A characterization of real matrix semigroups

ABSTRACT

We characterize all real matrix semigroups, indexed by the non-negative reals, which satisfy a mild boundedness assumption, without assuming continuity. Besides the continuous solutions of the semigroup functional equation, we give a description of solutions arising from non-measurable solutions of Cauchy’s functional equation. To do so, we discuss the primary decomposition and the Jordan—Chevalley decomposition of a matrix semigroup. Our motivation stems from a characterization of all multi-dimensional self-similar Gaussian Markov processes, which is given in a companion paper.

KEYWORDS:

• Matrix semigroup
• matrix exponential
• primary decomposition
• Jordan–chevalley decomposition
• Cauchy’s functional equation

MATHEMATICS SUBJECT CLASSIFICATION:

• 39B22
• 47D03
• 15A16

1. Introduction

It is a classical fact that all continuous matrix-valued functions $g:[0,\mathrm{\infty })\to {\mathbb{R}}^{n\times n}$ satisfying the semigroup property

(1.1) $\begin{array}{ll}g(x+y)& =g(x)g(y),x,y\ge 0,\end{array}$(1.1)

(1.2) $\begin{array}{ll}g(0)& =\mathrm{id},\end{array}$(1.2)

where $\mathrm{id}$ is the identity matrix, are given by the maps $g(x)=\exp (Mx)$, where $M\in {\mathbb{R}}^{n\times n}$ is arbitrary. See, e.g. Problem 2.1 in  (Engel & Nagel, 2000) and the subsequent discussion. To the best of our knowledge, few authors have considered non-continuous solutions of (1.1(1.1) $\begin{array}{ll}g(x+y)& =g(x)g(y),x,y\ge 0,\end{array}$(1.1) ) in the multidimensional case d > 1. Kuczma and Zajtz  (Kuczma & Zajtz, 1966) determine all matrix semigroups $(g(x){)}_{x\in \mathbb{R}}$ where $x\mapsto g(x)$ is measurable. Zajtz  (Zajtz, 1971) characterizes the matrix semigroups $(g(x){)}_{x\in \mathbb{Q}}$ indexed by rational numbers. The one-dimensional case, which is equivalent to Cauchy’s functional equation,

(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3)

is well studied, on the other hand (see  (Aczél, 1966; Bingham et al., 1987)). Our motivation to investigate non-continuous matrix semigroups stems from probability theory: In a companion paper, we develop a characterization of all self-similar Gaussian Markov processes. By self-similarity, the bivariate covariance function of such a multi-dimensional stochastic process can be transformed to a matrix function of a single argument, which must satisfy (1.1(1.1) $\begin{array}{ll}g(x+y)& =g(x)g(y),x,y\ge 0,\end{array}$(1.1) ). See Section 2 for some more details. Our main result (Theorem 2.6) determines all solutions of (1.1(1.1) $\begin{array}{ll}g(x+y)& =g(x)g(y),x,y\ge 0,\end{array}$(1.1) ) which satisfy a mild boundedness assumption. In Section 3, we prepare the proof by providing a decomposition of the vector space into invariant subspaces, and establish some useful properties of the decomposition. The main proofs (in Section 5) are preceded by a discussion of the Jordan—Chevalley decomposition of a semigroup, in Section 4. In Appendix A, we give two auxiliary results on Cauchy’s functional equation.

2. Preliminaries and main result

To motivate our investigation, we first recall some facts about Gaussian stochastic processes (see, e.g. Lifshits, 2012; Nourdin, 2012, for more information). Let $X=(X_t{)}_{t\ge 0}$ be a d-dimensional real centered Gaussian process. The covariance of X is a matrix-valued function $(s,t)\mapsto R(s,t)\in {\mathbb{R}}^{d\times d}$ satisfying

$v^{\mathrm{T}}R(s,t)u=\mathbb{E}[X_s^{\mathrm{T}}v{X}_t^{\mathrm{T}}u],s,t\ge 0,u,v\in {\mathbb{R}}^d,$

and uniquely characterizes the law of the process. Suppose that X is self-similar, and that $R(1,1)=\mathrm{id}$ is the identity matrix. In terms of the covariance function, self-similarity means that $R(as,at)=a^{2H}R(s,t)$ for $a,s,t>0$ and some self-similarity parameter H > 0. By a classical criterion  (Doob, 1953) [Theorem V.8.1], X is a Markov process if and only if its covariance function satisfies

(2.1) $\begin{array}{l}R(s,t)R(t,t{)}^{-1}R(t,u)=R(s,u),0\le s\le t\le u.\end{array}$(2.1)

Upon introducing $g(x):=R(e^{-x},1)$, self-similarity allows to reduce (2.1(2.1) $\begin{array}{l}R(s,t)R(t,t{)}^{-1}R(t,u)=R(s,u),0\le s\le t\le u.\end{array}$(2.1) ) to

$g(x)g(y)=g(x+y),x,y,\ge 0.$

This observation has been used in dimension d = 1 to prove that certain Gaussian processes do not have the Markov property (see, e.g. (Nourdin, 2012), [Theorem 2.3]). Unifying and generalizing these isolated results in a companion paper, we obtain a classification of all d-dimensional self-similar Gaussian Markov processes. A full classification requires finding all solutions of (1.1(1.1) $\begin{array}{ll}g(x+y)& =g(x)g(y),x,y\ge 0,\end{array}$(1.1) ), without assuming continuity. The boundedness assumption 2.5 stated below causes no problems, though. To state our main result, define the rotation matrix

$Q(\theta )=\left(\begin{array}{ll}\cos (\theta )& \sin (\theta )\\ -\sin (\theta )& \cos (\theta )\end{array}\right),\theta \in \mathbb{R},$

and, for even k and a function $\nu :\mathbb{R}\to \mathbb{R}$, the block-diagonal matrix

(2.2) $Q_k^{\nu }(x):=\left(\begin{array}{lll}Q(\nu (x))& & 0\\ & \ddots & \\ 0& & Q(\nu (x))\end{array}\right)\in {\mathbb{R}}^{k\times k}$(2.2)

consisting of $k/2$ rotation matrices. In our statements, ν will denote some non-measurable (equivalently, non-continuous) solution of (1.3(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3) ). The following assumption is in force throughout the paper.

Assumption 2.1.

In the following, V denotes a real d-dimensional vector space equipped with an inner product $⟨\cdot ,\cdot ⟩$.

Definition 2.2.

We write L(V) for the set of linear maps from V to itself. The operator norm induced by the inner product $⟨\cdot ,\cdot ⟩$ is denoted by $\Vert \cdot {\Vert }_{\mathrm{op}}$. If the basis is clear from the context, we will identify elements of L(V) and d × d matrices.

Definition 2.3.

We say that $g:{\mathbb{R}}_{\ge 0}\to L(V)$ is a semigroup if $g(0)=\mathrm{id}$ and

$g(x+y)=g(x)g(y)$

for all $x,y\ge 0$. Semigroups acting on $V^{\mathbb{C}}$, the complexification of V, are defined by the same property.

Definition 2.4.

We say that a semigroup $(g(x){)}_{x\ge 0}$ in L(V) is elementary if there exists an orthonormal basis such that

$\begin{array}{ll}g(x)=& \exp (Mx),x\ge 0,\text{or}\\ g(x)=& Q_d^{\nu }(x)\exp (Mx),x\ge 0,\end{array}$

for some non-continuous ν satisfying Cauchy’s Equation (1.3)(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3) and some matrix $M\in {\mathbb{R}}^{d\times d}$.

