# 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…