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Let $w\colon [0,T]\times\mathbb{T}^d \to \mathbb{R}^n$ be such that $$ \|w\|_{L^\infty(BMO)} := \sup_{t\in[0,T]}\|w(t,\cdot)\|_{BMO} \leq C $$ and $\int_{\mathbb{T}^d} w(t,x)\mathrm{d}x = 0 $ for all $t$.

The corollary from the John-Nirenberg inequality states that $$ \int_{\mathbb{T}^d} e^{p|w(t,x)|} \mathrm{d}x \leq C $$ for $p\leq \frac{c_2}{\|w\|_{L^\infty(BMO)}}$.

Moreover, let $x(t,y)\in C^1([0,T]\times\mathbb{T}^d)$ be such that $x(t,\cdot)$ is a diffemorphism for any $t$ and the Jacobian $J(t,y) = \det D_y x(t,y)$ satisfies $$ 0 < \frac{1}{c} \leq J(t,y) \leq c $$

My problem: I'm trying to show that for a fixed $p>1$ we can choose sufficiently small $t$ such that

\begin{equation}\label{int} \int_{\mathbb{T}^d} \exp\left(p\int_0^t |w(s,x(s,y))|\mathrm{d}s\right)\mathrm{d}y \tag{*} \end{equation} is finite.

So far I tried multiple things that didn't work out, but I believe they may be helpful:

  1. If we had in the exponent just $\int_0^t |w(s,x)|\mathrm{d}s$, then as $$ \left\|\int_0^t |w(s,\cdot)|\mathrm{d}s\right\|_{BMO} \leq \int_0^t \|w(s,\cdot)\|_{BMO} \mathrm{d}s \leq t\|w\|_{L^\infty(BMO)} \leq Ct, $$ we would have the integrability for all $p \leq \frac{c_2}{Ct}$, so choosing sufficiently small $t$ we can take $p$ as big as we want. However, the function $w(t,x(t,\cdot))$ may no longer be in $BMO$, so we cannot do it this way.

  2. The obvious way to deal with (*) is to use convexity of $\exp$ and get rid of the integral in the exponent. Then

$$\int_{\mathbb{T}^d} \exp\left(p\int_0^t |w(s,x(s,y))|\mathrm{d}s\right)\mathrm{d}y \leq \int_{\mathbb{T}^d}\int_0^t e^{p|w(s,x(s,y))|} \mathrm{d}s\mathrm{d} y $$ and then using Fubini and the bounds on $J(t,y)$, we get $$ \int_{\mathbb{T}^d}\int_0^t e^{p|w(s,x(s,y))|} \mathrm{d}s\mathrm{d} y \leq c\int_0^t \int_{\mathbb{T}^d} e^{p|w(s,x(s,y))|} J(s,y) \mathrm{d}y\mathrm{d}s = c\int_0^t \int_{\mathbb{T}^d} e^{p|w(s,x)|} \mathrm{d}x \mathrm{d}s. $$ Now we can apply the John-Nirenberg inequality, but we wouldn't get arbitrary large $p$. I was wondering if there are some functional inequalities, which would allow me to put the dependence of small $t$ again in the exponent, but so far I didn't find any.

  1. My third idea was also incorrect, but maybe it can be fixed somehow. It uses the fact that for nonnegative integrable functions there exist $\xi\in[0,t]$ such that $\int_0^t f(s)\mathrm{d}s \leq tf(\xi)$. In my case $w\in L^1([0,T]\times\mathbb{T}^d)$ and I would have for almost all $y$ $$ \int_0^t |w(s,x(s,y))|\mathrm{d}s \leq t|w(\xi,x(\xi,y))|. $$ Then $$\int_{\mathbb{T}^d} \exp\left(p\int_0^t|w(s,x(s,y))|\mathrm{d}s\right) \mathrm{d} y \leq \int_{\mathbb{T}^d} \exp\left(pt|w(\xi,x(\xi,y))|\right) \mathrm{d} y $$ and I could perform the change of variables in the same way as in 1. and eventually from John-Nirenberg obtain the integrability for $p \leq \frac{c_2}{Ct}$.

However, what I didn't take into account is that my $\xi$ depends on $y$, so instead I have $$ \int_o^t |w(s,x(s,y))|\mathrm{d}s \leq t |w(\xi(y), x(\xi(y),y))| $$ and I don't know if I can do something with it (after the change of variables I would still have the dependence of $x(t,y)$ inside $w$).

I would be really happy from the slightiest hints, as I got out of ideas. I'm also not 100% sure if this is even true, although it looks like it should work...

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I don't know much about BMO things, but I do know the following version of Jensen's inequality: $$\phi\left(\int f(s) d\mu(s)\right) \leq \int \phi(f(s)) d\mu(s),$$ provided $\mu$ is a probability measure. From that point of view, I'm not sure part 2 of your argument is correct. You should instead put $d\mu(s) = ds/t$ to arrive at: $$ \int_{\mathbb{T}^d} \exp\left(p \int_0^t |w(s,x(s,y)|ds \right) dy \leq \int_{\mathbb{T}^d} {1 \over t}\int_0^t e^{pt|w(s,x(s,y)|}ds dy. $$

According to your corollary, this is finite provided $pt$ is sufficiently small, so I think you're gucci.

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  • $\begingroup$ Thank you very much, I forgot about the length of the interval in Jensen's inequality $\endgroup$
    – M_S
    Jan 8, 2021 at 21:53

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