Geometric oscillation decay implies a Hölder modulus

Fix a ball $B_R(x_0)⋐\mathrm{\Omega }$. The claim is immediate when $x=y$, so assume $x\mathrm{\ne }y$ and put $d:=\mathrm{\mid }x-y\mathrm{\mid }>0$ and $z:=(x+y)\mathrm{/}2$. Since $x,y\in B_{R\mathrm{/}2}(x_0)$, the midpoint satisfies $z\in B_{R\mathrm{/}2}(x_0)$ and $d<R$. The oscillation of $u$ on $B_R(x_0)$ is finite by hypothesis. If $d\ge R\mathrm{/}2$, then $\mathrm{\mid }u(x)-u(y)\mathrm{\mid }\le {\mathrm{osc}}_{B_R(x_0)}u\le 4^{\alpha }(d\mathrm{/}R{)}^{\alpha }{\mathrm{osc}}_{B_R(x_0)}u$, so assume henceforth $d<R\mathrm{/}2$.

Let $k\ge 0$ be the largest integer with $2^{k+1}d\le R$; it exists because $d<R\mathrm{/}2$, and the set of admissible exponents is bounded above. For every $0\le j\le k$ one has $B_{2^{j}d}(z)⋐B_R(x_0)$: the midpoint $z$ is within $R\mathrm{/}2$ of $x_0$, while $2^{j}d\le R\mathrm{/}2$. Consequently the given oscillation hypothesis applies to the pair of radii $2^{j-1}d$ and $2^{j}d$ for every $1\le j\le k$.

Iterating the hypothesis, ${\mathrm{osc}}_{B_d(z)}u\le {\theta }^k{\mathrm{osc}}_{B_{2^{k}d}(z)}u$. Indeed the case $k=0$ is an equality, and if the claim holds for $k-1$ then it holds for $k$ by appending the one step ${\mathrm{osc}}_{B_{2^{k-1}d}(z)}u\le \theta {\mathrm{osc}}_{B_{2^{k}d}(z)}u$ supplied by step 1.2. Since $B_{2^{k}d}(z)\subseteq B_R(x_0)$, monotonicity of the oscillation gives ${\mathrm{osc}}_{B_d(z)}u\le {\theta }^k{\mathrm{osc}}_{B_R(x_0)}u$.

By maximality of $k$, $2^{k+2}d>R$, so $2^{-k}<4d\mathrm{/}R$. Since $2^{-k}\le 1$ and $\alpha \le {\alpha }_0$, [F2] gives ${\theta }^k=(2^{-k}{)}^{{\alpha }_0}\le (2^{-k}{)}^{\alpha }<(4d\mathrm{/}R{)}^{\alpha }=4^{\alpha }(d\mathrm{/}R{)}^{\alpha }$.

Since $x,y\in B_d(z)$, combining steps 2.1 and 2.2 gives $\mathrm{\mid }u(x)-u(y)\mathrm{\mid }\le {\mathrm{osc}}_{B_d(z)}u\le 4^{\alpha }(d\mathrm{/}R{)}^{\alpha }{\mathrm{osc}}_{B_R(x_0)}u$, which is the displayed inequality because $d=\mathrm{\mid }x-y\mathrm{\mid }$. Dividing by $\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha }$ and taking the supremum over $x\mathrm{\ne }y$ in $B_{R\mathrm{/}2}(x_0)$ yields $[u{]}_{0,\alpha ;B_{R\mathrm{/}2}(x_0)}\le 4^{\alpha }{R}^{-\alpha }{\mathrm{osc}}_{B_R(x_0)}u$ by [F3]; since every point of $\mathrm{\Omega }$ has a ball $B_R(x_0)⋐\mathrm{\Omega }$ about it and $R\mathrm{/}2$ is available, $u$ is locally $\alpha$-Hölder on $\mathrm{\Omega }$.

For the exponent clause, fix $0<{\alpha }^{\mathrm{\prime }}<\alpha$. If $d<R\mathrm{/}4$, then $4d\mathrm{/}R<1$, and ${\alpha }^{\mathrm{\prime }}<\alpha \le {\alpha }_0$ gives ${\theta }^k=(2^{-k}{)}^{{\alpha }_0}\le (2^{-k}{)}^{{\alpha }^{\mathrm{\prime }}}<(4d\mathrm{/}R{)}^{{\alpha }^{\mathrm{\prime }}}$. If $d\ge R\mathrm{/}4$, then $\mathrm{\mid }u(x)-u(y)\mathrm{\mid }\le {\mathrm{osc}}_{B_R(x_0)}u\le 4^{{\alpha }^{\mathrm{\prime }}}(d\mathrm{/}R{)}^{{\alpha }^{\mathrm{\prime }}}{\mathrm{osc}}_{B_R(x_0)}u$. Combining these cases proves the claimed ${\alpha }^{\mathrm{\prime }}$ estimate with constant $4^{{\alpha }^{\mathrm{\prime }}}$; no choice principle is used.