Global real evolution, cocycle and inverse

Choose a positive mesh length $h$ with $MKh<1$; if $K=0$ any positive $h$ suffices. Starting at $r$, apply [F2] on successive intervals of length at most $h$ toward $T$, and separately toward $0$, using each endpoint value as the next initial datum and restricting $f$ to the interval. There are finitely many intervals. Their continuous solutions agree at endpoints and give a continuous global function. To check the original formula, multiply the formula at an endpoint $q$ by $W(t-q)$ and add the integral from $q$ to $t$. The group law gives $W(t-q)W(q-r)=W(t-r)$ and turns the old integrand into $W(t-s)(P(s)u(s)+f(s))$. Additivity of oriented integrals then gives the integral from $r$ to $t$. Induction over the intervals proves the global equation.

The difference of two global solutions with the same data solves the homogeneous equation with zero data. Uniqueness on each successive interval from [F2], in both directions, proves that difference zero everywhere. For arbitrary $q$, splitting the global integral at $q$ gives the same reset formula $u(t)=W(t-q)u(q)+{\int }_q^{t}W(t-s)(P(s)u(s)+f(s))ds$, for all time orderings. This follows from the same group and oriented-additivity calculation as in step 1.1.

For zero forcing the norm satisfies $\mathrm{\parallel }u(t)\mathrm{\parallel }\le M\mathrm{\parallel }x\mathrm{\parallel }+MK{\int }_r^t\mathrm{\parallel }u(s)\mathrm{\parallel }ds$ when $t\ge r$, and the analogous bound with integral from $t$ to $r$ when $t\le r$. Gronwall in the two orientations yields $\mathrm{\parallel }u(t)\mathrm{\parallel }\le M\exp (MK\mathrm{\mid }t-r\mathrm{\mid })\mathrm{\parallel }x\mathrm{\parallel }$. For nonzero forcing use the continuous nondecreasing accumulated norm in the forward orientation, and its time reflection in the backward orientation; the coarse constant choice $M(\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }f{\mathrm{\parallel }}_1)$ for $a$ in [F3] gives the asserted uniform bound $C_0$.

Real linearity follows from local real linearity and global uniqueness. By the reset formula, $t\mapsto S(t,r)x$ is the unique zero-forcing solution based at $q$ with initial vector $S(q,r)x$, hence equals $S(t,q)S(q,r)x$. At $t=r$ and then with the two base times interchanged this proves the two-sided inverse formula.

Fix $r_0$ and $x$. The cocycle and the uniform norm bound give $\mathrm{\parallel }S(t,r)x-S(t,r_0)x\mathrm{\parallel }\le C_0\mathrm{\parallel }x-S(r,r_0)x\mathrm{\parallel }$, uniformly in $t$. The right side tends to zero as $r\to r_0$ because the trajectory based at $r_0$ is continuous and has value $x$ there. Combining this uniform estimate with continuity of $t\mapsto S(t,r_0)x$ proves joint strong continuity. All conclusions hold on the realification; no commutation with multiplication by $i$ was used. ∎