Exponential dichotomy for a hyperbolic endomorphism: invariant stable and unstable subspaces with exponential estimates

Let $\mathrm{\Lambda }$ be the set of eigenvalues of $T_{\mathbb{C}}$ and, for $\lambda \in \mathrm{\Lambda }$, let $e_{\lambda }\ge 1$ be its multiplicity as a root of the real polynomial ${\mu }_T$; by [F2] all elements of $\mathrm{\Lambda }$ have nonzero real part, and $\mathrm{\Lambda }$ is stable under conjugation with $e_{\bar{\lambda }}=e_{\lambda }$: conjugating ${\mu }_T=(x-\lambda {)}^{e_{\lambda }}q$ with $q(\lambda )\mathrm{\ne }0$ gives ${\mu }_T=(x-\bar{\lambda }{)}^{e_{\lambda }}\bar{q}$ with $\bar{q}(\bar{\lambda })\mathrm{\ne }0$. Define $p_s:={\prod }_{\mathrm{Re}\lambda <0}(x-\lambda {)}^{e_{\lambda }},p_u:={\prod }_{\mathrm{Re}\lambda >0}(x-\lambda {)}^{e_{\lambda }}\mathrm{.}$ Conjugation invariance of $\mathrm{\Lambda }$ with equal multiplicities shows $p_s,p_u\in \mathbb{R}[x]$; by construction ${\mu }_T=p_s{p}_u$, and $p_s,p_u$ have no common complex root, hence are coprime with a Bézout identity $a{p}_s+b{p}_u=1$ over $\mathbb{R}[x]$ [F4].

Exponential on a generalised eigenspace: fix $\lambda \in \mathrm{\Lambda }$ and put $A_{\lambda }:=\lambda I+N_{\lambda }$ on $G_{\lambda }$ with $N_{\lambda }^{e_{\lambda }}=0$ as in [F3]. For the partial sum through $m=M$, the binomial expansion and nilpotence of $N_{\lambda }$ give the finite rearrangement ${\sum }_{m=0}^M\frac{t^m}{m!}{A}_{\lambda }^m={\sum }_{k=0}^{e_{\lambda }-1}\frac{t^k}{k!}{N}_{\lambda }^k{\sum }_{j=0}^{M-k}\frac{(t\lambda {)}^j}{j!},$ with an empty inner sum when $M<k$. As $M\to \mathrm{\infty }$, the left side converges to $e^{t{A}_{\lambda }}$ by [F1], and each of the finitely many inner sums converges to $e^{t\lambda }$ by $\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$. Thus $e^{t{A}_{\lambda }}={\sum }_{k=0}^{e_{\lambda }-1}\frac{t^k}{k!}{N}_{\lambda }^k{e}^{t\lambda }\mathrm{.}$

Define $E_s:=\ker p_s(T)$ and $E_u:=\ker p_u(T)$. Both are $T$-invariant because $T$ commutes with $p_s(T)$ and $p_u(T)$. Every $v\in V$ satisfies $v=(a{p}_s+b{p}_u)(T)v=a(T)p_s(T)v+b(T)p_u(T)v\in E_u+E_s,$ the membership following from $p_u(T)a(T)p_s(T)=a(T){\mu }_T(T)=0$ on the first summand and $p_s(T)b(T)p_u(T)=b(T){\mu }_T(T)=0$ on the second, where ${\mu }_T(T)=0$ (The annihilator set $\mathrm{Ann}(T)=\{p\in F[x]:p(T)=0\}$; once existence is proved, its unique monic generator ${\mu }_T$ is the minimal polynomial). If $v\in E_s\cap E_u$, the same identity gives $v=a(T)p_s(T)v+b(T)p_u(T)v=0$. Hence $V=E_s\oplus E_u$.

