Gradient Hölder control by differentiated flux and one-dimensional lifting

For each $k$, put $z=D_{k}w$, $a(y)=P_p(y,Dw(y))$ and $F(y)=-P_{y_k}(y,Dw(y))$, with $P_{y_k}$ the partial derivative at fixed $p$. Differentiating the flux equation gives $\mathrm{div}(aDz)=\mathrm{div}F$. The matrix $a$ is measurable, symmetric and uniformly elliptic, $\mathrm{\mid }z\mathrm{\mid }\le p_0$ and $\mathrm{\mid }F\mathrm{\mid }\le K$. Smoothness justifies the differentiation without any uniform Hessian estimate.

Write $n$ for the coordinate dimension. Here is a local reduction of this inhomogeneous equation to the exact homogeneous source statement. If $M=\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}$ is zero there is nothing to prove. Otherwise put $Z(y,t)=z(y)\mathrm{/}M+t$ and $H=F\mathrm{/}M$. On a cylinder strictly inside the coordinate ball times $(-1,1)$ use the symmetric matrix $B=\left(\begin{array}{cc}a& -H\\ -H^T& L\end{array}\right)$, where $L=1+2\mathrm{/}\lambda$. Its quadratic form is at least $(\lambda \mathrm{/}2)\mathrm{\mid }\xi {\mathrm{\mid }}^2+\mathrm{\mid }\tau {\mathrm{\mid }}^2$, since $\mathrm{\mid }H\mathrm{\mid }\le 1$ and $2\mathrm{\mid }\tau H\cdot \xi \mathrm{\mid }\le (\lambda \mathrm{/}2)\mathrm{\mid }\xi {\mathrm{\mid }}^2+(2\mathrm{/}\lambda ){\tau }^2$. It has a uniform upper bound too. Direct distributional differentiation, using independence of $t$, gives ${\mathrm{div}}_{y,t}(BDZ)={\mathrm{div}}_y(aDz\mathrm{/}M-H)=0$.

Let $U$ be any $(n+1)$-dimensional ball with closure in this cylinder. The smooth function $Z$ and all its first classical derivatives are bounded on $\bar{U}$, so they are square-integrable on $U$; integration by parts against compactly supported tests identifies these classical derivatives as its weak derivatives. Thus $Z\in H^1(U)$, without requiring a uniform bound on its $H^1$ norm. For each $\phi \in C_c^{\mathrm{\infty }}(U)$, integration by parts in the divergence identity of step 2.1 gives ${\int }_U(BDZ)\cdot D\phi =0$. Hence $Z$ is a scalar homogeneous weak solution with exactly the convention and coefficient hypotheses needed in F20.

Apply the imported homogeneous theorem in dimension $n+1$ on the balls of step 3.1, after translation and dilation to the unit ball. Since $\mathrm{\mid }Z\mathrm{\mid }$ has a universal bound on the fixed cylinder, its $L^2$ norm is uniformly bounded on each such ball. Restrict the resulting Hölder estimate to $t=0$ and multiply by $M$. A finite cover of every smaller coordinate ball yields $\mathrm{\parallel }z{\mathrm{\parallel }}_{C^{0,\beta }}\le C(\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }})$. Doing this for the finitely many $k$ proves the assertion. The exponent may be decreased below one; no value of the prescribed later Schauder exponent is assumed here.