ F T^α forn → ∞, henceF f_n(T) =f_n(0)F → 0 forn→ ∞, thusF T^α = 0.

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(b) Part (a) shows:

F R^α = 0 ∧ F S^α = 0 ⇐⇒ F R= 0 ∧ F S = 0

=⇒ F(R+S) = 0

⇐⇒ F(R+S)^α = 0.

To prove the missing implication, suppose F(R + S) = 0. Then F RF^∗ + F SF^∗ = 0. Because F RF^∗ ≥ 0 and F SF^∗ ≥ 0, we getF RF^∗ = 0, thus

|F R^1/2|² = |F RF^∗| = 0andF R^1/2 = 0. Applying (a) again givesF R = 0.

Symmetry showsF S = 0.

We will also use the following well-known property of2×2matrices:

Lemma 4.2. A complex 2×2matrixM is positive semidefinite if and only if there exista, b∈[0,∞)andγ ∈Cwith|γ|² ≤absuch that

M =

a γ γ b

.

Lemma4.1(b) shows that, when checking Conjecture1, one may assume the denominator to be non-zero, or setting ⁰₀ := 0, in

q_c,p(F, R, S) :=

F (R+S)¹^c

c p

F R¹^c

c p

+ F S¹^c

c p

.

We are searching for the supremum ofq_c,p(F, R, S)over all complex2×2 matricesF, R, S withR, S ≥ 0. Forr ∈ [0,∞)andx ∈ Cdefiner∧x :=x

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if |x| ≤ rand r∧x := (r/|x|)xotherwise. Lemma4.2shows that Rhas the structure

R=

α² |α β| ∧γ

|α β| ∧γ β²

=:P(α, β, γ)

with α, β ∈ R and γ ∈ C, and a corresponding representation is valid for the matrix S. This means that we have to deal with six complex and four real variables, resulting in a 16-dimensional real optimisation problem: For λ = (λ₁, . . . , λ₁₆)∈R¹⁶we set

Fλ :=

λ1 +λ2i λ3+λ4i λ₅ +λ₆i λ₇+λ₈i

, R_λ :=P(λ₉, λ₁₀, λ₁₁+λ₁₂i), S_λ :=P(λ₁₃, λ₁₄, λ₁₅+λ₁₆i) and are asking for

σ(c, p) := sup

λ∈R¹⁶

q_c,p(F_λ, R_λ, S_λ).

To attack this problem, GNU Octave [5], version 2.1.57, was utilised. It offers a function for determining the singular values of a matrix, which can be employed for calculating the Schatten norms. For the optimisation task the implementation [6], version 2002/05/09, with standard parameters of the Down- hill Simplex Method of Nelder and Mead ([7], 10.4) was used. The results are in perfect agreement with Conjecture 1. For visualisation, approximations for σ(c, p)forc ∈ {1.2,1.4,1.6,1.8,2.0}have been calculated and plotted with a step size of0.01forp, see Figure1.

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Figure 1: Experimental approximations ofσ(c, p).

0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2

1 1.2 1.4 1.6 1.8 2

p

c = 1.2 c = 1.4 c = 1.6 c = 1.8 c = 2.0

The apparently smooth shape ofp 7→ σ(c, p)for p ≤ c, together with the fact that for eachpa new random starting pointλwas used for the Nelder-Mead algorihm, gives some confidence in the validity of the data.

A closer inspection of some of the calculated numerical values suggests σ(2,1) = 2, σ ³₂,1

=σ ⁹₅,⁶₅

= 2¹², σ ⁸₅,⁶₅

=σ 2,³₂

= 2¹³, σ ⁵₄,1

=σ ³₂,⁵₄

=σ ⁷₄,⁷₅

=σ 2,⁸₅

= 2¹⁴, σ ⁶₅,1

=σ ⁹₅,³₂

= 2¹⁵, which leads to the idea to look at log₂ σ(c, p). It seems there is a linear depen-

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dency oflog₂ σ(c, p)fromcifc≥p. This observation will be made precise in the next section.

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5. Generalisation of (MS)

It is natural to generalise (MS) and to ask for the smallestσ(c, p) ∈ [0,∞]for c∈(0,∞)andp∈(0,∞]such that

F (R+S)¹^c

c p

≤σ(c, p)

F R¹^c

c p

+ F S¹^c

c p

for all F ∈ c_p(H, K)and R, S ∈ L(H)₊ (and for all complex Hilbert spaces H and K). It is tempting to call σ(c, p) the Schatten-Minkowski constant for (c, p). By choosingF 6= 0and settingRto be the identity andS := 0it can be seen thatσ(c, p) ≥ 1. Now Conjecture1can be re-phrased usingσ(c, p), and, motivated by the numerical results, we add another conjecture:

Conjecture 2. (a) For1≤c≤2andp≥cwe haveσ(c, p) = 1.

