Lemma 6

Let E be a finite-dimensional real vector space and B a compact convex neighbourhood of 0 in E. Let $u$ be an endomorphism of E such that $u(B)\subset \mathring{B}$. Then the complex spectrum of $u$ is contained in the unit disk of $\mathbf{C}$ and, in particular, does not meet the circle with center 0 and radius 1 in $\mathbf{C}$.

Replacing B by $B\cap (-B)$, we are reduced to the case where B is balanced. The gauge $p$ of B (EVT, II, p. 28, ex. 3) is then a norm on E, which defines its topology (EVT, I, p. 14, th. 2). The hypothesis implies that $p(u(x))< p(x)$ for every element $x$ of E $-\{0\}$, hence that the spectral radius of $u$ is $<1$; as this is also the spectral radius of $u_{(\mathbf{C})}($cf. I, p. 86), the conclusion follows.

Let us now prove the proposition.

Let us denote by $x\preccurlyeq y$ the order relation on E associated with the convex cone C (EVT, II, p. 13, n$^o5$), that is to say, $x\preccurlyeq y$ if and only if $y-x\in C$. We have $u(x)\preccurlyeq u(y)$ if $x\preccurlyeq y$. Since $\mathring{C}$ is nonempty, the vector space $C-C$ generated by C (EVT, II, p. 12, cor. 1) contains a neighbourhood of 0, therefore the cone C is total.

Let us prove assertion a). Let $\varrho =\varrho (u)$. Let $y_0$ be an element of $\mathring{C}$. We have $y_0\not = 0$. Let $r >0$ be such that the closed ball with center $y_0$ and radius $r$ is contained in C. For every $y\in E-\{0\}$, we have $y_0-r\|y\|^{-1}y\in C$, hence $y\preccurlyeq r^{-1}\|y\|y_0$ for every $y\in E$.

Since $y_0\not = 0$, the hypotheses imply that $u(y_0)\in \mathring{C}$. There therefore exists $t >0$ such that $tu(y_0)-y_0\in C$. Let us put $v=tu$. It is a compact endomorphism of E such that $v(C)\subset C$ and $v(y_0)\succcurlyeq y_0$. For every integer $n\geqslant 1$, we have

therefore $(r^{-1}\|v^n\|\|y_0\| -1)y_0\in C$. Since C is salient, this implies that $t^n\|u^n\|=\|v^n\|\geqslant r/\|y_0\|$, whence $\varrho \geqslant t^{-1}>0 ($I, p. 20, prop. 1).

By the Krein-Rutman theorem (th. 4), the real number $\varrho$ is an eigenvalue of $u$, and there exists an eigenvector $x_0$ of $u$ in C for the eigenvalue $\varrho$. We have $\varrho x_0=u(x_0)\in \mathring{C}$ by hypothesis, therefore $x_0\in \mathring{C}$. This proves assertion a).

Let K be the intersection of the spectrum of $u_{(\mathbf{C})}$ and of the circle with center 0 and radius $\varrho$ in $\mathbf{C}$. Since $u$ is compact and $\varrho >0$, the set K is finite and the image of the spectral projector $e_K$ is a subspace of $E_{(\mathbf{C})}$ of finite dimension (prop. 2 of III, p. 83 and prop. 5 of III, p. 90). Since K is stable under complex conjugation, the image of $e_K$ is a subspace of $E_{(\mathbf{C})}$ rational over $\mathbf{R}$ (A, V, p.60, prop. 6) ; let F be the subspace of E such that $F_{(\mathbf{C})}$ is equal to the image of $e_K$. The space F is stable under $u$ and contains the eigenspace of $u$ corresponding to $\varrho$, therefore F is nonzero.

To prove assertions b) and c), it is enough to prove that F is of dimension 1.

Let $v$ denote the endomorphism of F deduced from $u$ by passing to subspaces. Let $C_F= C\cap F$. This is a closed salient convex cone with nonempty interior in F (since $x_0\in \mathring{C}\cap F$), hence total; it satisfies $v(C_F-\{0\})\subset \mathring{C}_F$, and in particular is stable under $v$. Since $\varrho (v)\leqslant \varrho$ and $x_0\in F$, we have $\varrho (v) =\varrho$. From the corollary to Theorem 4, applied to $C_F$ and $v$, there exists an eigenvector $\ell$ of $^tv$ in $C^{\circ}_F$ for the eigenvalue $\varrho (v) =\varrho >0$.

The subspace Ker($\ell$ ) of F is stable under $v$. Let $w$ denote the endomorphism of Ker($\ell$ ) deduced from $\varrho^{-1}v$ by passing to subspaces; its spectrum is contained in the unit circle. The set B of the $y\in$ Ker($\ell$ ) such that $x_0+y\in C_F$ is a closed convex neighbourhood of 0 in Ker($\ell$ ). For every $y\in B$ and every $z\in$ Ker($\ell$ ), we have

which belongs to $C_F$ if the norm of $z$ is sufficiently small, hence $w(y)\in \mathring{B}$.

The set B is bounded: in fact, if there existed a sequence $(y_n)_{n\in\mathbf{N}}$ in Ker($\ell$ ) such that $\|y_n\| \rightarrow +\infty$ and $x_0+y_n\in C_F$ then, by passing to a subsequence if necessary, we should have $y_n/\|y_n\| \rightarrow y$ where $y\in$ Ker($\ell$ ) is nonzero, and $y=$ lim $\|y_n\|^{-1}(x_0+y_n)$ would belong to $C_F$, contrary to Lemma 5. Hence the set B is compact.

It then follows from Lemma 6, applied to Ker($\ell$ ), to B and to $w$, that the complex spectrum of $w$ does not meet the circle with center 0 and radius 1; this implies that the spectrum of $w$ is empty, which means that Ker($\ell$ ) is reduced to $\{0\}$, hence that F has dimension 1. Assertions b) and c) are thus established.

The linear mapping $^tu$ is compact (Corollary 1 of III, p. 9) and $^tu(C^{\circ})\subset C^{\circ}$; moreover $\varrho (^tu) =\varrho >0$ (prop. 3 of I, p. 131). From the corollary to Theorem 4, there exists in $C^{\circ}$ an eigenvector $\ell \not = 0$ of $^tu$ for the eigenvalue $\varrho$. The kernel of $\ell$ is stable under $u$ and does not meet C $-\{0\}$ (Lemma 5). In particular, we have $x_0\notin$ Ker($\ell$ ).

Let F be a subspace of E stable under $u$. We have the decomposition $F = (F\cap \mathbf{R}x_0)\oplus (F\cap$ Ker($\ell$ )). If F does not contain $x_0$, we therefore have $F\subset$ Ker($\ell$ ), hence F does not meet C$-\{0\}$. This establishes d) and concludes the proof of the proposition.