Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 107 / 3

First replace $u$ by $u+\epsilon$ and later let $\epsilon \downarrow 0$. In the weak subsolution inequality use the admissible truncations approximating ${\eta }^2{u}^{\alpha -1}$. Uniform ellipticity, the coefficient bound, Cauchy-Schwarz inequality, and Young inequality giveand thereforeApply the Sobolev embedding theorem to $w=\eta u^{\alpha /2}$. The preceding estimate yields, for concentric balls $B_r\subset B_R\subset B_1$,Starting with $\alpha =p>1$, taking ${\alpha }_k=p{\sigma }^k$, and choosing radii decreasing to $1/2$, the product of constants converges because ${\sum }_k{\alpha }_k^{-1}<\infty$. Letting $k\to \infty$ provesThis exponent-raising argument is Moser iteration.

Use the Weak Harnack inequality: for some $q>1$ and every nonnegative weak supersolution,where $C$ and $q$ depend only on $n,\lambda ,\Lambda$. A weak solution is both a subsolution and a supersolution. Applying part (i), after rescaling from $B_1$ to $B_{1/2}$, and then the weak Harnack inequality givesThis is the Harnack inequality for uniformly elliptic divergence-form equations.

For a compactly supported variation $v+t\varphi$, differentiation of the area functional at $t=0$ gives the first variationAfter integration by parts, this is the minimal surface equation for a graphFor $v_R(x)=R^{-1}v(Rx)$ one has $D{v}_R(x)=Dv(Rx)$. The equation is invariant under this scaling, so $v_R$ solves it on $B_1$ for every $R>0$.

Differentiate the equation for $v_R$ with respect to $x_k$. The derivative $w=D_k{v}_R$ is a weak solution ofwhereThe eigenvalue in the direction of $p$ is $(1+|p{|}^2{)}^{-3/2}$ and every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. Thus a uniform bound on $|Dv|$ makes this a uniformly elliptic operator with constants independent of $R$.

Suppose $|Dv|\le M$. The coefficient matrices in part (ii) then have uniform ellipticity constants depending only on $M$. Applying the Harnack inequality for uniformly elliptic divergence-form equations to the nonnegative solutions $M+D_k{v}_R$ and $M-D_k{v}_R$ gives a scale-independent oscillation contractionScaling back,For fixed $r$, iterate this estimate with $R=4^{m}r$ and use the global bound $|D_{k}v|\le M$ to obtain ${\mathrm{osc}}_{B_r}{D}_{k}v\le 2M{\theta }^m\to 0$. Every partial derivative is therefore constant, so $v(x)=a\cdot x+b$ is an affine function. This is a bounded-gradient Bernstein theorem for entire minimal graphs.