The Schauder estimate fails at the H"older endpoint $\alpha =1$

The source is compactly supported and Lipschitz. For $r>0$ one has $x_1^3-3x_1{x}_2^2=r^3\cos 3\theta$, so $f(x)=\chi (r)r\cos 3\theta$, a product of the radial function $\chi$ with the degree-one homogeneous function $g(x):=r\cos 3\theta$. The function $g$ is smooth off the origin, satisfies $\mathrm{\mid }g\mathrm{\mid }\le r$ and, being $1$-homogeneous, has $\mathrm{\mid }\mathrm{\nabla }g\mathrm{\mid }\le C_0$ globally, while $\chi$ is Lipschitz with support in ${\bar{B}}_2$; hence $f$ is compactly supported, $\sup \mathrm{\mid }f\mathrm{\mid }\le 2$ and $[f{]}_{0,1}\le C_0\mathrm{\parallel }\chi {\mathrm{\parallel }}_{\mathrm{\infty }}+2[\chi {]}_{0,1}<\mathrm{\infty }$, that is $f\in C_c^{0,1}({\mathbb{R}}^2)$ with finite $C^{0,1}$ norm.

The explicit local solution. Put $P(x):=x_1^3-3x_1{x}_2^2$, so that $P=r^3\cos 3\theta$ and $p=-\frac{1}{6}P\log r$ for $0<r<1$. The polynomial $P$ is harmonic: ${\mathrm{\partial }}_{11}P=6x_1$, ${\mathrm{\partial }}_{22}P=-6x_1$, hence $\mathrm{\Delta }P=0$; moreover $\mathrm{\nabla }P\cdot \mathrm{\nabla }\log r=P_i{x}_i\mathrm{/}{r}^2$ with $P_i{x}_i=3(x_1^2-x_2^2)x_1+(-6x_1{x}_2)x_2=3x_1(x_1^2-3x_2^2)=3r^3\cos 3\theta$, so $\mathrm{\nabla }P\cdot \mathrm{\nabla }\log r=3r\cos 3\theta$; and $\mathrm{\Delta }\log r=0$ for $r>0$. Therefore, for $0<r<1$, $\mathrm{\Delta }p=-\frac{1}{6}(\mathrm{\Delta }P\log r+2\mathrm{\nabla }P\cdot \mathrm{\nabla }\log r+P\mathrm{\Delta }\log r)=-\frac{1}{6}\cdot 2\cdot 3r\cos 3\theta =-r\cos 3\theta =-f(x),$ because $\chi (r)=1$ there. At the origin $p$ is $C^2$ with $D^{2}p(0)=0$: indeed $p=O(r^3\mathrm{\mid }\log r\mathrm{\mid })$, ${\mathrm{\partial }}_{i}p=O(r^2\mathrm{\mid }\log r\mathrm{\mid })$ and ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}p=O(r\mathrm{\mid }\log r\mathrm{\mid })$ as $r\downarrow 0$, as the three terms of the product rule show, so $\mathrm{\Delta }p(0)=0=f(0)$ and the identity $-\mathrm{\Delta }p=f$ holds pointwise on all of the unit disc.

The potential differs from $p$ by a harmonic function. The source $f$ is Lipschitz, hence belongs to $C_c^{0,\alpha }({\mathbb{R}}^2)$ for every $0<\alpha <1$, so by [F2] the potential $Nf$ is $C^2$ with $-\mathrm{\Delta }Nf=f$ pointwise. Step 1.2 gives $-\mathrm{\Delta }p=f$ pointwise on the unit disc, so $\mathrm{\Delta }(Nf-p)=0$ there; by [F3] the function $H:=Nf-p$ is real analytic on the unit disc and its Hessian is Lipschitz on $B_{1\mathrm{/}2}(0)$, say $\mathrm{\mid }{D}^{2}H(x)-D^{2}H(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }$ for $x,y\in B_{1\mathrm{/}2}(0)$.

Failure of the Lipschitz bound. On the positive $x_1$-axis $p(x_1,0)=-\frac{1}{6}{x}_1^3\log x_1$ for $0<x_1<1$, so ${\mathrm{\partial }}_{11}p(x_1,0)=-\frac{1}{6}(6x_1\log x_1+5x_1)=-x_1\log x_1-\frac{5}{6}{x}_1,{\mathrm{\partial }}_{11}p(0,0)=0.$ By step 2.1, ${\mathrm{\partial }}_{11}Nf(x_1,0)=-x_1\log x_1-\frac{5}{6}{x}_1+{\mathrm{\partial }}_{11}H(x_1,0)$ and the last term deviates from its value at $0$ by at most $M{x}_1$. Hence for $0<x_1<\frac{1}{2}$, $\frac{\mathrm{\mid }{\mathrm{\partial }}_{11}Nf(x_1,0)-{\mathrm{\partial }}_{11}Nf(0,0)\mathrm{\mid }}{x_1}\ge \mathrm{\mid }\log x_1\mathrm{\mid }-\frac{5}{6}-M⟶+\mathrm{\infty }(x_1\downarrow 0)\mathrm{.}$ Therefore $[D^{2}Nf{]}_{0,1}={\sup }_{x\mathrm{\ne }y}\mathrm{\mid }{D}^{2}Nf(x)-D^{2}Nf(y)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }\ge {\mathrm{lim\hspace{0.17em}sup}}_{x_1\downarrow 0}(\dots )=+\mathrm{\infty }$, while $\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,1}}=\sup \mathrm{\mid }f\mathrm{\mid }+[f{]}_{0,1}<\mathrm{\infty }$ by step 1.1.

Conclusion. The compactly supported Lipschitz source $f$ of step 1.1 has finite $C^{0,1}$ norm, but its Newtonian potential has $[D^{2}Nf{]}_{0,1}=+\mathrm{\infty }$ by step 3.1; hence no finite constant $C$ can satisfy $[D^{2}Nf{]}_{C^{0,1}({\mathbb{R}}^2)}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,1}({\mathbb{R}}^2)}$, and the endpoint $\alpha =1$ version of the Schauder estimate is false. The example is consistent with the true sharp result: $D^{2}Nf$ is bounded and has the logarithmic modulus $\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }\log \mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }$, so the failure is exactly the loss of one logarithm, not a loss of boundedness.