# Integrability of $\exp\left(p\int_0^t |w(s,x(s,y))| \mathrm{d}s\right)$ for $w\in L^\infty(0,T;BMO(\mathbb{T}^d))$

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

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

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.