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

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.