Sum of squared logarithms - an inequality relating positive definite matrices and their matrix logarithm

Mircea Bîrsan; Patrizio Neff; Johannes Lankeit

doi:10.1186/1029-242X-2013-168

Impact Factor 0.791

Journal of Inequalities and Applications Cover Image

Research

Open Access

Sum of squared logarithms - an inequality relating positive definite matrices and their matrix logarithm

Mircea Bîrsan^{1, 2},
Patrizio Neff¹Email author and
Johannes Lankeit¹

Journal of Inequalities and Applications20132013:168

https://doi.org/10.1186/1029-242X-2013-168

Received: 21 January 2013

Accepted: 28 March 2013

Published: 12 April 2013

Abstract

Let $y_{1}, y_{2}, y_{3}, a_{1}, a_{2}, a_{3} \in (0, \infty)$ be such that $y_{1} y_{2} y_{3} = a_{1} a_{2} a_{3}$ and

y_{1} + y_{2} + y_{3} \geq a_{1} + a_{2} + a_{3}, y_{1} y_{2} + y_{2} y_{3} + y_{1} y_{3} \geq a_{1} a_{2} + a_{2} a_{3} + a_{1} a_{3} .

Then

{(log y_{1})}^{2} + {(log y_{2})}^{2} + {(log y_{3})}^{2} \geq {(log a_{1})}^{2} + {(log a_{2})}^{2} + {(log a_{3})}^{2} .

This can also be stated in terms of real positive definite $3 \times 3$ -matrices $P_{1}$ , $P_{2}$ : If their determinants are equal, $det P_{1} = det P_{2}$ , then

tr P_{1} \geq tr P_{2} and tr Cof P_{1} \geq tr Cof P_{2} ⟹ {∥ log P_{1} ∥}_{F}^{2} \geq {∥ log P_{2} ∥}_{F}^{2},

where log is the principal matrix logarithm and ${∥ P ∥}_{F}^{2} = \sum_{i, j = 1}^{3} P_{i j}^{2}$ denotes the Frobenius matrix norm. Applications in matrix analysis and nonlinear elasticity are indicated.

MSC:26D05, 26D07.

Keywords

matrix logarithm elementary symmetric polynomials inequality characteristic polynomial positive definite matrices means

1 Introduction

Convexity is a powerful source for obtaining new inequalities; see, e.g., [1, 2]. In applications coming from nonlinear elasticity, we are faced, however, with variants of the squared logarithm function; see the last section. The function ${(log (x))}^{2}$ is neither convex nor concave. Nevertheless, the sum of squared logarithms inequality holds. We will proceed as follows: In the first section, we will give several equivalent formulations of the inequality, for example, in terms of the coefficients of the characteristic polynomial (Theorem 1), in terms of elementary symmetric polynomials (Theorem 3), in terms of means (Theorem 5) or in terms of the Frobenius matrix norm (Theorem 7). A proof of the inequality will be given in Section 2, and some counterexamples for slightly changed variants of the inequality are discussed in Section 3. In the last section, an application of the sum of squared logarithms inequality in matrix analysis and in the mathematical theory of nonlinear elasticity is indicated.

2 Formulations of the problem

All theorems in this section are equivalent.

Theorem 1 For

n = 2

or

n = 3

let

P_{1}, P_{2} \in R^{n \times n}

be positive definite real matrices. Let the coefficients of the characteristic polynomials of

P_{1}

and

P_{2}

satisfy

tr P_{1} \geq tr P_{2} and tr Cof P_{1} \geq tr Cof P_{2} and det P_{1} = det P_{2} .

Then

{∥ log P_{1} ∥}_{F}^{2} \geq {∥ log P_{2} ∥}_{F}^{2} .

For $n = 3$ , we will now give equivalent formulations of this statement. The case $n = 2$ can be treated analogously. For its proof, see Remark 15. By orthogonal diagonalization of $P_{1}$ and $P_{2}$ , the inequalities can be rewritten in terms of the eigenvalues $y_{1}$ , $y_{2}$ , $y_{3}$ and $a_{1}$ , $a_{2}$ , $a_{3}$ , respectively.

Theorem 2 Let the real numbers

a_{1}, a_{2}, a_{3} > 0

and

y_{1}, y_{2}, y_{3} > 0

be such that

\begin{matrix} y_{1} + y_{2} + y_{3} \geq a_{1} + a_{2} + a_{3}, \\ y_{1} y_{2} + y_{2} y_{3} + y_{1} y_{3} \geq a_{1} a_{2} + a_{2} a_{3} + a_{1} a_{3}, \\ y_{1} y_{2} y_{3} = a_{1} a_{2} a_{3} . \end{matrix}

(1)

Then

{(log y_{1})}^{2} + {(log y_{2})}^{2} + {(log y_{3})}^{2} \geq {(log a_{1})}^{2} + {(log a_{2})}^{2} + {(log a_{3})}^{2} .

(2)

The elementary symmetric polynomials, see, e.g., [[3], p.178]

\begin{matrix} e_{0} (y_{1}, y_{2}, y_{3}) = 1, \\ e_{1} (y_{1}, y_{2}, y_{3}) = y_{1} + y_{2} + y_{3}, \\ e_{2} (y_{1}, y_{2}, y_{3}) = y_{1} y_{2} + y_{1} y_{3} + y_{2} y_{3}, \\ e_{3} (y_{1}, y_{2}, y_{3}) = y_{1} y_{2} y_{3} \end{matrix}

are known to have the Schur-concavity property (i.e., $- e_{k}$ is Schur-convex) [1, 4]; see (16). It is possible to express the problem in terms of these elementary symmetric polynomials as follows.

Theorem 3 Let

a_{1}, a_{2}, a_{3} > 0

and

y_{1}, y_{2}, y_{3} > 0

satisfy

\begin{matrix} e_{1} (y_{1}, y_{2}, y_{3}) \geq e_{1} (a_{1}, a_{2}, a_{3}), e_{2} (y_{1}, y_{2}, y_{3}) \geq e_{2} (a_{1}, a_{2}, a_{3}), \\ e_{3} (y_{1}, y_{2}, y_{3}) = e_{3} (a_{1}, a_{2}, a_{3}) . \end{matrix}

Then

e_{1} ({(log y_{1})}^{2}, {(log y_{2})}^{2}, {(log y_{3})}^{2}) \geq e_{1} ({(log a_{1})}^{2}, {(log a_{2})}^{2}, {(log a_{3})}^{2}) .

