Qid finite density recursion profile

Fix a nonempty finite graph $G$, $0<\eta <1$, and an integer $s\ge 0$. For real $a,b\ge 0$, define ${\beta }_s(a,b)=\min \{{\rho }_{G[W]}(a,b):W\subseteq V(G),\mathrm{\mid }W\mathrm{\mid }\ge {\eta }^s\mathrm{\mid }G\mathrm{\mid }\}\mathrm{.}$ Here $\rho$ is Induced copy density and homogeneous restriction parameter. The qualifying $W$ are nonempty because ${\eta }^s\mathrm{\mid }G\mathrm{\mid }>0$, and they form a finite family by $\mid \mathcal{P}(A)\mid =2^{\mid A\mid }$ for finite $A$. The family contains $V(G)$, since ${\eta }^s\le 1$. Thus its minimum is attained by finite comparison, and $0<{\beta }_s(a,b)\le 1$.

Every qualifying induced $F$ therefore has a nonempty $T$ with $\mathrm{\mid }T\mathrm{\mid }\ge {\beta }_s(a,b)\mathrm{\mid }F\mathrm{\mid }$ and $e(F[T])\le a\left(\genfrac{}{}{0px}{}{\mathrm{\mid }T\mathrm{\mid }}{2}\right)$ or $e(\bar{F}[T])\le b\left(\genfrac{}{}{0px}{}{\mathrm{\mid }T\mathrm{\mid }}{2}\right)$: choose a maximizing set for ${\rho }_F$. Conversely any uniform fractional guarantee over these $F$ is no larger than their minimum ${\rho }_F$, so this finite profile equals the largest uniform guarantee. If $a\ge 1$ or $b\ge 1$, every full $F$ qualifies and ${\beta }_s(a,b)=1$. At $s=0$, the only qualifying set is $V(G)$, so ${\beta }_0(a,b)={\rho }_G(a,b)$.

Proof/convention locator: Bucic, Nguyen, Scott and Seymour, Induced subgraph density I, 5.2, definition of beta_s.