Assumption 2.5.

The semigroup $g(x{)}_{x\ge 0}$ satisfies $\Vert g(x){\Vert }_{\mathrm{op}}\le f(x)$ for $x\ge 0$, where the function $f:[0,\mathrm{\infty })\to \mathbb{R}$ is locally bounded, right-continuous at 0 and satisfies $f(0)=1$.

We can now state our main theorem, which will be proven at the end of Section 5.

Theorem 2.6.

Let $(g(x){)}_{x\ge 0}$ be a semigroup satisfying Assumption 2.5. Then, there exists an orthogonal decomposition $V=\underset{i=1}{\overset{n}{⨁}}{V}_i$ such that each V_i is invariant under g(x), and either g(x) is elementary on V_i or $g(x){|}_{V_i}=0$ for x > 0.

We call $g((x){)}_{x\ge 0}$ degenerate if $g(x)=0$ for x > 0.

Example 2.7.

Let $c\in \mathbb{R}$, and let $f:\mathbb{R}\to \mathbb{R}$ be a non-continuous solution of Cauchy’s functional Equation (1.3)(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3) . Then, the semigroup given by

$g(x):=\left(\begin{array}{ll}\cos (f(x))& -\sin (f(x))\\ \sin (f(x))& \cos (f(x))\end{array}\right)e^{cx},x\ge 0,$

is an example of a semigroup covered by Theorem 2.6 (by putting d = 2 and $M=c\mathrm{id}$ in Definition 2.4), but not the previous results mentioned at the beginning of the introduction. It illustrates that, in contrast to dimension one, a two-dimensional locally bounded semigroup need not be continuous.

3. Primary decomposition for semigroups

In this section, we discuss a decomposition of V into subspaces which are invariant for the given semigroup.

Definition 3.1.

Let S be a linear operator on the vector space V. A subspace $U\subseteq V$ is called S-invariant if S maps U into U.

The primary decomposition theorem from linear algebra [O’Meara et al., 2011, Theorem 1.5.1] decomposes a vector space into invariant subspaces for a given operator. On each subspace, the operator has a single real eigenvalue or a pair of conjugate complex eigenvalues. Instead of a single operator, we need the following version for semigroups.

Theorem 3.2.

Primary decomposition For any semigroup $(g(x){)}_{x\ge 0}$ of linear maps acting on V, there exists a decomposition $V=\underset{i=1}{\overset{n}{⨁}}{V}_i$ with $\dim (V_i)\ge 1$ such that each V_i is g(x)-invariant for all $x\ge 0$, and for all i one of the following holds:

(1)	For all $x\ge 0$, g(x) has one eigenvalue $\lambda (x)\ge 0$ on Vi,
(2)	For all $x\ge 0$, g(x) has eigenvalues $\lambda (x),\bar{\lambda }(x)\in \mathbb{C}$ on Vi. They may coincide (and thus be real) for some values of x, but not for all $x\ge 0$.

Proof.

It is known that the primary decomposition extends to commuting sets of matrices. Indeed, the decomposition $V=\underset{i=1}{\overset{n}{⨁}}{V}_i$ into invariant subspaces follows from Theorem 5 on p. 40 in  (Jacobson, 1962), applied to the span of the semigroup $(g(x){)}_{x\ge 0}$. Thus, we only need to argue why (1) or (2) follows from the semigroup property. Assume that each g(x) has only one eigenvalue $\lambda (x)\in \mathbb{R}$ on some V_i. Since the g(x) commute, they share a common eigenvector v_i, and thus we have $\lambda (x)\lambda (y)=\lambda (x+y)$ for $x,y\ge 0$. Suppose $\lambda (x_0)<0$ for some $x_0>0$. Then, we have $\lambda (x_0)=\lambda (x_0/2{)}^2$, and hence $\lambda (x_0/2)\notin \mathbb{R}$, contradicting $\lambda (x)\in \mathbb{R}$ for all $x\ge 0$. Hence $\lambda (x)\in {\mathbb{R}}_{\ge 0}$ for all $x\ge 0$.

Theorem 3.3.

Consider a semigroup $g=(g(x){)}_{\ge 0}$ acting on $V^{\mathbb{C}}$, the complexification of V. Then, there exists a decomposition $V^{\mathbb{C}}=\underset{i=1}{\overset{n}{⨁}}{V}_i$ with $\dim (V_i)\ge 1$ such that, for each i and all $x\ge 0$, the space V_i is g(x)-invariant, and g(x) has only one eigenvalue $\lambda (x)\in \mathbb{C}$ on V_i.

Proof.

This is an immediate consequence of Theorem 5 on p. 40 in  (Jacobson, 1962) (cf. the preceding proof).

Definition 3.4.

We call the decomposition from Theorem 3.2 simultaneous real primary decomposition (SRPD) of V, omitting “w.r.t. g” if the semigroup is clear from the context. The component V_i is of first type in case (1), and of second type in case (2). Similarly, the simultaneous primary decomposition (SPD) is the decomposition from Theorem 3.3.

Lemma 3.5.

Let g be a semigroup acting on V, and let $V_i\subseteq V$ be a subspace of first type from the SRPD of V. Then, there exists a common eigenvector $v\in V$ such that $g(x)v=\lambda (x)v$ for all x > 0.

Proof.

We present an algorithm which yields the subspace of common eigenvectors. If $V_i=:V_i^{(0)}$ itself is this subspace, then we are done. Otherwise there exists $x_0>0$ such that the eigenspace $V_i^{(1)}$ of $g(x_0)$ is a strict subspace of V_i. For any y > 0 and any $v\in V_i^{(1)}$ we have

$\begin{array}{ll}0=g(y)0& =g(y)[\lambda (x_0)v-g(x_0)v]\\ & =\lambda (x_0)[g(y)v]-g(x_0)[g(y)v].\end{array}$

It follows that $V_i^{(1)}$ is g(x)-invariant for all x > 0. Now either $V_i^{(1)}$ consists of common eigenvectors, or we can again find $x_1>0$ such that the eigenspace $V_i^{(2)}$ of $g(x_1)$ in $V_i^{(1)}$ is a strict subspace of $V_i^{(1)}$. Repeating this argument yields a sequence of nontrivial subspaces $V_i^{(n)}$ whose dimensions are strictly decreasing, hence it has to terminate. Clearly, the final vector space in this sequence is the space of common eigenvectors of the semigroup.

Lemma 3.6.

Let g be a semigroup acting on V, and let V_i be a subspace from the SRPD of V such that there exists $x_0>0$ with $\lambda (x_0)=0$. Then, we have $\lambda (x)=0$ for all x > 0. Furthermore $V_i\subseteq \ker (g(x))$ for all x > 0.

Proof.

By the previous lemma, there exists a common eigenvector $v\in V_i$. (Since commuting matrices are simultaneously triangularizable it follows that they share a common eigenvector $v\in V^{\mathbb{C}}$.) We have

$\lambda (x+y)v=g(x+y)v=g(x)g(y)v=\lambda (x)\lambda (y)v.$

Hence λ satisfies $\lambda (x)\lambda (y)=\lambda (x+y)$. Since $\lambda (x_0)=0$ we have $\lambda (x_0/n)=0$ for all $n\in \mathbb{N}$ since $\lambda (x_0/n{)}^n=0$. Let y > 0, then there exists $n\in \mathbb{N}$ such that $x_0/n<y$. We obtain

$\lambda (y)=\lambda (x_0/n)\lambda (y-x_0/n)=0.$

Since $\lambda (y)=0$ for all y > 0 it follows that the characteristic polynomial of g(y) satisfies ${\chi }_{g(y)}(Z)=Z^{\dim (V_i)}$ and hence, by the Cayley—Hamilton theorem, $g(y{)}^{\dim (V_i)}\equiv 0$. For $v\in V_i$ we obtain

$g(y)v=g(y/\dim (V_i){)}^{\dim (V_i)}v=0.$

Corollary 3.7.