Because

y_{1} y_{2} y_{3} = a_{1} a_{2} a_{3} > 0

, we have

y_{1} y_{2} + y_{2} y_{3} + y_{1} y_{3} \geq a_{1} a_{2} + a_{2} a_{3} + a_{1} a_{3} \Leftrightarrow \frac{1}{y_{1}} + \frac{1}{y_{2}} + \frac{1}{y_{3}} \geq \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} .

Thus, we obtain the following theorem.

Theorem 4 Let the real numbers

a_{1}, a_{2}, a_{3} > 0

and

y_{1}, y_{2}, y_{3} > 0

be such that

\begin{matrix} y_{1} + y_{2} + y_{3} \geq a_{1} + a_{2} + a_{3}, \\ \frac{1}{y_{1}} + \frac{1}{y_{2}} + \frac{1}{y_{3}} \geq \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}}, \\ y_{1} y_{2} y_{3} = a_{1} a_{2} a_{3} . \end{matrix}

(3)

Then

{(log y_{1})}^{2} + {(log y_{2})}^{2} + {(log y_{3})}^{2} \geq {(log a_{1})}^{2} + {(log a_{2})}^{2} + {(log a_{3})}^{2} .

(4)

The conditions (3) are also simple expressions in terms of arithmetic, harmonic and geometric and quadratic mean

\begin{matrix} A (y_{1}, y_{2}, y_{3}) = \frac{y_{1} + y_{2} + y_{3}}{3}, H (y_{1}, y_{2}, y_{3}) = \frac{3}{\frac{1}{y_{1}} + \frac{1}{y_{2}} + \frac{1}{y_{3}}}, \\ G (y_{1}, y_{2}, y_{3}) = \sqrt[3]{y_{1} y_{2} y_{3}}, Q (y_{1}, y_{2}, y_{3}) = \sqrt{\frac{1}{3} (y_{1}^{2} + y_{2}^{2} + y_{3}^{2})} . \end{matrix}

Theorem 5 Let

a_{1}, a_{2}, a_{3} > 0

and

y_{1}, y_{2}, y_{3} > 0

. Then

A (y_{1}, y_{2}, y_{3}) \geq A (a_{1}, a_{2}, a_{3})

,

H (a_{1}, a_{2}, a_{3}) \geq H (y_{1}, y_{2}, y_{3})

(‘reverse!’) and

G (y_{1}, y_{2}, y_{3}) = G (a_{1}, a_{2}, a_{3})

imply

Q (log y_{1}, log y_{2}, log y_{3}) \geq Q (log a_{1}, log a_{2}, log a_{3}) .

We denote by

a_{i} = : d_{i}^{2}, y_{i} = : x_{i}^{2}

and arrive at

Theorem 6 Let the real numbers

d_{i}

and

x_{i}

be such that

d_{1}, d_{2}, d_{3} > 0

,

x_{1}, x_{2}, x_{3} > 0

and

\begin{matrix} x_{1}^{2} + x_{2}^{2} + x_{3}^{2} \geq d_{1}^{2} + d_{2}^{2} + d_{3}^{2}, \\ x_{1}^{2} x_{2}^{2} + x_{2}^{2} x_{3}^{2} + x_{1}^{2} x_{3}^{2} \geq d_{1}^{2} d_{2}^{2} + d_{2}^{2} d_{3}^{2} + d_{1}^{2} d_{3}^{2}, \\ x_{1} x_{2} x_{3} = d_{1} d_{2} d_{3} . \end{matrix}

(5)

Then

{(log x_{1})}^{2} + {(log x_{2})}^{2} + {(log x_{3})}^{2} \geq {(log d_{1})}^{2} + {(log d_{2})}^{2} + {(log d_{3})}^{2} .

(6)

If we again view $x_{i}$ and $d_{i}$ as eigenvalues of positive definite matrices, an equivalent formulation of the problem can be given in terms of their Frobenius matrix norms:

Theorem 7 For

n \in {2, 3}

, let

P_{1}, P_{2} \in R^{n \times n}

be positive definite real matrices. Let

{∥ P_{1} ∥}_{F}^{2} \geq {∥ P_{2} ∥}_{F}^{2} and {∥ P_{1}^{- 1} ∥}_{F}^{2} \geq {∥ P_{2}^{- 1} ∥}_{F}^{2} and det P_{1} = det P_{2} .

Then

{∥ log P_{1} ∥}_{F}^{2} \geq {∥ log P_{2} ∥}_{F}^{2} .

Let us reconsider the formulation from Theorem 5. If we denote

c_{i} : = log a_{i}, z_{i} : = log y_{i},

from

H (a_{1}, a_{2}, a_{3}) \geq H (y_{1}, y_{2}, y_{3})

, we obtain

e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} \geq e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}} .

Theorem 8 Let the real numbers

c_{1}

,

c_{2}

,

c_{3}

and

z_{1}

,

z_{2}

,

z_{3}

be such that

\begin{matrix} e^{z_{1}} + e^{z_{2}} + e^{z_{3}} \geq e^{c_{1}} + e^{c_{2}} + e^{c_{3}}, \\ e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} \geq e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}}, \\ z_{1} + z_{2} + z_{3} = c_{1} + c_{2} + c_{3} . \end{matrix}

(7)

Then

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} \geq c_{1}^{2} + c_{2}^{2} + c_{3}^{2} .

(8)

In order to prove Theorem 8, one can assume without loss of generality that

z_{1} + z_{2} + z_{3} = c_{1} + c_{2} + c_{3} = 0 .

(9)

Thus, we have the equivalent formulation

Theorem 9 Let the real numbers

{\bar{c}}_{1}

,

{\bar{c}}_{2}

,

{\bar{c}}_{3}

and

{\bar{z}}_{1}

,

{\bar{z}}_{2}

,

{\bar{z}}_{3}

be such that

\begin{matrix} e^{{\bar{z}}_{1}} + e^{{\bar{z}}_{2}} + e^{{\bar{z}}_{3}} \geq e^{{\bar{c}}_{1}} + e^{{\bar{c}}_{2}} + e^{{\bar{c}}_{3}}, \\ e^{- {\bar{z}}_{1}} + e^{- {\bar{z}}_{2}} + e^{- {\bar{z}}_{3}} \geq e^{- {\bar{c}}_{1}} + e^{- {\bar{c}}_{2}} + e^{- {\bar{c}}_{3}}, \\ {\bar{z}}_{1} + {\bar{z}}_{2} + {\bar{z}}_{3} = {\bar{c}}_{1} + {\bar{c}}_{2} + {\bar{c}}_{3} = 0 . \end{matrix}

(10)

Then

{\bar{z}}_{1}^{2} + {\bar{z}}_{2}^{2} + {\bar{z}}_{3}^{2} \geq {\bar{c}}_{1}^{2} + {\bar{c}}_{2}^{2} + {\bar{c}}_{3}^{2} .

