I am trying to understand how one can prove the following assertion using a continuity argument:

Let $0<\epsilon<\epsilon_0$. Let $I=[t_0,R]$ be a compact interval. Suppose that $S:I\to [0,\infty)$ is a continuous non-decreasing function such that $S(t_0)=0$ and $$S(T)\lesssim \epsilon_0(S(T)+\epsilon)^4+\epsilon_0^4(S(T)+\epsilon)+\epsilon_0(S(T)+\epsilon)^5$$ for all $T\in I$. (Here, $a\lesssim b$ means $a\le Cb$ for some constant $C>0$) Then if $\epsilon_0$ can be chosen to be sufficiently small, we in fact have $S(T)\le \epsilon$ for all $T\in I$.

From my understanding, the idea of continuity argument here is to consider the set $\Omega=\{T\in I: S(T)\le \epsilon\}$. Then $\Omega$ is closed and non-empty since $t_0\in\Omega$. If we can prove that $\Omega$ is also open (which I am stuck on) then the connectedness of $I$ would imply that $\Omega=I$, which is what we need.

As for the context of the problem, I am stuck at reading this paper by Tao and Visan (Proof of Theorem 1.3, page 17, the assertion is labelled as (3.6)) http://arxiv.org/abs/math/0507005

Apologies if this is actually trivial. I am completely new to this method of continuity argument.


I tend to think about such continuity argument in this way:

Let $T^*=\sup\{T\in I: S(t)\le\epsilon\}$ for all $t\le T$. Then it suffices to show $T^*=R$.

Suppose not. Then by continuity we have $S(T^*)=\epsilon$. Plugging this into the given inequality we get $\epsilon\ll \epsilon_0(\epsilon^4+\epsilon^5)+\epsilon_0^4\epsilon$. Then it is clear that if $0<\epsilon<\epsilon_0$ is small enough, then this inequality cannot hold.

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  • $\begingroup$ That's much easier than I thought, thanks! $\endgroup$ – Gawin Apr 28 '16 at 2:20

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