A transverse phase border makes the periodic BVP linearization boundedly invertible

The coefficient $A$ is continuous on $[0,1]$. For $0\le s\le t\le 1$, let $U(t,s)$ be the solution operator of $v^{\mathrm{\prime }}=A(t)v$. Existence on all of $[0,1]$ and uniqueness follow from [F3] with zero forcing. For any initial vector $\xi$, [F6] and the equation give $\mathrm{\parallel }U(t,s)\xi {\mathrm{\parallel }}_2\le \mathrm{\parallel }\xi {\mathrm{\parallel }}_2+a_0{\int }_s^t\mathrm{\parallel }U(r,s)\xi {\mathrm{\parallel }}_{2}dr$; Gronwall [F4] yields $\mathrm{\parallel }U(t,s){\mathrm{\parallel }}_2\le e^{a_0(t-s)}$. In particular $U$ exists on the full triangle $0\le s\le t\le 1$ with the stated bound.

The homogeneous solution with initial value $\xi$ is $U(t,0)\xi$. Apply [F5] to the $C^1$ augmented field $\tilde{f}(x,\theta )=(\theta f(x),0)$ on open $U\times (0,\mathrm{\infty })$. The given periodic solution exists through $s=1$, so its endpoint variation is the derivative of that flow on its open maximal domain; no convex set spanning the orbit is needed. The first $d$ endpoint coordinates differentiate to $Mv+f(\gamma (0))\tau$ in initial variation $(v,\tau )$: the $v$ column is monodromy, and the period column is ${\mathrm{\partial }}_T{\varphi }_T(\gamma (0))=f(\gamma (0))$ by [F5] and periodicity. Differentiating $B(u_0,u_1)=(x_1-x_0,\psi (x_0))$ by [F7] therefore gives $B_1+B_{2}U(1,0)=J$.

For data $(g,q)\in Z$, let $p$ be the unique solution of $p^{\mathrm{\prime }}=Ap+g$, $p(0)=0$. Existence on all of $[0,1]$ and uniqueness follow from [F3] with $C=A$ and $h=g$. Since $p^{\mathrm{\prime }}$ is continuous, [F6] gives $\mathrm{\parallel }p(t){\mathrm{\parallel }}_2=\mathrm{\parallel }{\int }_0^t{p}^{\mathrm{\prime }}(s)ds{\mathrm{\parallel }}_2\le {\int }_0^t\mathrm{\parallel }{p}^{\mathrm{\prime }}(s){\mathrm{\parallel }}_{2}ds$. Pointwise, $\mathrm{\parallel }{p}^{\mathrm{\prime }}(s){\mathrm{\parallel }}_2\le a_0\mathrm{\parallel }p(s){\mathrm{\parallel }}_2+\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$, so monotonicity in [F6] gives $\mathrm{\parallel }p(t){\mathrm{\parallel }}_2\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}+a_0{\int }_0^t\mathrm{\parallel }p(s){\mathrm{\parallel }}_{2}ds$, hence $\mathrm{\parallel }p{\mathrm{\parallel }}_{\mathrm{\infty }}\le E\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ by [F4]. Every solution with initial value $\xi$ equals $U(t,0)\xi +p(t)$, since their difference solves the homogeneous linear IVP with initial value $\xi$ and uniqueness [F3] applies.

The boundary condition for this solution is $J\xi +B_{2}p(1)=q$ by step 2.1. Thus its unique initial value is $\xi =J^{-1}(q-B_{2}p(1))$. Writing $D=\mathrm{\parallel }(g,q){\mathrm{\parallel }}_Z=\max (\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }},\mathrm{\parallel }q{\mathrm{\parallel }}_2)$ gives $\mathrm{\parallel }\xi {\mathrm{\parallel }}_2\le K_J(1+\mathrm{\parallel }{B}_2{\mathrm{\parallel }}_{2}E)D$ and, by steps 1.1–2.2, $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\le [E{K}_J(1+\mathrm{\parallel }{B}_2{\mathrm{\parallel }}_{2}E)+E]D$.

Since $v^{\mathrm{\prime }}=Av+g$, $\mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\le a_0\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}+D$. The definition of the $X$ norm now gives $\mathrm{\parallel }v{\mathrm{\parallel }}_X\le K_{\mathcal{L}}D$. Steps 2.1–2.2 construct a solution for every $(g,q)$; any solution is determined by $\xi$ and the invertible matrix $J$, so it is unique. Hence $\mathcal{L}$ is bijective with the stated inverse bound.