(11)

Let us prove that Theorem 8 can be reformulated as Theorem 9. Indeed, let us assume that Theorem 9 is valid and show that the statement of Theorem 8 also holds true. We denote by s the sum

s = z_{1} + z_{2} + z_{3} = c_{1} + c_{2} + c_{3}

and we designate

{\bar{z}}_{i} = z_{i} - \frac{s}{3}, {\bar{c}}_{i} = c_{i} - \frac{s}{3} (i = 1, 2, 3) .

Then the real numbers

{\bar{z}}_{i}

and

{\bar{c}}_{i}

satisfy the hypotheses of Theorem 9 and we obtain

{\bar{z}}_{1}^{2} + {\bar{z}}_{2}^{2} + {\bar{z}}_{3}^{2} \geq {\bar{c}}_{1}^{2} + {\bar{c}}_{2}^{2} + {\bar{c}}_{3}^{2}

. This inequality is equivalent to

\sum_{i = 1}^{3} {(z_{i} - \frac{s}{3})}^{2} \geq \sum_{i = 1}^{3} {(c_{i} - \frac{s}{3})}^{2},

which, by virtue of the condition (7)₃, reduces to

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} \geq c_{1}^{2} + c_{2}^{2} + c_{3}^{2} .

Thus, Theorem 8 is also valid.

By virtue of the logical equivalence

(A \land B \Rightarrow C) \Leftrightarrow (\neg C \Rightarrow \neg A \lor \neg B)

for any statements A, B, C, we can formulate the inequality (11) (i.e., Theorem 9) in the following equivalent manner.

Theorem 10 Let the real numbers

c_{1}

,

c_{2}

,

c_{3}

and

z_{1}

,

z_{2}

,

z_{3}

be such that

z_{1} + z_{2} + z_{3} = c_{1} + c_{2} + c_{3} = 0 and z_{1}^{2} + z_{2}^{2} + z_{3}^{2} < c_{1}^{2} + c_{2}^{2} + c_{3}^{2} .

(12)

Then one of the following inequalities holds:

\begin{aligned} e^{z_{1}} + e^{z_{2}} + e^{z_{3}} < e^{c_{1}} + e^{c_{2}} + e^{c_{3}} or \\ e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} < e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}} . \end{aligned}

(13)

We use the statement of Theorem 10 for the proof.

Before continuing, let us show that our new inequality is not a consequence of majorization and Karamata’s inequality [5]. Consider

z = (z_{1}, \dots, z_{n}) \in R_{+}^{n}

and

c = (c_{1}, \dots, c_{n}) \in R_{+}^{n}

arranged already in decreasing order

z_{1} \geq z_{2} \geq \dots \geq z_{n}

and

c_{1} \geq c_{2} \geq \dots \geq c_{n}

. If

\sum_{i = 1}^{k} z_{i} \geq \sum_{i = 1}^{k} c_{i} (1 \leq k \leq n - 1), \sum_{i = 1}^{n} z_{i} = \sum_{i = 1}^{n} c_{i},

(14)

we say that z majorizes c, denoted by

z ≻ c

. The following result is well known [[6], p.89], [4, 5]. If

f : R \to R

is convex, then

z ≻ c \Rightarrow \sum_{i = 1}^{n} f (z_{i}) \geq \sum_{i = 1}^{n} f (c_{i}) .

(15)

A function

g : R^{n} \mapsto R

which satisfies

z ≻ c \Rightarrow g (z_{1}, \dots, z_{n}) \geq g (c_{1}, \dots, c_{n})

(16)

is called Schur-convex. In Theorem 8, the convex function to be considered would be

f (t) = t^{2}

. Do conditions (7) (upon rearrangement of

z, c \in R_{+}^{3}

if necessary) yield already majorization

z ≻ c

? This is not the case, as we explain now. Let the real numbers

z_{1} \geq z_{2} \geq z_{3}

and

c_{1} \geq c_{2} \geq c_{3}

be such that

\begin{matrix} e^{z_{1}} + e^{z_{2}} + e^{z_{3}} \geq e^{c_{1}} + e^{c_{2}} + e^{c_{3}}, \\ e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} \geq e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}}, \\ z_{1} + z_{2} + z_{3} = c_{1} + c_{2} + c_{3} . \end{matrix}

(17)

These conditions do not imply the majorization

z ≻ c

,

z_{1} \geq c_{1}, z_{1} + z_{2} \geq c_{1} + c_{2}, z_{1} + z_{2} + z_{3} = c_{1} + c_{2} + c_{3} .

(18)

Therefore, our inequality (i.e., $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} \geq c_{1}^{2} + c_{2}^{2} + c_{3}^{2}$ ) does not follow from majorization in disguise.

Indeed, let

z_{1} = \frac{1}{2} + \frac{0.95}{2 \sqrt{3}}, z_{2} = \frac{1}{2} + \frac{0.85}{2 \sqrt{3}}, z_{3} = - 1 - \frac{0.9}{\sqrt{3}}

and

c_{1} = \frac{1}{2} + \frac{1}{2 \sqrt{3}}, c_{2} = - \frac{1}{2} + \frac{1}{2 \sqrt{3}}, c_{3} = - \frac{1}{\sqrt{3}} .

Then we have

z_{1} > z_{2} > z_{3}

and

c_{1} > c_{2} > c_{3}

, together with

\begin{matrix} e^{z_{1}} + e^{z_{2}} + e^{z_{3}} = 4.49497 \dots > 3.57137 \dots = e^{c_{1}} + e^{c_{2}} + e^{c_{3}}, \\ e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} = 5.50607 \dots > 3.47107 \dots = e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}}, \\ z_{1} + z_{2} + z_{3} = c_{1} + c_{2} + c_{3} = 0, \end{matrix}

but the majorization inequalities (18) are not satisfied, since $z_{1} < c_{1}$ .

