# Bounding a function above into a ball of radius $R$

Let $n:[0,T] \to \mathbb{R}$ be a non-negative function. It satisfies $$n(t) \leq C_1n(0) + C_2$$ where $C_1 > 1$ and $C_2 > 0$.

Is it possible to find a number $R >0$ such that if $n(0) \leq R$, then $$n(t) \leq R$$ too? $R$ can depend on $C_1$ or $C_2$ but not on $n(t)$ or $n(0)$.

• Why should $n(t) \leq C_1T+C_2 \leq T$? – matt.x Feb 1 '14 at 14:43
• @user2943324 no problem – matt.x Feb 1 '14 at 14:46
• Can $R$ depend on $T$? – you can call me Al Feb 1 '14 at 16:49
• @youcancallmeAl Yes it can – matt.x Feb 1 '14 at 18:46

No, this is not possible. Let's focus on the smaller class of functions $n$, namely those that satisfy $n(T)=C_1 n(0)$ and are linear in between $0$ and $T$. Within this class, your question can be stated as: does there exist $R\ge 0$ such that $n(0)\le R$ implies $C_1n(0)\le R$? To which the answer is obviously no: whichever $R$ we pick, the function with $n(0)=R$ gives a counterexample.