Complexified description: $(E_s{)}_{\mathbb{C}}=\ker p_s(T_{\mathbb{C}})={⨁}_{\mathrm{Re}\lambda <0}{G}_{\lambda }$ and $(E_u{)}_{\mathbb{C}}=\ker p_u(T_{\mathbb{C}})={⨁}_{\mathrm{Re}\lambda >0}{G}_{\lambda }$. Writing $z=\iota x+i\iota y$ with $x,y\in V$, one has $p_s(T_{\mathbb{C}})z=\iota (p_s(T)x)+i\iota (p_s(T)y)$, so $\ker p_s(T_{\mathbb{C}})=(\ker p_s(T){)}_{\mathbb{C}}$; the same holds for $p_u$. Indeed, on a summand $G_{\lambda }$ with $\mathrm{Re}\lambda <0$, the factor $(x-\lambda {)}^{e_{\lambda }}$ divides $p_s$, so $p_s(T_{\mathbb{C}})$ vanishes on $G_{\lambda }$; on a summand with $\mathrm{Re}\lambda >0$, $p_s$ and $(x-\lambda {)}^{e_{\lambda }}$ are coprime, and evaluating a Bézout identity on $G_{\lambda }$ (where $(T_{\mathbb{C}}-\lambda {)}^{e_{\lambda }}=0$) shows $p_s(T_{\mathbb{C}})$ is invertible there, so its kernel meets $G_{\lambda }$ trivially. Summing over [F3] gives the claim; the same argument with $p_u$ gives the second identity.

Estimates on one generalised eigenspace: fix a basis of $V_{\mathbb{C}}$ adapted to the decomposition $V_{\mathbb{C}}={⨁}_{\lambda }{G}_{\lambda }$ (for instance the union of Jordan bases of the summands, whose existence is [F3]), and let $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$ be the corresponding maximum norm; then every $z={\sum }_{\lambda }{z}_{\lambda }$ has its coordinate block in $G_{\lambda }$ equal to the coordinate block of $z_{\lambda }$, so $\mathrm{\parallel }{z}_{\lambda }{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $\lambda$. On each block use the induced operator norm $\mathrm{\parallel }{N}_{\lambda }{\mathrm{\parallel }}_{\mathrm{\infty }}:=\sup \{\mathrm{\parallel }{N}_{\lambda }z{\mathrm{\parallel }}_{\mathrm{\infty }}:z\in G_{\lambda },\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\}$; this is finite for a finite matrix. Put $P_{\lambda }(t):={\sum }_{k=0}^{e_{\lambda }-1}\frac{\mathrm{\mid }t{\mathrm{\mid }}^k}{k!}\mathrm{\parallel }{N}_{\lambda }{\mathrm{\parallel }}_{\mathrm{\infty }}^k,$ so that $\mathrm{\parallel }{e}^{t{A}_{\lambda }}z{\mathrm{\parallel }}_{\mathrm{\infty }}\le e^{(\mathrm{Re}\lambda )t}{P}_{\lambda }(t)\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$ for $z\in G_{\lambda }$ and all real $t$ by step 1.2. Choose $c:=\frac{1}{2}{\min }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }\mathrm{Re}\lambda \mathrm{\mid }>0.$ For $u\ge 0$ and $a>0$, the nonnegative exponential series gives $e^{au}\ge (au{)}^k\mathrm{/}k!$, hence $u^k{e}^{-au}\le k!a^{-k}$ for every integer $k\ge 0$. Applying this to each term of $P_{\lambda }(\pm u)$, with $a=-\mathrm{Re}\lambda -c>0$ on stable blocks and $a=\mathrm{Re}\lambda -c>0$ on unstable blocks, proves that the following suprema are finite: for $\mathrm{Re}\lambda <0$, $C_{\lambda }^s:={\sup }_{t\ge 0}{P}_{\lambda }(t)e^{(\mathrm{Re}\lambda +c)t}<+\mathrm{\infty },$ and for $\mathrm{Re}\lambda >0$, $C_{\lambda }^u:={\sup }_{t\le 0}{P}_{\lambda }(t)e^{(\mathrm{Re}\lambda -c)t}<+\mathrm{\infty }\mathrm{.}$