Let g be a semigroup acting on V, with SRPD $V=\underset{i=1}{\overset{n}{⨁}}{V}_i$. For any x > 0 we have $\ker (g(x))=\underset{i\in I}{⨁}{V}_i$, where

$I=\{i|V_i\text{of type 1 with }\lambda \equiv 0\}.$

Proof.

If $V_i\subseteq \ker (g(x))$, then $0=\lambda (x)\in \mathbb{R}$ for all $x\ge 0$ and hence V_i is of type 1. By the previous lemma, we have

$\ker (g(x))\supseteq \underset{i\in I}{⨁}{V}_i.$

If the inclusion was strict, then there would exist V_j with $j\notin I$ such that $\ker (g(x))\cap V_j\ne \mathrm{\varnothing }$. But since g(x) is invertible on V_j this gives a contradiction, hence we have equality.

Corollary 3.8.

Let g be a semigroup acting on V. Then, there exists a decomposition $V=V_1\oplus V_2$, such that $g(x){|}_{V_1}$ is invertible for all $x\ge 0$ and $g(x){|}_{V_2}\equiv 0$ for x > 0.

4. Multiplicative Jordan–Chevalley decomposition of semigroups

Due to Corollary 3.8, from now on we assume in most of our statements that g(x) is invertible for all $x\ge 0$. A standard result from linear algebra, the Jordan—Chevalley decomposition, asserts that any matrix A can be uniquely decomposed as $A=D+N$, where D is diagonalizable, N is nilpotent and D and N commute. If A is invertible, then we can express it as $A=D(\mathrm{id}+D^{-1}N):=DT$ with T unipotent and commuting with D.

Definition 4.1.

For an invertible linear map A on ${\mathbb{K}}^d$ with $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$, the multiplicative decomposition A = DT into commuting factors with D diagonalizable and T unipotent, is called the multiplicative Jordan—Chevalley decomposition.

For background on the (multiplicative) Jordan—Chevalley decomposition, we refer to Section 15.1 in  (Humphreys, 1975). We now analyze the structure of the multiplicative Jordan—Chevalley decomposition of a semigroup.

Theorem 4.2.

Let $(g(x){)}_{x\ge 0}$ be a semigroup of invertible linear maps acting on ${\mathbb{K}}^d$ with $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$ and let $g(x)=D(x)T(x)$ be the multiplicative Jordan—Chevalley decomposition of each g(x). Then $((D(x){)}_{x\ge 0})$ and $(T(x){)}_{x\ge 0}$ each form a semigroup, and the two families commute with each other, i.e. $T(x)D(y)=D(y)T(x)$ for all $x,y\ge 0$.

Proof.

We can w.l.o.g. assume that $\mathbb{K}=\mathbb{C}$, since by uniqueness of the Jordan—Chevalley decomposition $D(x)\in {\mathbb{R}}^{d\times d}$ and $T(x)\in {\mathbb{R}}^{d\times d}$ if $g(x)\in {\mathbb{R}}^{d\times d}$ for $x\ge 0$. Take the SPD from Theorem 3.3, $V=\underset{i=1}{\overset{k}{⨁}}{V}_i$, so that each g(x) has only one eigenvalue on V_i. Denote by $g(x){|}_{V_i}=D_i(x)T_i(x)$ the multiplicative Jordan—Chevalley decomposition g(x) restricted to V_i. Denote by ${\lambda }_i(x)$ the eigenvalue of g(x) on V_i. Clearly, $D_i(x)={\lambda }_i(x)\mathrm{id}$. Since the $(g_i(x){)}_{x\ge 0}$ are a commuting family of matrices, they share a common eigenvector $v_i\in V_i$. We have

${\lambda }_i(x){\lambda }_i(y)v_i=g_i(x)g_i(y)v_i=g_i(x+y)v_i={\lambda }_i(x+y)v_i,$

and hence $(D_i(x){)}_{x\ge 0}$ is a semigroup. Since each $D_i(x)$ is a multiple of the identity, it commutes with every linear map and hence

$\begin{array}{ll}{T}_i(x)T_i(y)=& \frac{1}{{\lambda }_i(x){\lambda }_i(y)}{g}_i(x)g_i(y)=\frac{1}{{\lambda }_i(x+y)}{g}_i(x+y)\\ =& T_i(x+y),\end{array}$

which shows that $(T_i(x){)}_{x\ge 0}$ is also a semigroup. The result then follows, since by uniqueness $(⨁D_i(x))\otimes (⨁T_i(x))$ is the multiplicative Jordan—Chevalley decomposition of g(x).

Theorem 4.3.

Let $(g(x){)}_{x\ge 0}$ be a semigroup of invertible linear maps acting on V and let $g(x)=D(x)T(x)$ be its multiplicative Jordan—Chevalley decomposition. Then, there exist commuting real diagonalizable linear maps J(x) and commuting real nilpotent linear maps N(x) satisfying

(1)	$J(x)+J(y)=J(x+y)$
(2)	$N(x)+N(y)=N(x+y)$
(3)	$J(x)N(y)=N(y)J(x)$ for all $x,y\ge 0$

such that

$D(x)=\exp (J(x)),T(x)=\exp (N(x))$

and

$g(x)=\exp (J(x)+N(x)).$

Proof.

Let $V=\underset{i=1}{\overset{n}{⨁}}{V}_i$ be the SRPD and let $g_i(x):=g(x){|}_{V_i}$. Assume first that V_i is of first type and ${\lambda }_i(x)$ is the single positive eigenvalue of $g_i(x)$. Then, the multiplicative Jordan—Chevalley decomposition on V_i is $g_i(x)={\lambda }_i(x)\mathrm{id}{T}_i(x)$. Define $J_i(x):=\log ({\lambda }_i(x))\mathrm{id}$. Since $T_i(x)-\mathrm{id}$ is nilpotent, we have

$N_i(x):=\log (T_i(x))=\sum _{k=1}^{d-1}\frac{1}{k}(\mathrm{id}-T_i(x){)}^k.$

Notice that since the logarithm converges for all unipotent matrices, the exponential map between the Lie algebra of nilpotent matrices and the Lie group of unipotent matrices is bijective (see p. 35 in  (Goodman & Wallach, 1998)). Since $T_i(x)$ is a semigroup we have

$\begin{array}{ll}\exp (N_i(x))\exp (N_i(y))=& \exp (N_i(y))\exp (N_i(x))\\ =& \exp (N_i(x+y)).\end{array}$

Rewrite this as

$\begin{array}{ll}& \exp (N_i(y){)}^{-1}\exp (N_i(x))\exp (N_i(y))\\ & =\exp (\exp (N_i(y){)}^{-1}{N}_i(x)\exp (N_i(y)))\\ & =\exp (N_i(x)).\end{array}$

Since $\exp (N_i(y){)}^{-1}{N}_i(x)\exp (N_i(y))$ is nilpotent and exp is bijective, we have

