3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.