3 Proof of the inequality

Of course, we may assume without loss of generality that $c_{1} \geq c_{2} \geq c_{3}$ and $z_{1} \geq z_{2} \geq z_{3}$ (and the same for $a_{i}$ , $d_{i}$ , $x_{i}$ , $y_{i}$ ).

The proof begins with the crucial lemma.

Lemma 11 Let the real numbers

a \geq b \geq c

and

x \geq y \geq z

be such that

a + b + c = x + y + z = 0, a^{2} + b^{2} + c^{2} = x^{2} + y^{2} + z^{2} .

(19)

Then the inequality

e^{a} + e^{b} + e^{c} \leq e^{x} + e^{y} + e^{z}

(20)

is satisfied if and only if the relation

a \leq x

(21)

holds, or equivalently, if and only if

c \leq z

(22)

holds.

Proof Let us denote by

r : = \sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2})} > 0

. Then, from (19), it follows

\begin{matrix} b + c = - a, b^{2} + c^{2} = \frac{3}{2} r^{2} - a^{2}, \\ y + z = - x, y^{2} + z^{2} = \frac{3}{2} r^{2} - x^{2}, \end{matrix}

and we find

\begin{aligned} b = \frac{1}{2} (- a + \sqrt{3 (r^{2} - a^{2})}), c = \frac{1}{2} (- a - \sqrt{3 (r^{2} - a^{2})}), \\ y = \frac{1}{2} (- x + \sqrt{3 (r^{2} - x^{2})}), z = \frac{1}{2} (- x - \sqrt{3 (r^{2} - x^{2})}) . \end{aligned}

(23)

In view of (19) and

a \geq b \geq c

,

x \geq y \geq z

, one can show that

a, x \in [\frac{r}{2}, r], b, y \in [- \frac{r}{2}, \frac{r}{2}], c, z \in [- r, - \frac{r}{2}] .

(24)

Indeed, let us verify the relations (24). We have

\begin{array}{rcl} \frac{r}{2} \leq a \leq r & \Leftrightarrow & \frac{1}{6} (a^{2} + b^{2} + c^{2}) \leq a^{2} \leq \frac{2}{3} (a^{2} + b^{2} + c^{2}) \\ \Leftrightarrow & b^{2} + c^{2} \leq 5 a^{2} and a^{2} \leq 2 (b^{2} + c^{2}) \\ \Leftrightarrow & b^{2} + {(a + b)}^{2} \leq 5 a^{2} and {(b + c)}^{2} \leq 2 (b^{2} + c^{2}) \\ \Leftrightarrow & 4 a^{2} - 2 a b - 2 b^{2} \geq 0 and b^{2} + c^{2} \geq 2 b c \\ \Leftrightarrow & 2 (a - b) (2 a + b) \geq 0 and {(b - c)}^{2} \geq 0, \end{array}

which hold true since

a \geq b

and

2 a + b \geq a + b + c = 0

. Similarly, we have

\begin{array}{rcl} - \frac{r}{2} \leq b \leq \frac{r}{2} & \Leftrightarrow & b^{2} \leq \frac{r^{2}}{4} \Leftrightarrow 4 b^{2} \leq \frac{2}{3} (a^{2} + b^{2} + c^{2}) \Leftrightarrow 5 b^{2} \leq a^{2} + c^{2} \\ \Leftrightarrow & 5 b^{2} \leq a^{2} + {(a + b)}^{2} \Leftrightarrow 2 a^{2} + 2 a b - 4 b^{2} \geq 0 \\ \Leftrightarrow & 2 (a - b) (a + 2 b) \geq 0, \end{array}

which holds true since

a \geq b

and

a + 2 b \geq a + b + c = 0

. Also, we have

\begin{array}{rcl} - r \leq c \leq - \frac{r}{2} & \Leftrightarrow & r^{2} \geq c^{2} \geq \frac{r^{2}}{4} \Leftrightarrow \frac{2}{3} (a^{2} + b^{2} + c^{2}) \geq c^{2} \geq \frac{1}{6} (a^{2} + b^{2} + c^{2}) \\ \Leftrightarrow & 2 (a^{2} + b^{2}) \geq c^{2} and 5 c^{2} \geq a^{2} + b^{2} \\ \Leftrightarrow & 2 (a^{2} + b^{2}) \geq {(a + b)}^{2} and 5 {(a + b)}^{2} \geq a^{2} + b^{2} \\ \Leftrightarrow & {(a - b)}^{2} \geq 0 and 4 a^{2} + 10 a b + 4 b^{2} \geq 0 \\ \Leftrightarrow & {(a - b)}^{2} \geq 0 and 2 (a + 2 b) (2 a + b) \geq 0, \end{array}

which hold true since $a + 2 b \geq a + b + c = 0$ and $2 a + b \geq a + b + c = 0$ . One can show in the same way that $x \in [\frac{r}{2}, r]$ , $y \in [- \frac{r}{2}, \frac{r}{2}]$ , $z \in [- r, - \frac{r}{2}]$ , so that (24) has been verified.

We prove now that the inequality (21) holds if and only if (22) holds. Indeed, using (23)_2,4 and (24) we get

\begin{array}{rcl} c \leq z & \Leftrightarrow & - a - \sqrt{3 (r^{2} - a^{2})} \leq - x - \sqrt{3 (r^{2} - x^{2})} \\ \Leftrightarrow & \frac{a}{r} + \sqrt{3 (1 - {(\frac{a}{r})}^{2})} \geq \frac{x}{r} + \sqrt{3 (1 - {(\frac{x}{r})}^{2})} \Leftrightarrow a \leq x, \end{array}

since the function $t \mapsto t + \sqrt{3 (1 - t^{2})}$ is decreasing for $t \in [\frac{1}{2}, 1]$ .

Let us prove next that the inequalities (20) and (21) are equivalent. To accomplish this, we introduce the function

f : [\frac{r}{2}, r] \to R

by

f (x) = e^{x} + e^{(- x + \sqrt{3 (r^{2} - x^{2})}) / 2} + e^{(- x - \sqrt{3 (r^{2} - x^{2})}) / 2} .

(25)

Taking into account (23) and (24)₁, the inequality (20) can be written equivalently as

f (a) \leq f (x),

(26)

which is equivalent to

a \leq x,

since the function f defined by (25) is monotone increasing on

[\frac{r}{2}, r]

, as we show next. To this aim, we denote by

cos φ : = \frac{x}{r} \in [\frac{1}{2}, 1], i.e. φ : = arccos (\frac{x}{r}) \in [0, \frac{π}{3}] .