Constants for the two invariant subspaces. By steps 2.2 and 2.3, for $z\in (E_s{)}_{\mathbb{C}}$ and $t\ge 0$ the decomposition $z={\sum }_{\mathrm{Re}\lambda <0}{z}_{\lambda }$ yields $\mathrm{\parallel }{e}^{t{T}_{\mathbb{C}}}z{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sum }_{\mathrm{Re}\lambda <0}{e}^{(\mathrm{Re}\lambda )t}{P}_{\lambda }(t)\mathrm{\parallel }{z}_{\lambda }{\mathrm{\parallel }}_{\mathrm{\infty }}\le e^{-ct}\left({\sum }_{\mathrm{Re}\lambda <0}{C}_{\lambda }^s\right)\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }},$ since, in the adapted basis of step 2.3, $\mathrm{\parallel }{z}_{\lambda }{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$, and $\mathrm{Re}\lambda +c\le -c$ for every $\mathrm{Re}\lambda <0$ (as $c=\frac{1}{2}\min \mathrm{\mid }\mathrm{Re}\lambda \mathrm{\mid }$). Similarly, for $z\in (E_u{)}_{\mathbb{C}}$ and $t\le 0$, $\mathrm{\parallel }{e}^{t{T}_{\mathbb{C}}}z{\mathrm{\parallel }}_{\mathrm{\infty }}\le e^{ct}\left({\sum }_{\mathrm{Re}\lambda >0}{C}_{\lambda }^u\right)\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Set $C:=\max (1,K^2{\sum }_{\mathrm{Re}\lambda <0}{C}_{\lambda }^s,K^2{\sum }_{\mathrm{Re}\lambda >0}{C}_{\lambda }^u)$ with $K$ as in [F5]. For $v\in E_s$ and $t\ge 0$, applying the first estimate to $z=\iota v$ and using [F5] in both directions, $\mathrm{\parallel }{e}^{tT}v{\mathrm{\parallel }}_2=\mathrm{\parallel }{e}^{t{T}_{\mathbb{C}}}\iota v{\mathrm{\parallel }}_{2,\mathbb{C}}\le K\mathrm{\parallel }{e}^{t{T}_{\mathbb{C}}}\iota v{\mathrm{\parallel }}_{\mathrm{\infty }}\le K{e}^{-ct}\left({\sum }_{\mathrm{Re}\lambda <0}{C}_{\lambda }^s\right)\mathrm{\parallel }\iota v{\mathrm{\parallel }}_{\mathrm{\infty }}\le K^2\left({\sum }_{\mathrm{Re}\lambda <0}{C}_{\lambda }^s\right)e^{-ct}\mathrm{\parallel }v{\mathrm{\parallel }}_2\le C{e}^{-ct}\mathrm{\parallel }v{\mathrm{\parallel }}_2$, which is clause 1; the same computation with the second estimate gives clause 2 for $v\in E_u$, $t\le 0$.

Clause 3: let $u\in E_u$ and $t\ge 0$. By [F1] $e^{tT}$ is invertible with inverse $e^{-tT}$, and $e^{-tT}u\in E_u$ because $E_u$ is $T$-invariant. Applying clause 2 with the time $-t\le 0$ to the vector $e^{tT}u$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2=\mathrm{\parallel }{e}^{-tT}{e}^{tT}u{\mathrm{\parallel }}_2\le C{e}^{-ct}\mathrm{\parallel }{e}^{tT}u{\mathrm{\parallel }}_2,$ hence $\mathrm{\parallel }{e}^{tT}u{\mathrm{\parallel }}_2\ge C^{-1}{e}^{ct}\mathrm{\parallel }u{\mathrm{\parallel }}_2$.

Clause 4: by clauses 1 and 3, every $v\in E_s$ satisfies $e^{tT}v\to 0$ as $t\to +\mathrm{\infty }$, and by clause 2 every $v\in E_u$ satisfies $e^{tT}v\to 0$ as $t\to -\mathrm{\infty }$. Conversely, let $v\in V$ satisfy $e^{tT}v\to 0$ as $t\to +\mathrm{\infty }$ and write $v=u+s$ with $u\in E_u$, $s\in E_s$; by clauses 3 and 1, $\mathrm{\parallel }{e}^{tT}v{\mathrm{\parallel }}_2\ge \mathrm{\parallel }{e}^{tT}u{\mathrm{\parallel }}_2-\mathrm{\parallel }{e}^{tT}s{\mathrm{\parallel }}_2\ge C^{-1}{e}^{ct}\mathrm{\parallel }u{\mathrm{\parallel }}_2-C{e}^{-ct}\mathrm{\parallel }s{\mathrm{\parallel }}_2\to +\mathrm{\infty }$ unless $u=0$. So convergence to $0$ forces $u=0$, that is $v\in E_s$. The characterisation of $E_u$ follows by the same argument with $t\to -\mathrm{\infty }$: clause 2 gives $\mathrm{\parallel }{e}^{tT}u{\mathrm{\parallel }}_2\le C{e}^{ct}\mathrm{\parallel }u{\mathrm{\parallel }}_2\to 0$ for $u\in E_u$, while for $s\mathrm{\ne }0$ the growth estimate $\mathrm{\parallel }{e}^{tT}s{\mathrm{\parallel }}_2\ge C^{-1}{e}^{-ct}\mathrm{\parallel }s{\mathrm{\parallel }}_2$ for $t\le 0$, obtained from clause 1 at time $-t\ge 0$ and invertibility of $e^{tT}$, forces divergence.