Freezing cannot absorb a fixed oscillation on arbitrarily small balls

The coefficient field. The function $\mathrm{sign}(x_1)$ is bounded and measurable (it is piecewise constant with a single jump), so $a$ is bounded and measurable, and $A=a{I}_n$ is a bounded symmetric measurable matrix field. For every $x$, $a(x)\in [a_0-b,a_0+b]$ (with the standard convention $\mathrm{sign}(0)=0$ if used), and off the interface it equals one of the endpoint values. Thus the eigenvalues of $A(x)$ lie in $[a_0-b,a_0+b]$, so $A$ is uniformly elliptic with $\lambda =a_0-b>0$ and $\mathrm{\Lambda }=a_0+b$; no continuity or Hölder regularity is available at $x_1=0$.

The oscillation is fixed and never small. Let $x_0$ satisfy $x_{0,1}=0$ and let $\rho >0$. The two points $x_0\pm \frac{\rho }{2}{e}_1$ lie in $B_{\rho }(x_0)$ and have first coordinate relative to the interface equal to $\pm \rho \mathrm{/}2$, so $a$ takes both values $a_0+b$ and $a_0-b$ on the ball: ${\mathrm{osc}}_{B_{\rho }(x_0)}a=2b$, independent of $\rho$. Consequently the freezing modulus of continuity $ϵ(\delta ):={\sup }_{\mathrm{\mid }x-y\mathrm{\mid }\le \delta }\mathrm{\mid }a(x)-a(y)\mathrm{\mid }$ satisfies $ϵ(\delta )=2b$ for every $\delta >0$, because pairs straddling the hyperplane $x_1=0$ are at arbitrarily small distance.

The freezing error is not Hölder at any scale. Let $0<\alpha <1$ and consider the pair $x_0+t{e}_1$, $x_0-t{e}_1$ with $t>0$: both lie in $B_{2t}(x_0)$ and $\mathrm{\mid }a(x_0+t{e}_1)-a(x_0-t{e}_1)\mathrm{\mid }=2b$, so $\frac{\mathrm{\mid }a(x_0+t{e}_1)-a(x_0-t{e}_1)\mathrm{\mid }}{\mathrm{\mid }(x_0+t{e}_1)-(x_0-t{e}_1){\mathrm{\mid }}^{\alpha }}=\frac{2b}{(2t{)}^{\alpha }}⟶+\mathrm{\infty }(t\downarrow 0)\mathrm{.}$ Hence $[a{]}_{0,\alpha ;B_{\rho }(x_0)}=+\mathrm{\infty }$ for every $\rho >0$ and every $\alpha \in (0,1)$: the coefficient is bounded but nowhere near Hölder on any ball centred on its interface.

No radius absorbs the frozen error. Take $L:=a(x)\mathrm{\Delta }$ with frozen part $L_0:=a_0\mathrm{\Delta }$ and $u(x):=x_1^2\mathrm{/}2$, which is $C^{\mathrm{\infty }}$ with $\mathrm{\Delta }u=1$ and finite scaled norm $\mathrm{\parallel }u{\mathrm{\parallel }}_{2,\alpha ;B_{\rho }(x_0)}^{\ast }\le C(1+\rho +{\rho }^2)$ on every ball of radius $\rho \le 1$. Then $(L-L_0)u=(a-a_0)\mathrm{\Delta }u=a-a_0$ on $B_{\rho }(x_0)$ (up to the measure-zero hyperplane), so ${\rho }^2\mathrm{\parallel }(L-L_0)u{\mathrm{\parallel }}_{0,\alpha ;B_{\rho }(x_0)}^{\ast }\ge {\rho }^{2+\alpha }[a-a_0{]}_{0,\alpha ;B_{\rho }(x_0)}=+\mathrm{\infty }$ by step 3.1, while the right-hand side $\epsilon \mathrm{\parallel }u{\mathrm{\parallel }}_{2,\alpha ;B_{\rho }(x_0)}^{\ast }+C_{\epsilon }{\sup }_{B_{\rho }}\mathrm{\mid }u\mathrm{\mid }$ is finite for every $\epsilon >0$ and every finite $C_{\epsilon }$. Hence the freezing estimate cannot hold for this coefficient for any choice of the freezing radius $\rho$, and no radius can make the coefficient-oscillation term absorbable.

Conclusion. A bounded measurable (indeed piecewise constant) uniformly elliptic coefficient with a fixed jump has oscillation $2b$ at every scale and freezing error of infinite $\alpha$-Hölder seminorm; the freezing and Schauder arguments therefore genuinely need the vanishing small-scale oscillation provided by $C^{0,\alpha }$ (or continuity) hypotheses, and this is not a technical convenience. The statement above is refuted, while the freezing lemma with its stated $C^{0,\alpha }$ hypothesis is untouched by this example.