$\exp (N_i(y){)}^{-1}{N}_i(x)\exp (N_i(y))=N_i(x).$

For any $t\in \mathbb{R}$ set $N_i^t(x):=t\mathrm{id}+N_i(x)$, which is invertible for $t\ne 0$. It is clear that

$N_i^t(x)\exp (N_i(y))=\exp (N_i(y))N_i^t(x).$

Applying the same idea again yields

$\exp (N_i^t(x{)}^{-1}{N}_i(y)N_i^t(x))=\exp (N_i(y)).$

Again by uniqueness we obtain

$N_i^t(x)N_i(y)=N_i(y)N_i^t(x)$

or equivalently

$N_i(x)N_i(y)=N_i(y)N_i(x).$

By commutativity we obtain

$\begin{array}{ll}\exp (N_i(x)+N_i(y))=& \exp (N_i(x))\exp (N_i(y))\\ =& \exp (N_i(x+y))\end{array}$

and hence, again by uniqueness, we have $N_i(x)+N_i(y)=N_i(x+y)$. It follows that $J_i(x)$ and $N_i(x)$ have the desired properties.

In the second case V_i is of second type. By Theorem 3.3, and since $g_i(x)$ is real, V_i decomposes over $\mathbb{C}$ as $V_i=U_1\oplus U_2$, where $g_i(x)$ has only one eigenvalue on each U_j and U₁ is isomorphic to U₂ with isomorphism given by $u\in U_1\mapsto \bar{u}\in U_2$. By taking the principal branch of the logarithm, in the same manner as in the real case, we obtain commuting $J,N$ on U₁ and $\bar{J},\bar{N}$ on U₂ satisfying Cauchy’s equation such that

$g_i(x)=\exp (J(x)+N(x))\oplus \exp (\bar{J}(x)+\bar{N}(x)).$

Notice that $J(x)=(\mu (x)+\mathbf{i}\nu (x))\mathrm{id}$, where $\mathbf{i}=\sqrt{-1}$ and $\nu (x)$ satisfies Cauchy’s functional equation on $\mathbb{R}/[-\pi ,\pi )$. By Lemma A.1, we can lift any solution on $\mathbb{R}/[-\pi ,\pi )$ to a solution on $\mathbb{R}$ such that linearity is preserved. From now on denote this lift by $\nu (x)$. Hence, if we choose any basis of U₁ and its complex conjugate on U₂ we obtain that $g_i(x)$ is similar to

$g_i(x)\sim \exp \left[\left(\begin{array}{ll}J(x)+N(x)& 0\\ 0& \bar{J}(x)+\bar{N}(x)\end{array}\right)\right].$

Taking the similarity transform with the matrix

$A:=\frac{1}{\sqrt{2}}\left(\begin{array}{ll}\mathrm{id}& \mathbf{i}\cdot \mathrm{id}\\ \mathbf{i}\cdot \mathrm{id}& \mathrm{id}\end{array}\right),$

where $\mathrm{id}={\mathrm{id}}_{\dim (U_1)}$, we obtain

$\begin{array}{ll}& A\left(\begin{array}{ll}J(x)+N(x)& 0\\ 0& \bar{J}(x)+\bar{N}(x)\end{array}\right)A^{-1}\\ & =\left(\begin{array}{ll}\mathrm{Re}(J(x)+N(x))& \mathrm{Im}(J(x)+N(x))\\ -\mathrm{Im}(J(x)+N(x))& \mathrm{Re}(J(x)+N(x))\end{array}\right)\in L(V_i).\end{array}$

Since matrix similarity over $\mathbb{C}$ is equivalent to matrix similarity over $\mathbb{R}$ for two real matrices, there exists a real matrix B_i on V_i such that

(4.1) $\begin{array}{ll}{g}_i(x)=& \exp [B_i\left(\begin{array}{ll}\mathrm{Re}(J(x))& \mathrm{Im}(J(x))\\ -\mathrm{Im}(J(x))& \mathrm{Re}(J(x))\end{array}\right)B_i^{-1}\\ & +B_i\left(\begin{array}{ll}\mathrm{Re}(N(x))& \mathrm{Im}(N(x))\\ -\mathrm{Im}(N(x))& \mathrm{Re}(N(x))\end{array}\right)B_i^{-1}].\end{array}$(4.1)

Setting

$\begin{array}{ll}{J}_i(x):=& B_i\left(\begin{array}{ll}\mathrm{Re}(J(x))& \mathrm{Im}(J(x))\\ -\mathrm{Im}(J(x))& \mathrm{Re}(J(x))\end{array}\right)B_i^{-1}\\ =& B_i\left(\begin{array}{ll}\mathrm{id}\mu (x)& \mathrm{id}\nu (x)\\ -\mathrm{id}\nu (x)& \mathrm{id}\mu (x)\end{array}\right)B_i^{-1}\end{array}$

and

$N_i(x):=B_i\left(\begin{array}{ll}\mathrm{Re}(N(x))& \mathrm{Im}(N(x))\\ -\mathrm{Im}(N(x))& \mathrm{Re}(N(x))\end{array}\right)B_i^{-1},$

we see that J_i and N_i satisfy the desired conditions, with $D_i(x)=\exp (J_i(x))$ and $T_i(x)=\exp (N_i(x))$ by uniqueness of the Jordan—Chevalley decomposition. The direct sums $\underset{i=1}{\overset{n}{⨁}}{J}_i(x)$ and $\underset{i=1}{\overset{n}{⨁}}{N}_i(x)$ give the required matrices. Furthermore, in the case where, there exists x > 0 for which $g_i(x)$ has a complex eigenvalue $\lambda (x)=\mu (x)+\mathbf{i}\nu (x)$ we have

$\begin{array}{ll}{g}_i(x)=& B_i\underset{U_i(x)}{\underbrace{\left(\begin{array}{ll}\cos (\nu (x))\mathrm{id}& \sin (\nu (x))\mathrm{id}\\ -\sin (\nu (x))\mathrm{id}& \cos (\nu (x))\mathrm{id}\end{array}\right)}}{B}_i^{-1}\exp (\mu (x)\mathrm{id}\\ & +B_{i}N(x)B_i^{-1}),\end{array}$

where $U_i(x)\in SO(\dim (V_i))$. Recall the matrix defined in (2.2(2.2) $Q_k^{\nu }(x):=\left(\begin{array}{lll}Q(\nu (x))& & 0\\ & \ddots & \\ 0& & Q(\nu (x))\end{array}\right)\in {\mathbb{R}}^{k\times k}$(2.2) ). By changing the order of the basis, we have that U_i is similar to the block diagonal matrix (recall the notation (2.2(2.2) $Q_k^{\nu }(x):=\left(\begin{array}{lll}Q(\nu (x))& & 0\\ & \ddots & \\ 0& & Q(\nu (x))\end{array}\right)\in {\mathbb{R}}^{k\times k}$(2.2) ))

(4.2) $\begin{array}{l}{U}_i(x)\sim \left(\begin{array}{lll}Q(\nu (x))& & \\ & \ddots & \\ & & Q(\nu (x))\end{array}\right)=Q_{\dim (V_i)}^{\nu }(x).\end{array}$(4.2)

Hence

(4.3) $g_i(x)={\tilde{B}}_i{Q}_{\dim (V_i)}^{\nu }(x){\tilde{B}}_i^{-1}\exp (\mu (x)\mathrm{id}+B_{i}N(x)B_i^{-1}),$(4.3)

where ${\tilde{B}}_i$ is the composition of B_i with some permutation matrix P.