Then the function (25) can be written as

\begin{aligned} f (x) = h (r, φ), where h : (0, \infty) \times [0, \frac{π}{3}] \to R, \\ h (r, φ) = e^{r cos φ} + e^{r cos (φ + 2 π / 3)} + e^{r cos (φ - 2 π / 3)} . \end{aligned}

(27)

We have to show that

h (r, φ)

is decreasing with respect to

φ \in [0, \frac{π}{3}]

. We compute the first derivative

\begin{matrix} \frac{\partial}{\partial φ} h (r, φ) \\ = - r [e^{r cos φ} sin φ + e^{r cos (φ + 2 π / 3)} sin (φ + \frac{2 π}{3}) + e^{r cos (φ - 2 π / 3)} sin (φ - \frac{2 π}{3})] . \end{matrix}

(28)

The function (28) has the same sign as the function

F (r, φ) : = \frac{1}{r} e^{- r cos φ} \frac{\partial}{\partial φ} h (r, φ),

(29)

i.e., the function

F : (0, \infty) \times [0, \frac{π}{3}] \to R

given by

F (r, φ) = - sin φ - e^{- r \sqrt{3} sin (φ + π / 3)} sin (φ + \frac{2 π}{3}) - e^{r \sqrt{3} sin (φ - π / 3)} sin (φ - \frac{2 π}{3}) .

(30)

In order to show that

F (r, φ) \leq 0

for all

(r, φ) \in (0, \infty) \times [0, \frac{π}{3}]

, we remark that

{lim}_{r ↘ 0} F (r, φ) = 0

for fixed

φ \in [0, \frac{π}{3}]

and we compute

\begin{array}{rcl} \frac{\partial}{\partial r} F (r, φ) & = & \sqrt{3} [e^{- r \sqrt{3} sin (φ + π / 3)} sin (φ + \frac{π}{3}) sin (φ + \frac{2 π}{3}) \\ - e^{r \sqrt{3} sin (φ - π / 3)} sin (φ - \frac{π}{3}) sin (φ - \frac{2 π}{3})] \\ = & \sqrt{3} [e^{- r \sqrt{3} sin (φ + π / 3)} \frac{1}{2} (- cos (2 φ + π) + cos \frac{π}{3}) \\ - e^{r \sqrt{3} sin (φ - π / 3)} \frac{1}{2} (- cos (2 φ - π) + cos \frac{- π}{3})] \\ = & \frac{\sqrt{3}}{2} (cos 2 φ + \frac{1}{2}) [e^{- r \sqrt{3} sin (φ + π / 3)} - e^{r \sqrt{3} sin (φ - π / 3)}] \leq 0, \end{array}

since $φ \in [0, \frac{π}{3}]$ implies $cos 2 φ \geq - \frac{1}{2}$ and $- sin (φ + \frac{π}{3}) \leq sin (φ - \frac{π}{3})$ .

Consequently, the function

F (r, φ)

is decreasing with respect to r and for any

(r, φ) \in (0, \infty) \times [0, \frac{π}{3}]

we have that

F (r, φ) \leq lim_{r ↘ 0} F (r, φ) = 0 .

(31)

From (29) and (31), it follows that $h (r, φ)$ is decreasing with respect to $φ \in [0, \frac{π}{3}]$ . This means that $f (x)$ is increasing as a function of $x \in [\frac{r}{2}, r]$ , i.e., the relation (26) is indeed equivalent to $a \leq x$ and the proof is complete. □

Consequence 12 Let the real numbers

a \geq b \geq c

and

x \geq y \geq z

be such that

a + b + c = x + y + z = 0, a^{2} + b^{2} + c^{2} = x^{2} + y^{2} + z^{2} .

Then one of the following inequalities holds:

e^{a} + e^{b} + e^{c} \leq e^{x} + e^{y} + e^{z},

(32)

or

e^{- a} + e^{- b} + e^{- c} \leq e^{- x} + e^{- y} + e^{- z} .

(33)

The inequalities (32) and (33) are satisfied simultaneously if and only if $a = x$ , $b = y$ and $c = z$ .

Proof According to Lemma 11, the inequality (32) is equivalent to

a \leq x,

(34)

while the inequality (33) is equivalent to

- a \leq - x .

(35)

Since one of the relations (34) and (35) must hold, we have proved that one of the inequalities (32) and (33) is satisfied. They are simultaneously satisfied if and only if both (34) and (35) hold true, i.e., $a = x$ (and consequently $b = y$ , $c = z$ ). □

Consequence 13 Let the real numbers

a \geq b \geq c

and

x \geq y \geq z

be such that

\begin{matrix} a + b + c = x + y + z = 0, a^{2} + b^{2} + c^{2} = x^{2} + y^{2} + z^{2} \\ and e^{a} + e^{b} + e^{c} = e^{x} + e^{y} + e^{z} . \end{matrix}

Then we have $a = x$ , $b = y$ and $c = z$ .

Proof Since by hypothesis $e^{a} + e^{b} + e^{c} \leq e^{x} + e^{y} + e^{z}$ holds, we can apply Lemma 11 to deduce $a \leq x$ and $c \leq z$ .

On the other hand, by virtue of the inverse inequality $e^{x} + e^{y} + e^{z} \leq e^{a} + e^{b} + e^{c}$ and Lemma 11, we obtain $x \leq a$ and $z \leq c$ . In conclusion, we get $a = x$ , $c = z$ and $b = y$ . □

Proof of Theorem 10 In order to prove (13), we define the real numbers

t_{i} = k z_{i} (i = 1, 2, 3), where k = \sqrt{\frac{c_{1}^{2} + c_{2}^{2} + c_{3}^{2}}{z_{1}^{2} + z_{2}^{2} + z_{3}^{2}}} > 1 .

(36)

Then we have

t_{1} + t_{2} + t_{3} = c_{1} + c_{2} + c_{3} = 0 and t_{1}^{2} + t_{2}^{2} + t_{3}^{2} = c_{1}^{2} + c_{2}^{2} + c_{3}^{2} .

