A half-line first-order operator with a hyperbolic limit has a right inverse

Use an equivalent norm from the inner product in the statement. By [F1], ${\mathbb{R}}^d=E^s\oplus E^u$ with spectral projections $P_s,P_u$ and some $\lambda >0$ such that $\mathrm{\parallel }{e}^{tL}{P}_s\mathrm{\parallel }\le e^{-\lambda t}$ for $t\ge 0$ and $\mathrm{\parallel }{e}^{tL}{P}_u\mathrm{\parallel }\le e^{\lambda t}$ for $t\le 0$. For the constant path $L$ define $(R_{L}v)(t)={\int }_0^t{e}^{(t-s)L}{P}_{s}v(s)ds-{\int }_t^{\mathrm{\infty }}{e}^{(t-s)L}{P}_{u}v(s)ds\mathrm{.}$ Both integrals converge. Their norms are bounded by $2{\lambda }^{-1}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$; splitting the integrals into a compact initial interval and a tail where $v$ is small shows $R_{L}v(t)\to 0$. Differentiation gives $(R_{L}v{)}^{\mathrm{\prime }}=L{R}_{L}v+v$, so the derivative also tends to zero. Thus $R_L:C_0^0\to C_0^1$ is bounded and $F_L{R}_L=I$.

Write $B(t)=A(t)-L$. If $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}$ is sufficiently small, $F_A{R}_L=I-B{R}_L$ on $C_0^0$, and the Neumann series gives the bounded right inverse $R_A:=R_L(I-B{R}_L{)}^{-1}$. It tends to $R_L$ in operator norm as $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$.

For this small-perturbation case choose $0<\beta <\lambda$ and, for $a\in E^s$, consider the integral equation $u(t)=e^{tL}a+{\int }_0^t{e}^{(t-s)L}{P}_{s}B(s)u(s)ds-{\int }_t^{\mathrm{\infty }}{e}^{(t-s)L}{P}_{u}B(s)u(s)ds\mathrm{.}$ On the Banach space of continuous paths with ${\sup }_{t\ge 0}{e}^{\beta t}\mathrm{\mid }u(t)\mathrm{\mid }<\mathrm{\infty }$, the integral operator has norm at most $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}((\lambda -\beta {)}^{-1}+(\lambda +\beta {)}^{-1})$. Shrink the permitted $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}$ so this is below one. Iterating from $e^{tL}a$ converges geometrically to a unique solution $u_a$; differentiating the equation gives $u_a^{\mathrm{\prime }}=A{u}_a$. Its initial value is $a+h_A(a)$, where $h_A(a)\in E^u$ and $\mathrm{\parallel }{h}_A\mathrm{\parallel }\to 0$ as $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Conversely, variation of constants for any homogeneous solution tending to zero gives this integral equation with $a=P_{s}u(0)$: the unwanted unstable term is fixed by its terminal value at infinity. Uniqueness also holds in the unweighted bounded-path norm under the same smallness condition. Hence $S_A$ is exactly the graph of $h_A:E^s\to E^u$, has dimension $\dim E^s$, and all its solutions decay at least as $e^{-\beta t}$.

Choose a smooth scalar $\rho$ supported in $(0,\mathrm{\infty })$ with integral one. For each $a\in E^u$ put $v_a(t)=-\rho (t)e^{tL}a$. Then $v_a(0)=0$ and step 1.1 gives $R_L{v}_a(0)=a$. For a basis of $E^u$, step 2.1 makes the vectors $R_A{v}_a(0)$ close to that basis, while step 3.1 makes $S_A$ a graph close to $E^s$. Thus these vectors together with $S_A$ span ${\mathbb{R}}^d$ for sufficiently small $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}$; this follows by invertibility of the finite matrix close to the identity in the fixed splitting $E^s\oplus E^u$.

For general $A(t)\to L$, choose $T$ so large that the shifted path $A_T(s)=A(T+s)$ meets the smallness bounds of steps 2.1–4.1. Apply those steps on $[T,\mathrm{\infty })$. For $v\in C_0^0([0,\mathrm{\infty }))$, set the solution's value at $T$ to $R_{A_T}(v(T+\cdot ))(0)$ and solve $u^{\mathrm{\prime }}=Au+v$ backwards on $[0,T]$ by [F2]. This constructs a bounded linear right inverse $R_A:C_0^0\to C_0^1$; boundedness on the finite interval follows from the variation-of-constants formula and the finite maximum of $\mathrm{\parallel }A(t)\mathrm{\parallel }$ there. Homogeneous evolution carries $S_A$ isomorphically onto $S_{A_T}$, so $\dim S_A=\dim E^s$ and exponential decay persists. To obtain the spanning assertion at $0$, choose $w$ on the shifted tail with $w(0)=0$ and extend it by zero to $[0,T]$. The right inverse so constructed has $R_{A}v(0)={\mathrm{\Phi }}_A(0,T)R_{A_T}w(0)$, while $S_A={\mathrm{\Phi }}_A(0,T)S_{A_T}$; step 4.1 then spans all initial values. Finally $t\mapsto -t$ changes $u^{\mathrm{\prime }}-A(t)u=v$ on a negative half-line into $z^{\mathrm{\prime }}-(-A(-s))z=-v(-s)$ on a positive half-line; negative eigenvalues of $-A(-\mathrm{\infty })$ are positive eigenvalues of $A(-\mathrm{\infty })$.