Theorem 4 (Krein–Rutman)

Let E be a complete normed space over $\mathbf{R}$. Let C be a closed total salient convex cone in E and let $u\in \mathscr{L}(E)$ be a compact linear mapping such that $u(C)\subset C$. If the spectral radius $\varrho (u)$ is $>0$, then $\varrho (u)$ is an isolated point of the spectrum of $u$, and there exists a nonzero eigenvector $x\in C$ of $u$ for the eigenvalue $\varrho (u)$.

Let $E_{(\mathbf{C})}$ be the complexified space of E and $u_{(\mathbf{C})}$ the endomorphism of $E_{(\mathbf{C})}$ obtained by extension of the scalars from $u$. The spectral radius of $u_{(\mathbf{C})}$ is equal to $\varrho (u) ($I, p. 86); we denote it simply by $\varrho$. Since $\varrho >0$, the complex spectrum of $u$ is not reduced to 0. There therefore exists $\lambda_0\in$ Sp$_s(u_{(\mathbf{C})})$ such that $|\lambda_0|=\varrho$ (Prop. 5 of III, p. 90). Let $e_0$ be the spectral projector of $u_{(\mathbf{C})}$ associated with $\lambda_0$.

The resolvent $\lambda \mapsto R(u_{(\mathbf{C})}, \lambda ) = (\lambda -u_{(\mathbf{C})})^{-1}$ is holomorphic on the complement of the spectrum of $u_{(\mathbf{C})}$ (Theorem 1 of I, p. 24). The complex number $\lambda_0$ is a pole of the resolvent and its residue is the spectral projector $e_0($I, p. 131).

Let $y$ and $y'$ be elements of E such that $y+iy'\in E_{(\mathbf{C})}$ is an eigenvector of $u_{(\mathbf{C})}$ for the eigenvalue $\lambda_0$. Since $e_0(y+iy') =y+iy'\not = 0$, there exists an element $x_0\in C$ such that $e_0(x_0)\not = 0$ (Lemma 4). Since C is closed and salient, its polar $C^{\circ}$ is total (EVT, II, p. 48, Corollary 1), and there then exists a linear form $\ell_0\in C^{\circ}$ such that $\langle e_0(x_0), \ell_0\rangle \not = 0$.

Consider the function $f$ defined by $f(0) = 0$ and by

for $z\in \mathbf{C}$ such that $z^{-1}\notin$ Sp($u_{(\mathbf{C})}$). This function satisfies

for $|z|<1/\varrho$ (Theorem 1 of I, p. 24, d)) and is therefore holomorphic in the disc with centre 0 and radius $1/\varrho$.

There exists a holomorphic function $\widetilde{R}$ defined on an open neighbourhood U of $\lambda_0$ and with values in $\mathscr{L}(E)$ such that for $z$ belonging to U$-\{\lambda_0\}$, one has

(Prop. 17 of I, p. 83). For $z$ such that $z^{-1}\in U$ and $z\not = 1/\lambda_0$, one therefore has

The term of the series corresponding to $n= 0$ is $(z^{-1}-\lambda_0)^{-1}\langle e_0(x_0), \ell_0\rangle$. Since $\langle e_0(x_0), \ell_0\rangle \not = 0$, the uniqueness of the Laurent expansion (VAR, R1, p. 30, 3.3.9) implies that $f$ cannot be prolonged to a holomorphic function in a neighbourhood of $1/\lambda_0$. In particular, the radius of convergence of the power series expansion (1) of the function $f$ at the point 0 is equal to $1/\varrho$.

The coefficients of the power series (1) are $\langle u^n(x_0), \ell_0\rangle \geqslant 0$ since $u(C)\subset C$ and $\ell_0\in C^{\circ}$. By Lemma 3, the function $f$ cannot be extended to a holomorphic function in a neighbourhood of $1/\varrho$. The resolvent of $u_{(\mathbf{C})}$ therefore cannot be extended to a holomorphic function in a neighbourhood of $\varrho$, that is to say, $\varrho \in$ Sp($u$). This implies that $\varrho$ is an eigenvalue of $u_{(\mathbf{C})}$ (prop. 5 of III, p. 90). Since $\varrho$ is real, it is also an eigenvalue of $u$.

Let $d\geqslant 1$ be the order of the pole of the resolvent of $u_{(\mathbf{C})}$ at $\varrho$. Let $e$ be the endomorphism

$e=$ lim$_{z\rightarrow\varrho}(z-\varrho )^dR(u_{(\mathbf{C})}, z)$.

It is nonzero and commutes with $u_{(\mathbf{C})}$, and its image is contained in the eigenspace of $u_{(\mathbf{C})}$ corresponding to $\varrho ($cf. No.$^o3$ of I, p. 131). Let now $\ell$ be an element of $C^{\circ}$ and $x$ an element of C. By Theorem 1 of I, p. 24, d), one has

$\langle e(x), \ell \rangle =$ lim$_{zz>\varrho\rightarrow\varrho}(z-\varrho )^d\sum_{n\geqslant 0}\langle u^n(x), \ell \rangle z^{-n-1}\geqslant 0$.

The bipolar theorem (EVT, II, p 48, th. 1) implies that $e(x)\in C$ for all $x\in C$. Since C is total, there exists $x\in C$ such that $e(x)\not = 0$ (Lemma 4), and then $e(x)$ belongs to C and is an eigenvector of $u$ for the eigenvalue $\varrho$, as required.