(37)

If we apply the Consequence 12 for the numbers

c_{1} \geq c_{2} \geq c_{3}

and

t_{1} \geq t_{2} \geq t_{3}

, then we obtain that

\begin{aligned} e^{t_{1}} + e^{t_{2}} + e^{t_{3}} \leq e^{c_{1}} + e^{c_{2}} + e^{c_{3}} or \\ e^{- t_{1}} + e^{- t_{2}} + e^{- t_{3}} \leq e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}} . \end{aligned}

(38)

In what follows, let us show that

e^{z_{1}} + e^{z_{2}} + e^{z_{3}} < e^{t_{1}} + e^{t_{2}} + e^{t_{3}} .

(39)

Using the notations

ρ : = \sqrt{\frac{2}{3} (z_{1}^{2} + z_{2}^{2} + z_{3}^{2})}

and

cos ζ : = \frac{z_{1}}{ρ} \in [\frac{1}{2}, 1], i.e., ζ : = arccos (\frac{z_{1}}{ρ}) \in [0, \frac{π}{3}],

we have

k ρ : = \sqrt{\frac{2}{3} (t_{1}^{2} + t_{2}^{2} + t_{3}^{2})}

and

cos ζ = \frac{t_{1}}{k ρ}

. With the help of the function h defined in (27), we can write the inequality (39) in the form

\begin{aligned} e^{ρ cos ζ} + e^{ρ cos (ζ + 2 π / 3)} + e^{ρ cos (ζ - 2 π / 3)} < e^{k ρ cos ζ} + e^{k ρ cos (ζ + 2 π / 3)} + e^{k ρ cos (ζ - 2 π / 3)}, or \\ h (ρ, ζ) < h (k ρ, ζ), \forall (ρ, ζ) \in (0, \infty) \times [0, \frac{π}{3}], k > 1 . \end{aligned}

(40)

The relation (40) asserts that the function h defined in (27) is increasing with respect to the first variable

r \in (0, \infty)

. To show this, we compute the derivative

\frac{\partial}{\partial r} h (r, φ) = e^{r cos φ} cos φ + e^{r cos (φ + 2 π / 3)} cos (φ + \frac{2 π}{3}) + e^{r cos (φ - 2 π / 3)} cos (φ - \frac{2 π}{3}) .

(41)

By virtue of the Chebyshev’s sum inequality, we deduce from (41) that

\frac{\partial}{\partial r} h (r, φ) > 0 .

(42)

Indeed, the Chebyshev’s sum inequality [[6], 2.17] asserts that: if

a_{1} \geq a_{2} \geq \dots \geq a_{n}

and

b_{1} \geq b_{2} \geq \dots \geq b_{n}

then

n \sum_{k = 1}^{n} a_{k} b_{k} \geq (\sum_{k = 1}^{n} a_{k}) (\sum_{k = 1}^{n} b_{k}) .

In our case, we derive the following result: for any real numbers x, y, z such that

x + y + z = 0

, the inequality

x e^{x} + y e^{y} + z e^{z} \geq \frac{1}{3} (x + y + z) (e^{x} + e^{y} + e^{z}) = 0,

(43)

holds true, with equality if and only if $x = y = z = 0$ .

Applying the result (43) to the function (41), we deduce the relation (42). This means that $h (r, φ)$ is an increasing function of r, i.e. the inequality (40) holds, and hence, we have proved (39).

One can show analogously that the inequality

e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} < e^{- t_{1}} + e^{- t_{2}} + e^{- t_{3}}

(44)

is also valid. From (38), (39) and (44), it follows that the assertion (13) holds true. Thus, the proof of Theorem 10 is complete. □

Since the statements of the Theorems 8 and 10 are equivalent, we have proved also the inequality (8).

Remark 14 The inequality (8) becomes an equality if and only if $z_{i} = c_{i}$ , $i = 1, 2, 3$ .

Proof Indeed, assume that

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = c_{1}^{2} + c_{2}^{2} + c_{3}^{2}

. Then we can apply the Consequence 12 and we deduce that

e^{z_{1}} + e^{z_{2}} + e^{z_{3}} \leq e^{c_{1}} + e^{c_{2}} + e^{c_{3}} or e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} \leq e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}} .

(45)

Taking into account (7)_1,2 in conjunction with (45), we find

e^{z_{1}} + e^{z_{2}} + e^{z_{3}} = e^{c_{1}} + e^{c_{2}} + e^{c_{3}} or e^{- z_{1}} + e^{- z_{2}} + e^{- z_{3}} = e^{- c_{1}} + e^{- c_{2}} + e^{- c_{3}} .

(46)

By virtue of (46), we can apply the Consequence 13 to derive $z_{1} = c_{1}$ , and consequently $z_{2} = c_{2}$ , $z_{3} = c_{3}$ . □

Let us prove the following version of the inequality (6) for two pairs of numbers $d_{1}$ , $d_{2}$ and $x_{1}$ , $x_{2}$ :

Remark 15 If the real numbers

d_{1} \geq d_{2} > 0

and

x_{1} \geq x_{2} > 0

are such that

x_{1}^{2} + x_{2}^{2} \geq d_{1}^{2} + d_{2}^{2} and x_{1} x_{2} = d_{1} d_{2} = 1,

(47)

then the inequality

{(log x_{1})}^{2} + {(log x_{2})}^{2} \geq {(log d_{1})}^{2} + {(log d_{2})}^{2}

(48)

holds true. Note that the additional condition

\frac{1}{x_{1}^{2}} + \frac{1}{x_{2}^{2}} \geq \frac{1}{d_{1}^{2}} + \frac{1}{d_{2}^{2}}

is automatically fulfilled.

Proof Since

x_{1} x_{2} = d_{1} d_{2} = 1

and

d_{1} \geq d_{2} > 0

,

x_{1} \geq x_{2} > 0

, we have

x_{1} \geq 1

,

d_{1} \geq 1

and

log x_{1} = - log x_{2} \geq 0, log d_{1} = - log d_{2} \geq 0,

so that the inequality (48) is equivalent to $log x_{1} \geq log d_{1}$ , i.e., we have to show that $x_{1} \geq d_{1}$ .

