Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 107 / 1

The operator is strictly elliptic when its symmetric principal matrix $A(x)=(a^{ij}(x))$ is positive definite at every point:It is a uniformly elliptic operator when some $\lambda >0$ satisfiesfor every $x$ and $\xi$. Here the least eigenvalue of the continuous matrix $A(x)$ is a positive continuous function on the compact set ${\overline{B}}_1$. It therefore has a positive minimum, which supplies $\lambda$ and proves uniform ellipticity.

A scaled Interpolation inequality in Holder spaces says that for $u\in C^{l,\alpha }(B_R(x_0))$, $0\le j<l$, and every $\delta >0$,where $C=C(n,l,\alpha )$. Corresponding interpolation between any two Hölder orders follows by applying this estimate to derivatives and rescaling the ball.

One useful form of the Simon absorption lemma is the following. Let $S$ be nonnegative and subadditive on balls contained in $B_R$, and suppose that for every such $B_r$,where $A$ is also subadditive and $\epsilon$ is smaller than a constant depending only on $n$ and $\gamma$. ThenMore generally, an additional controlled remainder on the right remains with a dimensional constant. The proof covers $B_{R/2}$ by finitely many smaller balls, sums by subadditivity, and iterates so that the $S$ term is absorbed.

The Interior Schauder estimate iswhere $C$ depends only on $n$, $\alpha$, the ellipticity constant, and the stated coefficient bounds.

It is enough first to prove the estimate for the Hölder seminorm of $D^{2}u$. The Interpolation inequality in Holder spaces controls $Du$ and $D^{2}u$ by a small multiple of $[D^{2}u{]}_{\alpha }$ plus $\Vert u{\Vert }_0$; the small term is later absorbed. Freezing $a^{ij}$ at the centre of each ball and moving coefficient differences and lower-order terms to the right reduces the local estimate toOnce this is known, the Simon absorption lemma gives the desired estimate on $B_{1/2}$.

For completeness, prove the frozen-coefficient estimate by contradiction. If it failed, choose solutions $u_k$, points $x_k$, and scales ${\rho }_k\downarrow 0$ at which the scale-invariant Hölder quotient is almost maximal. Let $q_k$ be the quadratic Taylor polynomial of $u_k$ at $x_k$ and defineThen $v_k(0)=D{v}_k(0)=D^2{v}_k(0)=0$, the Hessians have uniformly bounded local $C^{0,\alpha }$ seminorms, and the normalization makes their oscillation nonzero on a fixed ball. The localized equation iswhere the coefficient matrices converge locally uniformly to one constant positive-definite matrix and the given normalized error $g_k$ tends locally uniformly to zero.

The Arzela-Ascoli theorem produces a locally $C^2$ convergent subsequence with limit $v$ satisfying a constant-coefficient elliptic equation on ${\mathbb{R}}^n$. A linear rotation and scaling turn it into a harmonic function. The normalization gives $D^{2}v(0)=0$ but a nonconstant Hessian, while maximality of the scaled quotient gives growth at most $C(1+|y{|}^{2+\alpha })$. Applying the Liouville theorem to derivatives shows that every second derivative is constant because $0<\alpha <1$. This contradicts the normalized Hessian oscillation. The frozen estimate follows, and interpolation plus absorption completes the proof.

At $\alpha =0$, bounded Hessians have no uniform modulus of continuity, so the Arzela-Ascoli theorem does not supply the compact $C^2$ limit used in the contradiction. At $\alpha =1$, the blow-up has cubic growth, and the Liouville theorem no longer forces its Hessian to be constant: nonzero harmonic cubic polynomials are possible. These are the two endpoint failures behind the restriction $0<\alpha <1$; ordinary Schauder theory replaces neither endpoint by a general $C^{2,\alpha }$ estimate of the displayed form.