fa.functional analysis – Integrability of $expleft(pint_0^t |w(s,x(s,y))| mathrm{d}sright)$ for $win L^infty(0,T;BMO(mathbb{T}^d))$

Let $wcolon (0,T)timesmathbb{T}^d to mathbb{R}^n$ be such that
$$ |w|_{L^infty(BMO)} := sup_{tin(0,T)}|w(t,cdot)|_{BMO} leq C $$
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 $pleq frac{c_2}{|w|_{L^infty(BMO)}}$.

Moreover, let $x(t,y)in C^1((0,T)timesmathbb{T}^d)$ 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} expleft(pint_0^t |w(s,x(s,y))|mathrm{d}sright)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
    $$ left|int_0^t |w(s,cdot)|mathrm{d}sright|_{BMO} leq int_0^t |w(s,cdot)|_{BMO} mathrm{d}s leq t|w|_{L^infty(BMO)} leq Ct, $$
    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

$$int_{mathbb{T}^d} expleft(pint_0^t |w(s,x(s,y))|mathrm{d}sright)mathrm{d}y leq int_{mathbb{T}^d}int_0^t e^{p|w(s,x(s,y))|} mathrm{d}smathrm{d} y $$
and then using Fubini and the bounds on $J(t,y)$, we get
$$ int_{mathbb{T}^d}int_0^t e^{p|w(s,x(s,y))|} mathrm{d}smathrm{d} y leq cint_0^t int_{mathbb{T}^d} e^{p|w(s,x(s,y))|} J(s,y) mathrm{d}ymathrm{d}s = cint_0^t int_{mathbb{T}^d} e^{p|w(s,x)|} mathrm{d}x mathrm{d}y. $$
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 $xiin(0,t)$ such that $int_0^t f(s)mathrm{d}s leq tf(xi)$. In my case $win L^1((0,T)timesmathbb{T}^d)$ 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
    $$int_{mathbb{T}^d} expleft(pint_0^t|w(s,x(s,y))|mathrm{d}sright) mathrm{d} y leq int_{mathbb{T}^d} expleft(pt|w(xi,x(xi,y))|right) mathrm{d} y $$
    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…