Indeed, if we insert

x_{2} = \frac{1}{x_{1}}

and

d_{2} = \frac{1}{d_{1}}

into the inequality (47)₁ then we find

x_{1}^{2} + \frac{1}{x_{1}^{2}} \geq d_{1}^{2} + \frac{1}{d_{1}^{2}},

which means that $x_{1} \geq d_{1}$ since the function $t \mapsto t^{2} + \frac{1}{t^{2}}$ is increasing for $t \in [1, \infty)$ . This completes the proof. □

Alternative proof of Remark 15 Let

x_{3} = d_{3} = 1

. Then (47) implies

x_{1}^{2} + x_{2}^{2} + x_{3}^{2} \geq d_{1}^{2} + d_{2}^{2} + d_{3}^{2}

and

x_{1} x_{2} x_{3} = d_{1} d_{2} d_{3} = 1

as well as

x_{1}^{2} x_{2}^{2} + x_{2}^{2} x_{3}^{2} + x_{1}^{2} x_{3}^{2} = 1 + x_{2}^{2} + x_{1}^{2} \geq 1 + d_{2}^{2} + d_{1}^{2} = d_{1}^{2} d_{2}^{2} + d_{2}^{2} d_{3}^{2} + d_{1}^{2} d_{3}^{2},

(49)

because $x_{1}^{2} x_{2}^{2} = 1 = d_{1}^{2} d_{2}^{2}$ , and Theorem 6 provides the assertion. □

4 Some counterexamples for weakened assumptions

Example 16 Unlike in the 2D case in Remark 15, for two triples of numbers the second condition (18)₂ of Theorem 2, namely

y_{1} y_{2} + y_{2} y_{3} + y_{1} y_{3} \geq a_{1} a_{2} + a_{2} a_{3} + a_{1} a_{3}

, cannot be removed. Let

y_{1} = e^{6}, y_{2} = 1, y_{3} = e^{- 6}, a_{1} = e^{4}, a_{2} = e^{4}, a_{3} = e^{- 8} .

Then

y_{1} y_{2} y_{3} = a_{1} a_{2} a_{3} = 1

and

y_{1} + y_{2} + y_{3} > e^{6} \geq e^{2} e^{4} > 3 e^{4} > a_{1} + a_{2} + a_{3},

but

\begin{array}{rcl} {(log y_{1})}^{2} + {(log y_{2})}^{2} + {(log y_{3})}^{2} & = & 36 + 0 + 36 \\ < & 16 + 16 + 64 = {(log a_{1})}^{2} + {(log a_{2})}^{2} + {(log a_{3})}^{2} . \end{array}

Example 17 The condition

y_{1} y_{2} y_{3} = a_{1} a_{2} a_{3}

cannot be weakened to

y_{1} y_{2} y_{3} \geq a_{1} a_{2} a_{3}

. Indeed, let

y_{2} = y_{3} = a_{1} = a_{2} = 1

,

y_{1} = e

,

a_{3} = e^{- 2}

. Then

\begin{matrix} y_{1} + y_{2} + y_{3} = e + 1 + 1 \geq 1 + 1 + e^{- 2} = a_{1} + a_{2} + a_{3}, \\ y_{1} y_{2} + y_{1} y_{3} + y_{2} y_{3} = e + e + 1 \geq 1 + e^{- 2} + e^{- 2} = a_{1} a_{2} + a_{1} a_{3} + a_{2} a_{3}, \\ y_{1} y_{2} y_{3} = e \geq e^{- 2} = a_{1} a_{2} a_{3} . \end{matrix}

But nevertheless

{(log y_{1})}^{2} + {(log y_{2})}^{2} + {(log y_{3})}^{2} = 1 + 0 + 0 < 0 + 0 + 4 = {(log a_{1})}^{2} + {(log a_{2})}^{2} + {(log a_{3})}^{2} .

A counterexample for the two variable case can be constructed analogously.

Example 18 Even with an analogous condition, the inequality (4) does not hold for

n = 4

numbers (without further assumptions). Indeed, let

y_{1} = e, y_{2} = y_{3} = e^{7}, y_{4} = e^{- 15}, a_{1} = a_{2} = e^{6}, a_{3} = e^{7}, a_{4} = e^{- 19} .

Then

y_{1} y_{2} y_{3} y_{4} = a_{1} a_{2} a_{3} a_{4} = 1

. Also,

y_{1} + y_{2} + y_{3} + y_{4} = e + e^{7} + e^{7} + e^{- 15} > 0 + e^{7} + 2 e^{6} + e^{- 19} = a_{1} + a_{2} + a_{3} + a_{4} .

Furthermore,

y_{1} y_{2} + y_{1} y_{3} + y_{1} y_{4} + y_{2} y_{3} + y_{2} y_{4} + y_{3} y_{4} = e^{8} + e^{8} + e^{- 14} + e^{14} + e^{- 8} + e^{- 8}

and

a_{1} a_{2} + a_{1} a_{3} + a_{1} a_{4} + a_{2} a_{3} + a_{2} a_{4} + a_{3} a_{4} = e^{12} + e^{13} + e^{- 13} + e^{13} + e^{- 13} + e^{- 12} .

Since

e^{2} > 2 e + 1

, we have

e^{14} > e^{13} + e^{13} + e^{12}

and, therefore,

y_{1} y_{2} + y_{1} y_{3} + y_{1} y_{4} + y_{2} y_{3} + y_{2} y_{4} + y_{3} y_{4} \geq a_{1} a_{2} + a_{1} a_{3} + a_{1} a_{4} + a_{2} a_{3} + a_{2} a_{4} + a_{3} a_{4} .

Nevertheless, for the sum of squared logarithms, the ‘reverse’ inequality

\begin{array}{rcl} {(log y_{1})}^{2} + {(log y_{2})}^{2} + {(log y_{3})}^{2} + {(log y_{4})}^{2} & = & 1 + 49 + 49 + 225 = 324 \\ < & 482 = 36 + 36 + 49 + 361 \\ = & {(log a_{1})}^{2} + {(log a_{2})}^{2} + {(log a_{3})}^{2} + {(log a_{4})}^{2} \end{array}

holds true.

Example 19 The inequality (4) does not remain true either, if the function

log (y)

is replaced by its linearization

(y - 1)

. Indeed, let

y_{1} = 9

,

y_{2} = 5

,

y_{3} = \frac{1}{45}

,

a_{1} = 10

,

a_{2} = 1

,

a_{3} = \frac{1}{10}

. Then

y_{1} + y_{2} + y_{3} > 14 > 11.1 = a_{1} + a_{2} + a_{3}

and

y_{1} y_{2} + y_{1} y_{3} + y_{2} y_{3} > 45 \geq 11.1 = a_{1} a_{2} + a_{1} a_{3} + a_{2} a_{3} .

