Age-cutoff survival, outflow and infinite-age tail bounds

For an initial cohort at age $a$, the survival factor is the exponential of minus the integrated mortality along $a(s)=a+s$ by [F1]. Since ${\mu }_{\ast }\le \mu (r)\le \mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}$ almost everywhere, integrating these inequalities over $[a,a+s]$ and exponentiating proves the two survival bounds.

A newborn has age at most $t\le T<A$ by time $t$, so it cannot contribute to outflow at $A$. An initial individual of age $a$ exits by time $t$ exactly when $A-t\le a\le A$, and its survival to the exit age is $S(a,A-a)$. Integrating over those initial ages gives the formula for $O_A(t)$; using $S\le 1$ and $A-t\ge A-T$ gives the stated upper bound.

Split $n_0=n_0{\mathbf{1}}_{[0,A-T]}+n_0{\mathbf{1}}_{(A-T,\mathrm{\infty })}$. By linearity of the Volterra equations in [F2] and [F3], it suffices to compare the evolutions of these two parts. Every initial cohort in the first part remains below age $A$ through time $T$, and its newborn descendants have age at most $T<A$; the characteristic formulas therefore make the infinite-age and finite-age evolutions identical for this part. The infinite-age evolution of the second part has mass at most $e^{{\beta }_{\mathrm{\infty }}T}{\int }_{A-T}^{\mathrm{\infty }}{n}_0$ by [F4]. The finite-age model starts this part only with its restriction to $[0,A]$, whose mass is no larger than the same tail; its mass is at most that same exponential factor by [F3]. The triangle inequality gives the factor-two bound in the Statement.

Under the smooth hypotheses of [F5], the piecewise-constant reconstruction differs in $L^1$ from the finite-age solution by at most $C_T(\mathrm{\Delta }a+\mathrm{\Delta }t+{ϵ}_Q)$ at each time level. Add this to the infinite-versus-cutoff error from step 3.1 by the triangle inequality. This is the displayed fully discrete bound, with the mesh/time/quadrature and age-tail contributions separate.