5. Proof of the main result

After providing some final preparatory results, this section ends with the proof of Theorem 2.6. Consider again the SRPD from Theorem 3.2. Now on each V_i, g(x) has either one positive eigenvalue ${\lambda }_i(x)=e^{{\mu }_i(x)}$ or two complex conjugate eigenvalues ${\lambda }_i(x)=e^{{\mu }_i(x)+\mathbf{i}{\nu }_i(x)}$ and $\overline{{\lambda }_i}(x)=e^{{\mu }_i(x)-\mathbf{i}{\nu }_i(x)}$, where it is possible that ${\nu }_i(x)=\pi$ for some values of x. If V_i is of first type, set ${\nu }_i(x)=0$. Each ν_i is a solution to Cauchy’s functional Equation (1.3)(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3) . Consider then the set $\{f|f={\nu }_i\text{or }f=-{\nu }_i\text{for some }1\le i\le n\}$ and partition it into equivalent solutions, according to Definition A.2. This then gives a partition of the index set $\{1,\dots ,n\}=\bigcup _{l=1}^k{I}_l$ in the following manner: If ${\nu }_i\sim {\nu }_j$ or ${\nu }_i\sim -{\nu }_j$, then $i,j$ are in the same subset of the partition. This is well-defined, since if $f\sim g$ then $-f\sim -g$. Set $W_l:=\underset{i\in I_l}{⨁}{V}_i$.

Definition 5.1.

We call the decomposition $V=\underset{i=1}{\overset{k}{⨁}}{W}_i$ the partitioned SRPD of V.

Furthermore associate with each W_j one solution η_j of Cauchy’s equation such that ${\eta }_j\sim {\nu }_i$ with $i\in I_j$. If η_j is linear we always take ${\eta }_j\equiv 0$. Notice that for i ≠ j we have ${\eta }_i\nsim {\eta }_j$, hence there can be at most one W_i with ${\eta }_i=0$, and furthermore this is the only W_i which can have odd dimension since it contains all V_j of type 1 (recall Definition 3.4).

Theorem 5.2.

Let $(g(x){)}_{x\ge 0}$ be a semigroup of invertible linear maps acting on V, and denote by $V=\underset{i=1}{\overset{k}{⨁}}{W}_i$ its partitioned SRPD. Denote by η_i the solution associated with W_i. If η_i is non-continuous, then there exists a change of basis A_i on W_i such that

$g(x){|}_{W_i}=A_i{Q}_{d_i}^{{\eta }_i}(x)A_i^{-1}\exp (G_i(x)),$

where $d_i=\dim (W_i)$, and

$A_i{Q}_{d_i}^{{\eta }_i}(x)A_i^{-1}{G}_i(y)=G_i(y)A_i{Q}_{d_i}^{{\nu }_i}(x)A_i^{-1}$

for all $x,y\ge 0$. If ${\eta }_i=0$, then

$g(x){|}_{W_i}=\exp (G_i(x)).$

In both cases $G_i(x):W_i\to W_i$ is a commuting family of matrices on W_i satisfying Cauchy’s functional equation. Furthermore, if v is a common eigenvector of $(G_i(x){)}_{x\ge 0}$ such that $G_i(x)v=[\mu (x)+\mathbf{i}\nu (x)]v$, then ν is linear in x.

Proof.

Assume first that η_i, the associated solution of Cauchy’s functional equation, is non-continuous. For each W_i we have the decomposition $W_i=\underset{j\in I_i}{⨁}{V}_j$ where V_j are subspaces from the SRPD. Since η_i is non-continuous, each V_j in the direct sum has to be of second type, hence by (4.3(4.3) $g_i(x)={\tilde{B}}_i{Q}_{\dim (V_i)}^{\nu }(x){\tilde{B}}_i^{-1}\exp (\mu (x)\mathrm{id}+B_{i}N(x)B_i^{-1}),$(4.3) ) on each V_j we have

$g_j(x)={\tilde{B}}_j{Q}_{\dim (V_j)}^{{\nu }_j}(x){\tilde{B}}_j^{-1}\exp (\mu (x)\mathrm{id}+B_{j}N(x)B_j^{-1}).$

By definition of η_i we have ${\eta }_i\sim {\nu }_j$ and hence there exists $c_j\in \mathbb{R}$ such that ${\nu }_j(x)={\eta }_i(x)+c_{j}x$. Let

$C_j:=c_j\left(\begin{array}{lll}L& & \\ & \ddots & \\ & & L\end{array}\right)$

with

$L:=\left(\begin{array}{ll}0& 1\\ -1& 0\end{array}\right),$

where C_j is of dimension $\dim (V_j)\times \dim (V_j)$. Then, we have

$\begin{array}{ll}{g}_j(x)& ={\tilde{B}}_j{Q}_{\dim (V_j)}^{{\nu }_j}(x){\tilde{B}}_j^{-1}\exp ({\mu }_j(x)\mathrm{id}+B_j{N}_j(x)B_j^{-1})\\ & ={\tilde{B}}_j{Q}_{\dim (V_j)}^{{\eta }_i}(x)\exp (C_{j}x){\tilde{B}}_j^{-1}\exp ({\mu }_j(x)\mathrm{id}+B_j{N}_j(x)B_j^{-1})\\ & ={\tilde{B}}_j{Q}_{\dim (V_j)}^{{\eta }_i}(x){\tilde{B}}_j^{-1}\exp ({\tilde{B}}_j{C}_{j}x{\tilde{B}}_j^{-1}+{\mu }_j(x)\mathrm{id}+B_j{N}_j(x)B_j^{-1}).\end{array}$

Set ${\tilde{G}}_j(x):={\tilde{B}}_j{C}_{j}x{\tilde{B}}_j^{-1}+{\mu }_j(x)\mathrm{id}+B_j{N}_j(x)B_j^{-1}$. Then

(5.1) $A_i:=\underset{j\in I_i}{⨁}{\tilde{B}}_j\text{and}{G}_i(x):=\underset{j\in I_i}{⨁}{\tilde{G}}_j(x)$(5.1)

satisfy

$g(x){|}_{W_i}=\underset{j\in I_i}{⨁}{g}_j(x)=A_i{Q}_{d_i}^{{\eta }_i}(x)A_i^{-1}\exp (G_i(x)).$

By construction the common eigenvalues of $\exp (G_i(x))$ satisfy $\lambda (x)=e^{{\mu }_j(x)\pm \mathbf{i}{c}_{j}x}$. If ${\eta }_i\equiv 0$, then by Theorem 4.3 there exists $G_i(x)$ such that $g(x){|}_{W_i}=\exp (G_i(x))$. By the definition of W_i, it follows that the imaginary parts of the eigenvalues of $G_i(x)$ have to be equivalent to ${\eta }_i\equiv 0$, hence they have to be linear.

Theorem 5.3.

Let g be a non-degenerate semigroup acting on V which satisfies Assumption 2.5. Then, there exists a matrix M and a semigroup $S(x)\in SO(d)$, where SO(d) is the set of special orthogonal matrices, such that

$g(x)=S(x)\exp (Mx),x\ge 0,$

with M commuting with S(x).

Proof.