But

\begin{array}{rcl} {(y_{1} - 1)}^{2} + {(y_{2} - 1)}^{2} + {(y_{3} - 1)}^{2} & = & 64 + 16 + {(\frac{44}{45})}^{2} < 81 < 9^{2} + 0 + {(\frac{9}{10})}^{2} \\ = & {(a_{1} - 1)}^{2} + {(a_{2} - 1)}^{2} + {(a_{3} - 1)}^{2} . \end{array}

5 Conjecture for arbitrary n

The structure of the inequality in dimensions $n = 2$ and $n = 3$ and extensive numerical sampling strongly suggest that the inequality holds for all $n \in N$ if the n corresponding conditions are satisfied. More precisely, in terms of the elementary symmetric polynomials, we expect the following:

Conjecture 20 Let

n \in N

and

y_{i}, a_{i} > 0

for

i = 1, \dots, n

. If for all

i = 1, \dots, n - 1

we have

e_{i} (y_{1}, \dots, y_{n}) \geq e_{i} (a_{1}, \dots, a_{n}) and e_{n} (y_{1}, \dots, y_{n}) = e_{n} (a_{1}, \dots, a_{n}),

then

\sum_{i = 1}^{n} {(log y_{i})}^{2} \geq \sum_{i = 1}^{n} {(log a_{i})}^{2} .

6 Applications

The investigation in this paper has been motivated by some recent applications. The new sum of squared logarithms inequality is one of the fundamental tools in deducing a novel optimality result in matrix analysis and the conditions in the form (3) had been deduced in the course of that work. Optimality in the matrix problem suggested the sum of squared logarithms inequality. Indeed, based on the present result in [7], it has been shown that for all invertible

Z \in C^{3 \times 3}

and for any definition of the matrix logarithm as possibly multivalued solution

X \in C^{3 \times 3}

of

exp X = Z

it holds

\begin{aligned} min_{Q^{*} Q = I} {∥ log Q^{*} Z ∥}_{F}^{2} = {∥ log U_{p}^{*} Z ∥}_{F}^{2} = {∥ log H ∥}_{F}^{2}, \\ min_{Q * Q = I} {∥ sym log Q^{*} Z ∥}_{F}^{2} = {∥ sym log U_{p}^{*} Z ∥}_{F}^{2} = {∥ log H ∥}_{F}^{2}, \end{aligned}

(50)

where

sym X = \frac{1}{2} (X + X^{*})

is the Hermitian part of

X \in C^{3 \times 3}

and

U_{p}

is the unitary factor in the polar decomposition of Z into unitary and Hermitian positive definite matrix H

Z = U_{p} H .

(51)

This result (50) generalizes the fact that for any complex logarithm and for all

z \in C ∖ {0}

min_{ϑ \in (- π, π]} {| {log}_{C} [e^{- i ϑ} z] |}^{2} = {| {log}_{R} | z | |}^{2}, min_{ϑ \in (- π, π]} {| Re {log}_{C} [e^{- i ϑ} z] |}^{2} = {| {log}_{R} | z | |}^{2} .

(52)

The optimality result (50) can now also be viewed as another characterization of the unitary factor in the polar decomposition. In addition, in a forthcoming contribution [8], we use (50) to calculate the geodesic distance of the isochoric part of the deformation gradient

\frac{F}{det F^{\frac{1}{3}}} \in SL (3, R)

to

SO (3, R)

in the canonical left-invariant Riemannian metric on

SL (3, R)

, to the effect that

{dist}_{geod}^{2} (\frac{F}{det F^{\frac{1}{3}}}, SO (3, R)) = {∥ {dev}_{3} log \sqrt{F^{T} F} ∥}_{F}^{2},

(53)

where ${dev}_{3} X = X - \frac{1}{3} (tr X) I$ is the orthogonal projection of $X \in R^{3 \times 3}$ to trace free matrices. Thereby, we provide a rigorous geometric justification for the preferred use of the Hencky-strain measure ${∥ log \sqrt{F^{T} F} ∥}_{F}^{2}$ in nonlinear elasticity and plasticity theory [9].

Declarations

Acknowledgements

The first author (MB) was supported by the German state grant: ‘Programm des Bundes und der Länder für bessere Studienbedingungen und mehr Qualität in der Lehre’.

Authors’ Affiliations

(1)

Lehrstuhl für Nichtlineare Analysis und Modellierung, Fakultät für Mathematik, Universität Duisburg-Essen

(2)

Department of Mathematics, University ‘A.I. Cuza’ of Iaşi

References

Guan K: Schur-convexity of the complete elementary symmetric functions. J. Inequal. Appl. 2006., 2006: Article ID 67624. doi:10.1155/JIA/2006/67624Google Scholar
Roventa I: A note on Schur-concave functions. J. Inequal. Appl. 2012, 159: 1–9.MathSciNetGoogle Scholar
Steele JM: The Cauchy-Schwarz Master Class: an Introduction to the Art of Mathematical Inequalities. Cambridge University Press, Cambridge; 2004.View ArticleGoogle Scholar
Khan AR, Latif N, Pečarić J: Exponential convexity for majorization. J. Inequal. Appl. 2012, 105: 1–13.Google Scholar
Karamata J: Sur une inégalité relative aux fonctions convexes. Publ. Math. Univ. Belgrad 1932, 1: 145–148.Google Scholar
Hardy GH, Littlewood JE, Pólya G: Inequalities. The University Press, Cambridge; 1934.Google Scholar
Neff, P, Nagatsukasa, Y, Fischle, A: The unitary polar factor $Q = U_{p}$ minimizes ${∥ Log (Q^{*} Z) ∥}^{2}$ and ${∥ {sym}_{*} Log (Q^{*} Z) ∥}^{2}$ in the spectral norm in any dimension and the Frobenius matrix norm in three dimensions (2013, submitted)Google Scholar
Neff, P, Eidel, B, Osterbrink, F, Martin, R: The isotropic Hencky strain energy ${∥ log U ∥}^{2}$ measures the geodesic distance of the deformation gradient $F \in {GL}^{+} (n)$ to $SO (n)$ in the unique left-invariant Riemannian metric on ${GL}^{+} (n)$ which is also right $O (n)$ -invariant (2013, in preparation)Google Scholar
Hencky H: Über die Form des Elastizitätsgesetzes bei ideal elastischen Stoffen. Z. Techn. Physik 1928, 9: 215–220.Google Scholar

Copyright

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.