Bounded kernels give a local positive measure-valued solution

For a positive path $\nu$ in a ball of radius $R$, define $a_{\nu }(t,x)={\int }_{E}K(x,y){\nu }_t(dy)+\beta (x)$ and the positive gain $G_C({\nu }_t)=\frac{1}{2}\iint K(x,y){\delta }_{x+y}{\nu }_t(dx){\nu }_t(dy)$, $G_F({\nu }_t)={\int }_E\beta (x)B_x{\nu }_t(dx)$, and $G=G_C+G_F$. Then $0\le a_{\nu }\le K_{\ast }R+{\beta }_{\ast }$. Since $w(x+y)\le 2w(x)w(y)$, $\mathrm{\parallel }{G}_C(\mu ){\mathrm{\parallel }}_w\le K_{\ast }\mathrm{\parallel }\mu {\mathrm{\parallel }}_w^2$; bilinearity gives its local Lipschitz bound. Also $\mathrm{\parallel }{G}_F(\mu ){\mathrm{\parallel }}_w\le {\beta }_{\ast }(D+1)\mathrm{\parallel }\mu {\mathrm{\parallel }}_w$ by the daughter estimate in [F5]. Thus $G$ is bounded and Lipschitz on the ball, and $a_{\nu }-a_{\eta }$ is bounded by $K_{\ast }\mathrm{\parallel }\nu -\eta {\mathrm{\parallel }}_w$.

On $C([0,T];{\mathcal{M}}_w(E))$ define $\mathrm{\Phi }(\nu {)}_t=e^{-{\int }_0^t{a}_{\nu }(s,\cdot )ds}{\mu }_0+{\int }_0^t{e}^{-{\int }_s^t{a}_{\nu }(r,\cdot )dr}G({\nu }_s)ds$. The loss rate is continuous in time in $L^{\mathrm{\infty }}(E)$, since $\mathrm{\parallel }{a}_{\nu }(t,\cdot )-a_{\nu }(s,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}\le K_{\ast }\mathrm{\parallel }{\nu }_t-{\nu }_s{\mathrm{\parallel }}_w$; the gain path is continuous in ${\mathcal{M}}_w(E)$ by the bounded-operator estimates in [F4]–[F5]. For bounded $b$, multiplication $M_b\sigma =b\sigma$ satisfies $\mathrm{\parallel }{M}_b\mathrm{\parallel }\le \mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}$, so the survival multipliers are continuous in operator norm and $\mathrm{\Phi }\nu$ is a continuous path. For positive $\nu$ it is positive. On the closed positive path ball of radius $R>2\mathrm{\parallel }{\mu }_0{\mathrm{\parallel }}_w$, the gain bound gives $\mathrm{\parallel }\mathrm{\Phi }\nu {\mathrm{\parallel }}_{C{\mathcal{M}}_w}\le \mathrm{\parallel }{\mu }_0{\mathrm{\parallel }}_w+T{C}_R\le R$ for sufficiently small $T$. The bounds $\mathrm{\mid }{a}_{\nu }-a_{\eta }\mathrm{\mid }\le K_{\ast }\mathrm{\parallel }\nu -\eta {\mathrm{\parallel }}_w$ and $\mathrm{\mid }{e}^{-u}-e^{-v}\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$ give $\mathrm{\parallel }\mathrm{\Phi }\nu -\mathrm{\Phi }\eta {\mathrm{\parallel }}_{C{\mathcal{M}}_w}\le (T{L}_R+T{K}_{\ast }\mathrm{\parallel }{\mu }_0{\mathrm{\parallel }}_w+T^2{K}_{\ast }{C}_R)\mathrm{\parallel }\nu -\eta {\mathrm{\parallel }}_{C{\mathcal{M}}_w}$. Choose $T$ so this coefficient is below $1$.

The closed positive path ball is complete: a Cauchy sequence converges uniformly by [F3], its uniform limit is continuous, and the positive cone is closed because ${\nu }^{-}(E)\le \mathrm{\parallel }\nu -{\nu }_j{\mathrm{\parallel }}_{TV}$ when ${\nu }_j\ge 0$. It is nonempty because it contains the constant path ${\nu }_t^0={\mu }_0$. If $q<1$ is the contraction coefficient in step 2.1, iteration ${\nu }^{j+1}=\mathrm{\Phi }({\nu }^j)$ from ${\nu }^0$ satisfies $\mathrm{\parallel }{\nu }^{j+1}-{\nu }^j\mathrm{\parallel }\le q^j\mathrm{\parallel }{\nu }^1-{\nu }^0\mathrm{\parallel }$. For $m>n$, $\mathrm{\parallel }{\nu }^m-{\nu }^n\mathrm{\parallel }\le \mathrm{\parallel }{\nu }^1-{\nu }^0\mathrm{\parallel }{\sum }_{j=n}^{m-1}{q}^j\le q^n\mathrm{\parallel }{\nu }^1-{\nu }^0\mathrm{\parallel }\mathrm{/}(1-q)\to 0$, so completeness gives a limit. The contraction bound makes $\mathrm{\Phi }$ continuous, hence the limit is fixed; if two fixed points existed, their distance would be at most $q$ times itself, hence zero. For a fixed point, $S(t,s):=M_{e^{-{\int }_s^t{a}_{\nu }(r,\cdot )dr}}$ has operator-norm derivative ${\mathrm{\partial }}_{t}S(t,s)=-M_{a_{\nu }(t,\cdot )}S(t,s)$, since the scalar exponential difference quotients converge uniformly in $x$. Splitting the difference quotient of the source integral at $s=t$ gives its endpoint term $G({\nu }_t)$ and its integral term $-M_{a_{\nu }(t,\cdot )}{\int }_0^{t}S(t,s)G({\nu }_s)ds$; thus the fixed point solves $\dot{\nu }=G(\nu )-a_{\nu }\nu$. Conversely, for any positive solution $\zeta$, the inverse multiplier $A(t)=M_{e^{{\int }_0^t{a}_{\zeta }(s,\cdot )ds}}$ is operator-norm differentiable with $A^{\mathrm{\prime }}=M_{a_{\zeta }}A$. The product difference quotient gives $(A\zeta {)}^{\mathrm{\prime }}=AG(\zeta )$; integrating this continuous derivative by telescoping partitions and multiplying by $A(t{)}^{-1}$ recovers the same mild formula. Continuity at $0$ puts any competing solution in this ball on a sufficiently short interval, where fixed-point uniqueness makes it equal to the constructed solution.

The derivative established in step 3.1 is ${\dot{\mu }}_t=G({\mu }_t)-a_{\mu }(t,\cdot ){\mu }_t$. Symmetry of $K$ combines the two coagulation loss marginals into $-\int K(x,y){\delta }_x{\mu }_t(dx){\mu }_t(dy)$; the fragmentation loss is $-\int \beta (x){\delta }_x{\mu }_t(dx)$. Hence this is precisely the weak equation in [F1], proving the claim.