1
$\begingroup$

First of all, I am sorry for the ''not clear title' for this question but I cannot find a better way to describe this seemingly very simple and standard inequality,

So.. I am reading a paper 'Two-dimensional Navier-Stokes Equation Driven by a space time white noise' by Daprato and Debussche. And I came across an inequality regarding a probability measure.(It is in the proof of main theorem p.198) It seems very strandard but I cannot see why it is.

$$\mathbb{P}\left[ \sup_{t\in [0,T]} \left| u_N(t,u_0) \right|_{\mathcal{B}^\sigma_{p,\rho}} \geq M \right]\leq \sum_{k=0}^{[T/t^*_M]} \mathbb{P}\left[\sup_{t\in [kt^*_M,(k+1)t^*_M]} \left| u_N(t,u_0) \right|_{\mathcal{B}^\sigma_{p,\rho}}\geq M \right]$$

Here, $\mathcal{B}^\sigma_{p,\rho}$ is a Besov space and $\mathbb{P}$ is a probability measure.

I initially thought it is a typo and the $M$ in the right hand side should be replaced with $M/[T/t^*_M]$ but I realized that there is a possibility that I am missing something. So I wanted to hear from someone else.

I thank in advance for any help with this.

$\endgroup$
1
$\begingroup$

It looks fine to me.

If we let $I_k = [k t^\ast_M, (k+1) t^\ast_M]$ be the relevant subintervals of $[0,T]$, then the supremum of $|u_n|$ over $[0,T]$ must be almost attained along some sequence of points, and by pigeonhole infinitely many of them must be in one of the $I_k$, call it $I_{k_0}$, so that $\sup_{I_{k_0}} |u_n| = \sup_{[0,T]} |u_n|$. So if the sup over $[0,T]$ is at least $M$, then the sup over some $I_k$ must also be at least $M$ (and conversely). In other words, the event $\{\sup_{[0,T]} |u_n| \ge M\}$ equals the union of the events $A_k = \{\sup_{I_k} |u_n| \ge M\}$.

Now we just use the union bound.

$\endgroup$
1
  • $\begingroup$ Thank you for the answer! I got your point and it seems clear to me. $\endgroup$ – Lev Bahn Aug 10 '20 at 0:40

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.