Consider the SRPD $V=\underset{i=1}{\overset{n}{⨁}}{V}_i$ from Theorem 3.2, and define $g_i(x):=g(x){|}_{V_i}$. Clearly, $\Vert g_i(x){\Vert }_{\mathrm{op}}\le \Vert g(x){\Vert }_{\mathrm{op}}\le f(x)$. For each eigenvalue ${\lambda }_i(x)=e^{{\mu }_i(x)+\mathbf{i}{\nu }_i(x)}$ on V_i, we have

$|{\lambda }_i(x)|\le \Vert g_i(x){\Vert }_{\mathrm{op}}\le f(x).$

Since ${\lambda }_i(x){\lambda }_i(y)={\lambda }_i(x+y)$, we have $|{\lambda }_i(x)||{\lambda }_i(y)|=|{\lambda }_i(x+y)|$. Moreover, ${\mu }_i(x)=\log (|{\lambda }_i(x)|)$, and so ${\mu }_i(x)\le \log (f(x))$ for $x\ge 0$ and ${\mu }_i(x)+{\mu }_i(y)={\mu }_i(x+y)$. It follows that µ_i is locally bounded and hence that ${\mu }_i(x)=a_{i}x$ for some $a_i\ge 0$. As in the proof of Theorem 4.3, on V_i we have $g_i(x)=\exp (J_i(x))T_i(x)$ where $J_i(x)$ is diagonalizable with eigenvalues $a_{i}x\pm \mathbf{i}{\nu }_i(x)$, and thus

$\Vert T_i(x){\Vert }_{\mathrm{op}}\le f(x)e^{-a_{i}x}.$

As in Theorem 4.3, set

$N_i(x):=\log (T_i(x))=\sum _{k=1}^{d-1}\frac{1}{k}(\mathrm{id}-T_i(x){)}^k.$

Then, since $\Vert \mathrm{id}-T_i(x){\Vert }_{\mathrm{op}}\le 1+f(x)e^{-a_{i}x}$, we obtain

(5.2) $\begin{array}{ll}\Vert N_i(x){\Vert }_{\mathrm{op}}\le & \sum _{k=1}^{d-1}\frac{1}{k}\Vert \mathrm{id}-T_i(x){\Vert }_{\mathrm{op}}^k\\ \le & \sum _{k=1}^{d-1}\frac{1}{k}(1+f(x)e^{-a_{i}x}{)}^k=:F(x),\end{array}$(5.2)

where F(x) is again locally bounded. Since the operator norm in finite dimensions is equivalent to the $L^{\mathrm{\infty }}$-norm, it follows that each entry of $N_i(x)$ is locally bounded and satisfies Cauchy’s functional equation. Thus, there exists a nilpotent linear map P_i such that $N_i(x)=P_{i}x$. Let $V=\underset{i=1}{\overset{k}{⨁}}{W}_i$ be the partitioned SRPD of V. Assume first that the solution associated with W_i is ${\eta }_i\equiv 0$. Since $W_i=\underset{j\in I_i}{⨁}{V}_j$ we have $g(x){|}_{W_i}=\underset{j\in I_i}{⨁}\exp (J_j(x)+P_{j}x)$. The real part of the eigenvalues of $J_j(x)$ is linear in x and by Theorem 5.2 also the complex parts have to be linear since ${\nu }_i\equiv 0$. Since $J_j(x)$ is diagonalizable and all its eigenvalues are continuous in x, $J_j(x)$ is continuous in x and hence $J_j(x)={\tilde{M}}_{j}x$. Setting $M_i:=\underset{j\in I_i}{⨁}({\tilde{M}}_j+P_j)$ and $S_i(x)=\mathrm{id}$ yields $g(x){|}_{W_i}=S_i(x)\exp (M_{i}x)$. By Theorem 5.2 and (5.1(5.1) $A_i:=\underset{j\in I_i}{⨁}{\tilde{B}}_j\text{and}{G}_i(x):=\underset{j\in I_i}{⨁}{\tilde{G}}_j(x)$(5.1) ), we have for W_i with η_i non-continuous

$g(x){|}_{W_i}=A_i{Q}_{d_i}^{{\eta }_i}(x)A_i^{-1}\exp (G_i(x))$

with

$\begin{array}{ll}{G}_i(x)& =\underset{j\in I_i}{⨁}({\tilde{B}}_j{C}_{j}x{\tilde{B}}_j^{-1}+{\mu }_j(x)\mathrm{id}+B_j{N}_j(x)B_j^{-1})\\ & =\underset{j\in I_i}{⨁}({\tilde{B}}_j{C}_j{\tilde{B}}_j^{-1}+a_j\mathrm{id}+B_j{P}_j{B}_j^{-1})x.\end{array}$

Hence, setting $M_i:=\underset{j\in I_i}{⨁}({\tilde{B}}_j{C}_j{\tilde{B}}_j^{-1}+a_j\mathrm{id}+B_j{P}_j{B}_j^{-1})$, we have $G_i(x)=M_{i}x$. Thus

$g(x){|}_{W_i}=A_i{Q}_{d_i}^{{\eta }_i}(x)A_i^{-1}\exp (M_{i}x).$

Next we show that $S_i(x):=A_i{Q}_{d_i}^{{\eta }_i}(x)A_i^{-1}$ is an isometry on W_i. Since $\Vert g(x){\Vert }_{\mathrm{op}}\le f(x)$, we have

(5.3) $\begin{array}{l}⟨v,v⟩-\frac{1}{f(x{)}^2}⟨g(x)v,g(x)v⟩\ge 0\end{array}$(5.3)

for any $v\in V$. Hence, for $v\in W_i$ we have

$⟨v,v⟩-\frac{1}{f(x{)}^2}⟨\exp (M_{i}x)S_i(x)v,\exp (M_{i}x)S_i(x)v⟩\ge 0.$

Fix an arbitrary $x_0\ge 0$, and set ${\theta }_0:={\eta }_i(x_0)$. The graph of η_i is dense in $\mathbb{R}\times \mathbb{R}$ (see Appendix A), and so there is a sequence of positive reals x_n with $\underset{n\to \mathrm{\infty }}{lim}{x}_n=0$ such that $lim{\eta }_i(x_n)={\theta }_0$. Then, for $v\in W_i$ we obtain

$\begin{array}{ll}0& \le lim(⟨v,v⟩-\frac{1}{f(x_n{)}^2}⟨\exp (M_i{x}_n)S_i(x_n)v,\\ & \exp (M_i{x}_n)S_i(x_n)v⟩)\\ & =⟨v,v⟩-⟨S_i(x_0)v,S_i(x_0)v⟩,\end{array}$

where $lim\frac{1}{f(x_n)}=1$ since f is right-continuous at 0. Choose $({\tilde{x}}_n)$ such that $lim{\tilde{x}}_n=0$ and $lim{\eta }_i({\tilde{x}}_n)=-{\theta }_0$. Then, since $S(x_0)v\in W_i$, we obtain