(b) For0≤c≤2andp≤cwe haveσ(c, p) = 2^c^p⁻¹. Again, the casesc= 1andc= 2are not too difficult to prove:

Theorem 5.1.

(a)σ(1, p) =

(1 forp≥1 2¹^p⁻¹ forp≤1 (b)σ(2, p) =

(1 forp≥2 2²^p⁻¹ forp≤2.

Proof. σ(1, p) ≤ 1 for p ≥ 1 and σ(2, p) ≤ 1 for p ≥ 2 is the subject of Theorem 2.1, while σ(c, p) ≥ 1 is noted above. Example 3.1 tells us that

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σ(c, p)≥2^c/p−1for0< p ≤c <∞, yielding

σ(1, p)≥2¹^p⁻¹ forp≤1 and σ(2, p)≥2^p²⁻¹ forp≤2.

Now for the missing ‘≤’ inequalities. For the casec = 1, recall the inequality between the power means of degrees p ≤ 1 and 1, see e.g. [8], 8.12, which reads

α^p+β^p 2

¹_p

≤ α+β

2 or equivalently α^p+β^p ≤2^1−p(α+β)^p forα, β ∈[0,∞). Together with the quasi-norm inequality of| · |_p this gives

|F(R+S)|^p_p ≤ |F R|^p_p+|F S|^p_p ≤2^1−p(|F R|_p+|F S|_p)^p and thus|F(R+S)|_p ≤2¹^p⁻¹(|F R|_p+|F S|_p).

For the casec = 2, start with the power means inequality for the degrees p≤2and2,

α^p+β^p 2

¹_p

≤

α²+β² 2

¹₂

or equivalently α^p+β^p ≤2¹⁻^p² (α²+β²)^p² forα, β ∈[0,∞). Together with the quasi-norm inequality of| · |^p

2 this gives

F(R+S)¹²

p p

=|F(R+S)F^∗|

p 2p 2

≤ |F RF^∗|

p p2 2

+|F SF^∗|

p 2p 2

= F R¹²

p p+

F S¹²

p

p ≤2¹⁻^p²

F R¹²

2 p+

F S¹²

2 p

^p₂

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and consequently

F(R+S)¹²

2 p

≤2²^p⁻¹

F R¹²

2 p

+ F S¹²

2 p

.

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6. Conclusion

Starting with Conjecture 1, which we proved for the cases c = 1 and c = 2 in Theorem 2.1, a numerical study of 2 ×2 matrices led to the generalised Conjecture2, which we also proved forc= 1andc= 2in Theorem5.1.

The given proofs make use of the (quasi-) triangle inequality of the Schatten (quasi-) norm. Another ingredient to Theorem 5.1is the power means inequality. Presumably, a combination of these inequalities shall also be central when dealing with the case c 6= 1,2. However, it is unclear how to apply the triangle inequality in this situation, because there is no obvious way to get from F(R+S)^1/cto an expression whereRandScan be separated.

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References

[1] C.A. MCCARTHY,c_p, Israel J. Math., 5 (1967), 249–271.

[2] J. WEIDMANN, Linear Operators in Hilbert Spaces, Springer, New York 1980.

[3] R. MEISE ANDD. VOGT, Introduction to Functional Analysis, Clarendon Press, Oxford 1997.

[4] M. SIGG, Zur sesquilinearen Integration Schatten-operatorwertiger Funk- tionen, Konstanzer Schriften in Mathematik und Informatik, 200 (2004), http://www.inf.uni-konstanz.de/Preprints/.

[5] J.W. EATON et al,http://www.octave.org.

[6] E. GROSSMANN, http://octave.sourceforge.net/index/

f/nelder_mead_min.html.

[7] W.H. PRESS, B.P. FLANNERY, S.A. TEUKOLSKYANDW.T. VETTER- LING, Numerical Recipes in C: The Art of Scientific Computing, Cam- bridge University Press 1992,http://www.nr.com.

[8] J. HERMAN, R. KU ˇCERA AND J. ŠIMŠA, Equations and Inequalities, Springer, New York (2000).