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

For $v=\log u_{\epsilon }$, Cauchy-Schwarz inequality and part (a.ii) implyThe John-Nirenberg inequality therefore supplies $p_0=p_0(n,\mu )>0$ such that, with $v_{B_1}$ denoting the average,Since $e^{p_0(v-v_{B_1})}$ and $e^{-p_0(v-v_{B_1})}$ are each bounded by this integrand,

If $u=0$ on $\Sigma$ and $|\Sigma |>0$, thenPart (b.i) consequently givesLetting $\epsilon \downarrow 0$ and using Fatou lemma yields ${\int }_{B_1}{u}^{p_0}=0$. Since $u\ge 0$ is continuous,

Suppose a nonzero solution $w$ did not change sign. Replacing it by $-w$ if necessary gives $w\ge 0$. The argument of part (b.ii) is local and invariant under translation and scaling: it applies on every ball whose concentric double lies in $B_2$. If the zero set of $w$ had positive measure, a density point and this local result would make $w$ vanish on one ball. Applying the same result successively on overlapping balls would then give $w\equiv 0$ throughout the connected ball $B_2$, a contradiction. Thus the zero set has measure zero in $B_2$. Applying part (a.i) to $u=w$ givesfor every compactly supported $C^1$ function $\zeta$; the omitted zero set has measure zero. But $\inf q>{\lambda }_1$, while the variational characterization of the First Dirichlet eigenvalue provides a $\zeta$ withThis contradiction proves that every nonzero solution takes both positive and negative values.