For $v=\log u_\varepsilon$, <Cauchy-Schwarz inequality> and part (a.ii) imply
$$
\rho^{1-n}\int_{B_\rho(z)\cap B_1}|Dv|
\leq C\left(\rho^{2-n}\int|Dv|^2\right)^{1/2}
\leq M(n,\mu).
$$
The <John-Nirenberg inequality> therefore supplies $p_0=p_0(n,\mu)>0$ such that, with $v_{B_1}$ denoting the average,
$$
\int_{B_1}e^{p_0|v-v_{B_1}|}\leq C(n).
$$
Since $e^{p_0(v-v_{B_1})}$ and $e^{-p_0(v-v_{B_1})}$ are each bounded by this integrand,
$$
\left(\int_{B_1}u_\varepsilon^{p_0}\right)
\left(\int_{B_1}u_\varepsilon^{-p_0}\right)
=\left(\int e^{p_0(v-v_{B_1})}\right)
\left(\int e^{-p_0(v-v_{B_1})}\right)
\leq C.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 107 / 4 / b / i / Solution

{c}
{wiki}

A function of bounded mean oscillation has exponentially integrable oscillation. A scale-invariant bound on $\rho^{1-n}\int_{B_\rho}|Dv|$ implies, for some dimensional $p>0$,
$$
\int_{B_1}\exp\bigl(p|v-v_{B_1}|/M\bigr)\leq C,
$$
where $M$ is the scale-invariant gradient bound.


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 Incoming links: John-Nirenberg inequality