$\begin{array}{ll}0\le & lim(⟨S(x_0)v,S(x_0)v⟩-\frac{1}{f({\tilde{x}}_n{)}^2}\\ & ⟨\exp (M_i{\tilde{x}}_n)S_i({\tilde{x}}_n)S_i(x_0)v,\exp (M_i{\tilde{x}}_n)S_i({\tilde{x}}_n)S_i(x_0)v⟩)\\ =& ⟨S_i(x_0)v,S_i(x_0)v⟩-⟨v,v⟩,\end{array}$

since $Q_{d_i}(-{\theta }_0)Q_{d_i}({\theta }_0)=\mathrm{id}$. Hence, we obtain

$⟨v,v⟩=⟨S(x_0)v,S(x_0)v⟩,$

which implies that $S(x_0)$ is an isometry on W_i. Next we show that all W_i are pairwise orthogonal. Let $v\in W_i$ and $u\in W_j$ for i ≠ j. By Lemma A.3, w.l.o.g. there is a sequence ${\hat{x}}_n$ such that $lim{\hat{x}}_n=0$, $lim{\eta }_i({\hat{x}}_n)=\pi$ and $lim{\eta }_j({\hat{x}}_n)=\theta \in (-\pi ,\pi )$. Hence $limS({\hat{x}}_n){|}_{W_i}=-\mathrm{id}$ and $limS({\hat{x}}_n){|}_{W_j}=A_j{Q}_{d_j}(\theta )A_j^{-1}$ with $\theta \in (-\pi ,\pi )$. We have the identity $Q_{d_j}(\varphi )+Q_{d_j}(-\varphi )=2\cos (\varphi )\mathrm{id}$. Applying this to $\varphi :=\theta /2$ we obtain

$\mathrm{id}+Q_{d_j}(\theta )=2\cos (\theta /2)Q_{d_j}(\theta /2).$

Hence we have $(\mathrm{id}+Q_{d_j}(\theta ){)}^{-1}=(2\cos (\theta /2){)}^{-1}{Q}_{d_j}(-\theta /2)$. Set $\tilde{u}:=A_j^{-1}(\mathrm{id}+Q_{d_j}(\theta ){)}^{-1}{A}_{j}u$. Then $\tilde{u}\in W_j$, and applying (5.3(5.3) $\begin{array}{l}⟨v,v⟩-\frac{1}{f(x{)}^2}⟨g(x)v,g(x)v⟩\ge 0\end{array}$(5.3) ) to $v+\tilde{u}$ and taking the limit along ${\hat{x}}_n$ yields

$\begin{array}{ll}0& \le ⟨v+\tilde{u},v+\tilde{u}⟩-⟨limS({\hat{x}}_n)(v+\tilde{u}),limS({\hat{x}}_n)(v+\tilde{u})⟩\\ & =\Vert v{\Vert }^2+\Vert \tilde{u}{\Vert }^2-\Vert limS({\hat{x}}_n)v{\Vert }^2-\Vert limS({\hat{x}}_n)\tilde{u}{\Vert }^2+2⟨v,\tilde{u}⟩\\ & -2⟨limS({\hat{x}}_n)v,limS({\hat{x}}_n)\tilde{u}⟩\\ & =2⟨v,\tilde{u}⟩-2⟨limS({\hat{x}}_n)v,limS({\hat{x}}_n)\tilde{u}⟩,\end{array}$

where the last equality follows from the fact that each $S({\hat{x}}_n)$ is an isometry on W_i and W_j. Since the inequality above also holds for $v-\tilde{u}$, we obtain the equality

$\begin{array}{ll}0=& ⟨v,\tilde{u}⟩-⟨limS({\hat{x}}_n)v,limS({\hat{x}}_n)\tilde{u}⟩\\ =& ⟨v,\tilde{u}⟩+⟨v,A_j{Q}_{d_j}(\theta )A_j^{-1}\tilde{u}⟩=⟨v,A_j(\mathrm{id}+Q_{d_j}(\theta ))A_j^{-1}\tilde{u}⟩\\ =& ⟨v,u⟩.\end{array}$

Since $v,u$ were arbitrary, this shows orthogonality. Hence, the decomposition $V=\underset{i=1}{\overset{k}{⨁}}{W}_i$ is orthogonal, and since $S_i(x)$ is an isometry on W_i, it follows that $S(x):=\underset{i=1}{\overset{k}{⨁}}{S}_i(x)$ is in SO(d). Setting $M=\underset{i=1}{\overset{k}{⨁}}{M}_i$, we obtain

$g(x)=S(x)\exp (Mx).$

Corollary 5.4.

Assume that $S(x)=A{Q}_d^{\nu }(x)A^{-1}$ is a semigroup with A being an invertible matrix such that S(x) is an isometry for each $x\ge 0$. Then, there exists an orthogonal matrix U such that

$S(x)=U{Q}_d^{\nu }(x)U^{\mathrm{T}}.$

Proof.

One can easily verify the identities $S(x)+S(x{)}^{\mathrm{T}}=2\cos (\nu (x))\mathrm{id}$ and

$\sin (\nu (y))S(x)-\sin (\nu (x))S(y)=\sin (\nu (x)-\nu (y))\mathrm{id}.$

Choose x > 0 such that $\sin (\nu (x))\ne 0$. Such an x clearly exists since ν is linear on $\mathbb{Q}x$. Choose any $v\in V$ of unit length and set

$u:=\frac{S(x)v-\cos (\nu (x))v}{\sin (\nu (x))}.$

The definition of u is invariant under the choice of x as long as $\sin (\nu (x))\ne 0$. This can be seen by noticing that

$\begin{array}{ll}\sin (\nu (y))S(x)-\sin (\nu (x))S(y)& =\sin (\nu (x)-\nu (y))\mathrm{id}\\ ⟺\frac{S(x)v-\cos (\nu (x))v}{\sin (\nu (x))}& =\frac{S(y)v-\cos (\nu (y))v}{\sin (\nu (y))}.\end{array}$

Hence we obtain $S(x)v=\cos (\nu (x))v+\sin (\nu (x))u$. We have

$\begin{array}{ll}⟨u,v⟩=& \frac{1}{\sin (\nu (x))}⟨S(x)v-\cos (\nu (x))v,v⟩\\ =& \frac{1}{\sin (\nu (x))}(⟨S(x)v,v⟩-\cos (\nu (x)))=0,\end{array}$

where the last equality follows from

$\begin{array}{ll}2⟨S(x)v,v⟩=& ⟨S(x)v,v⟩+⟨S(x{)}^{\mathrm{T}}v,v⟩\\ =& ⟨S(x)v+S(x{)}^{\mathrm{T}}v,v⟩=2\cos (\nu (x)).\end{array}$

Similarly, we can show that u is also of unit length. Set $H:=\mathrm{span}(u,v)$. Clearly, H is invariant under $(S(x){)}_x$ and hence so is $H^{\perp }$ since each $S(x)\in SO(d)$. In this manner, we can construct an orthonormal basis, and we denote by U the matrix associated with this change of basis. Then, we have

$\begin{array}{ll}Ug(x)U^{\mathrm{T}}=& UAS(x)A^{-1}{U}^{\mathrm{T}}\exp (UM{U}^{\mathrm{T}}x)\\ =& Q_d^{\nu }(x)\exp (UM{U}^{\mathrm{T}}x).\end{array}$

Lemma 5.5.

Suppose that the semigroup g, acting on V, satisfies Assumption 2.5. Let $V=V_1\oplus V_2$ be the decomposition of Corollary 3.8 such that $g{|}_{V_1}$ is invertible. Then, for any $v\in V_1$ we have

$\underset{x\to 0}{lim}\frac{\Vert g(x)v\Vert }{\Vert v\Vert }=1.$

Proof.

By Theorem 5.3, we have $g(x){|}_{V_1}=S(x)\exp (Mx)$. Since $S(x)\in SO(V_1)$, we obtain

$⟨g(x)v,g(x)v⟩=⟨\exp (Mx)v,\exp (Mx)⟩,v\in V_1,$

which is continuous in x. Hence

$\underset{x\to 0}{lim}\Vert g(x)v\Vert =\underset{x\to 0}{lim}\Vert \exp (Mx)v\Vert =\Vert v\Vert .$

Corollary 5.6.

Suppose that the semigroup g, acting on V, satisfies Assumption 2.5. Then, the decomposition from Corollary 3.8 is orthogonal.

Proof.

Let $V=V_1\oplus V_2$ with $g(x){|}_{V_1}$ being non-degenerate and $g(x){|}_{V_2}\equiv 0$. Assume V₁ is not orthogonal to V₂. Then, there exists $v\in V_1$ such that $p_{V_2}(v)\ne 0$, where $p_{V_2}$ denotes the orthogonal projection onto V₂. By Lemma 5.5, we have

$\underset{x\to 0}{lim}\frac{\Vert g(x)v\Vert }{\Vert v\Vert }=1.$

Calculating the same limit for $v-p_{V_2}(v)$ we obtain

$\begin{array}{ll}\underset{x\to 0}{lim}\frac{\Vert g(x)(v-p_{V_2}(v))\Vert }{\Vert v-p_{V_2}(v)\Vert }=& \underset{x\to 0}{lim}\frac{\Vert g(x)v\Vert }{\Vert v\Vert }\frac{\Vert v\Vert }{\Vert v-p_{V_2}(v)\Vert }\\ =& \frac{\Vert v\Vert }{\Vert v-p_{V_2}(v)\Vert }.\end{array}$

Since $\Vert v{\Vert }^2=\Vert v-p_{V_2}(v){\Vert }^2+\Vert p_{V_2}(v){\Vert }^2$ and $\Vert p_{V_2}(v)\Vert \ne 0$, we have $\frac{\Vert v\Vert }{\Vert v-p_{V_2}(v)\Vert }>1$. Hence

$\underset{x\to 0}{lim}\frac{\Vert g(x)(v-p_{V_2}(v))\Vert }{\Vert v-p_{V_2}(v)\Vert }>1,$

but this contradicts $\Vert g(x){\Vert }_{\mathrm{op}}\le f(x)$. Hence $V_1\perp V_2$.

Proof.

Proof of Theorem 2.6

By Corollary 5.6, we have the orthogonal decomposition $V={\tilde{V}}_1\oplus {\tilde{V}}_2$ with $g(x){|}_{{\tilde{V}}_2}\equiv 0$ and $g(x){|}_{{\tilde{V}}_1}$ non-degenerate. Applying Theorem 5.3 to $g(x){|}_{{\tilde{V}}_1}$ yields the result.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Supplemental data

Supplemental data for this article can be accessed online at https://doi.org/10.1080/27684830.2023.2289203.

Additional information

Funding

Financial support from the Austrian Science Fund (FWF) under grants P 30750 and Y 1235 is gratefully acknowledged.

Appendix A.

Cauchy’s functional equation

It is classical that all continuous solutions of the Equation (1.3)(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3) , $f(x)+f(y)=f(x+y)$, are linear, and that the non-linear solutions are not continuous, even not Lebesgue measurable, and have dense graphs. For this, and further references, we refer to [2, Section 1.1]. In this section, we provide two auxiliary results on Cauchy’s equation. They concern lifting solutions from an interval to the real line, resp. the joint behavior of two solutions that differ by a non-linear function.

Lemma A.1.

Let $f:\mathbb{R}\to \mathbb{R}/[-a,a)$ be a solution to Cauchy’s functional Equation (1.3)(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3) on $\mathbb{R}/[-a,a)$ with a > 0. Then, there exists a solution $\tilde{f}:\mathbb{R}\to \mathbb{R}$ of (1.3(1.3) $f(x)+f(y)=f(x+y),f:\mathbb{R}\to \mathbb{R},$(1.3) ) such that $f(x)\equiv \tilde{f}(x)\mathrm{mod}[-a,a)$. The solution $\tilde{f}$ is linear if and only if f is linear.

Proof.

Take a Hamel basis $(r_i{)}_{i\in I}$ of $\mathbb{R}$ such that $r_i\in [-a,a)$ for every $i\in I$. This is clearly possible by rescaling every basis element if necessary. For any $x\in \mathbb{R}$ there exists a finite subset $I_x\subset I$ and $c_i\in \mathbb{Q}$ such that $x=\sum _{i\in I_x}{c}_i{r}_i$. The function $\tilde{f}(x)=\sum _{i\in I_x}{c}_{i}f(r_i)$ satisfies $f(x)\equiv \tilde{f}(x)\mathrm{mod}[-a,a)$. If f is linear, then clearly $\tilde{f}$ is linear as well. If f is not linear then there exist two basis elements r₁ and r₂ such that $f(r_1)-\frac{r_1}{r_2}f(r_2)\ne 0$, and hence $\tilde{f}$ is also not linear.

Definition A.2.

We say that two solutions ν and η of Cauchy’s functional equation are equivalent if $\nu -\eta$ is linear.

Lemma A.3.

Let $f,g:\mathbb{R}\to \mathbb{R}$ be two non-equivalent solutions of Cauchy’s functional equation. Then there exists a sequence $(x_n{)}_{n\ge 0}$ in ${\mathbb{R}}_{\ge 0}$, converging to 0, such that either $\underset{n}{lim}f(x_n)=\pi$ and $\underset{n}{lim}g(x_n)=\theta$ with $|\theta |<\pi$ or vice versa.

Proof.

Choose $x,y\in {\mathbb{R}}_{>0}$ such that the two vectors $v_1=(x,f(x),g(x))$ and $v_2=(y,f(y),g(y))$ are linearly independent and $\frac{f(x)}{x}-\frac{f(y)}{y}\ne \frac{g(x)}{x}-\frac{g(y)}{y}$. This is possible, since

$\frac{f(x)}{x}-\frac{f(y)}{y}=\frac{g(x)}{x}-\frac{g(y)}{y}$

for all $x,y>0$ would imply that f − g is linear. Assume w.l.o.g. that $|\frac{f(x)}{x}-\frac{f(y)}{y}|>|\frac{g(x)}{x}-\frac{g(y)}{y}|$. Since f and g are both linear on $\mathbb{Q}x$ and $\mathbb{Q}y$ and v₁ and v₂ are linearly independent, there exist sequences $q_n\in \mathbb{Q}$ and $r_n\in \mathbb{Q}$ such that $q_{n}x-r_{n}y\ge 0$ for every n, $lim{q}_{n}x-r_{n}y=0$ and $\underset{n}{lim}f(q_{n}x-r_{n}y)=\underset{n}{lim}{q}_{n}f(x)-r_{n}f(y)=\pi$. We show that $x_n=q_{n}x-r_{n}y$ has the desired property. Clearly $lim\frac{r_n}{q_n}=\frac{x}{y}$ and

$\begin{array}{ll}\pi =& \underset{n}{lim}{q}_{n}f(x)-r_{n}f(y)=\underset{n}{lim}{q}_n(f(x)-\frac{r_n}{q_n}f(y))\\ =& (f(x)-\frac{x}{y}f(y))lim{q}_n.\end{array}$

Hence

$\begin{array}{ll}& |lim{q}_{n}g(x)-r_{n}g(y)|\\ & =|(g(x)-\frac{x}{y}g(y))lim{q}_n|\\ & =\left|\pi (\frac{g(x)}{x}-\frac{g(y)}{y})/(\frac{f(x)}{x}-\frac{f(y)}{y})\right|<\pi .\end{array}$