Interior estimate for the Poisson equation with Hölder data

Work under [F10]. Rescale to the unit ball: put $v(z):=u(a+rz)$ and $F(z):=r^{2}f(a+rz)$ for $z\in B_1(0)$. Differentiating twice with the chain rule [F8] and Laplacian convention [F7] gives $\mathrm{\Delta }v(z)=r^2(\mathrm{\Delta }u)(a+rz)=-r^{2}f(a+rz)=-F(z)$, so $-\mathrm{\Delta }v=F$ pointwise on $B_1(0)$; moreover $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty };B_1(0)}=\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty };B_r(a)}$, $\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty };B_1(0)}=r^2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty };B_r(a)}$ and $[F{]}_{0,\alpha ;B_1(0)}=r^{2+\alpha }[f{]}_{0,\alpha ;B_r(a)}$ by [F1].

Cutoff. By [F4] fix a smooth $\eta :{\mathbb{R}}^n\to [0,1]$ with $\eta =1$ on ${\bar{B}}_{3\mathrm{/}4}(0)$ and $\mathrm{supp}\eta \subseteq B_{7\mathrm{/}8}(0)$, and let $G:=\eta F$ on $B_{7\mathrm{/}8}(0)$, extended by $0$ to all of ${\mathbb{R}}^n$. Then $G$ is continuous and compactly supported, and its global Hölder seminorm satisfies $[G{]}_{\alpha ;{\mathbb{R}}^n}\le C_{n,\alpha }(\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty };B_1(0)}+[F{]}_{0,\alpha ;B_1(0)})$: for $x,y$ in the support one uses $\mathrm{\mid }G(x)-G(y)\mathrm{\mid }\le \mathrm{\mid }\eta (x)\mathrm{\mid }\mathrm{\mid }F(x)-F(y)\mathrm{\mid }+\mathrm{\mid }\eta (x)-\eta (y)\mathrm{\mid }\mathrm{\mid }F(y)\mathrm{\mid }$ and the smoothness of the fixed cutoff, while if one point lies outside the support the estimate follows from $\mathrm{\mid }G\mathrm{\mid }\le \mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}$, the vanishing of $\eta$ at the support boundary and $\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha }\ge \mathrm{\mid }x-y\mathrm{\mid }$ for $\mathrm{\mid }x-y\mathrm{\mid }\le 1$; the constant depends only on the fixed cutoff, hence only on $n$ and $\alpha$.

The Newtonian potential. Put $w:=NG$ using [F2]; by [F3] the potential is $C^2$ on ${\mathbb{R}}^n$ with $-\mathrm{\Delta }w=G$, and on the compact set $K:={\bar{B}}_{3\mathrm{/}4}(0)$ its size is controlled: ${\sum }_{j=0}^2{\max }_{\mathrm{\mid }\gamma \mathrm{\mid }=j}{\sup }_K\mathrm{\mid }{D}^{\gamma }w\mathrm{\mid }+{\max }_{\mathrm{\mid }\gamma \mathrm{\mid }=2}[D^{\gamma }w{]}_{0,\alpha ;K}\le C_{n,\alpha }(\mathrm{\parallel }G{\mathrm{\parallel }}_{\mathrm{\infty }}+[G{]}_{\alpha ;{\mathbb{R}}^n})\le C_{n,\alpha }^{\mathrm{\prime }}(\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty };B_1(0)}+[F{]}_{0,\alpha ;B_1(0)})$ by step 2.1 and the size bounds on $\eta$.

The remainder is harmonic. Since $\eta =1$ on ${\bar{B}}_{3\mathrm{/}4}(0)$, we have $G=F$ on $B_{3\mathrm{/}4}(0)$, so $-\mathrm{\Delta }(v-w)=-F+G=0$ there by steps 1.1 and 3.1; thus $h:=v-w$ is harmonic on $B_{3\mathrm{/}4}(0)$. Moreover $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty };B_{3\mathrm{/}4}(0)}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty };B_1(0)}+\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty };B_{3\mathrm{/}4}(0)}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty };B_r(a)}+C_{n,\alpha }^{\mathrm{\prime }}(\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty };B_1}+[F{]}_{0,\alpha ;B_1})$ by step 3.1.

Estimates for the harmonic part. For every $y\in B_{1\mathrm{/}2}(0)$, the closed ball ${\bar{B}}_{1\mathrm{/}8}(y)$ lies in $B_{5\mathrm{/}8}(0)\subset B_{3\mathrm{/}4}(0)$, where $h$ is harmonic. Applying [F5] with radius $1\mathrm{/}8$ gives, for every multi-index $\gamma$ with $\mathrm{\mid }\gamma \mathrm{\mid }\le 3$, $\mathrm{\mid }{D}^{\gamma }h(y)\mathrm{\mid }\le C_{n,\gamma }{8}^{n+\mathrm{\mid }\gamma \mathrm{\mid }}{\int }_{B_{1\mathrm{/}8}(y)}\mathrm{\mid }h\mathrm{\mid }\le C_n^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty };B_{3\mathrm{/}4}}$, using [F6] to bound the ball's volume. If $\mathrm{\mid }\gamma \mathrm{\mid }=2$ and $x,y\in B_{1\mathrm{/}2}(0)$, their segment stays in $B_{1\mathrm{/}2}(0)$. Apply the real mean value theorem [F9] separately to the real and imaginary parts of $t\mapsto D^{\gamma }h(x+t(y-x))$ on $[0,1]$ (only the real part is needed when $h$ is real); the chain rule [F8] and the bounds just obtained for derivatives of order three then give $\mathrm{\mid }{D}^{\gamma }h(x)-D^{\gamma }h(y)\mathrm{\mid }\le C_n^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }x-y\mathrm{\mid }\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty };B_{3\mathrm{/}4}}$. Since $0<\mathrm{\mid }x-y\mathrm{\mid }<1$ implies $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha }$ for $0<\alpha <1$, this bounds $[D^{\gamma }h{]}_{0,\alpha ;B_{1\mathrm{/}2}}$ by $C_n^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty };B_{3\mathrm{/}4}}$; the radius factors for the scaled norm on $B_{1\mathrm{/}2}$ only change the constant.

Combining on the half ball. By step 3.1 the derivatives of $w$ through order two are bounded on ${\bar{B}}_{1\mathrm{/}2}(0)\subset K$ by $C_{n,\alpha }^{\mathrm{\prime }}(\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}+[F{]}_{0,\alpha })$, and its second derivatives have $\alpha$-Hölder seminorm on ${\bar{B}}_{1\mathrm{/}2}(0)$ bounded by the same quantity; step 5.1 gives the corresponding bounds for $h$ by $C_n^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty };B_{3\mathrm{/}4}}$, which step 4.1 bounds by $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty };B_r(a)}+C_{n,\alpha }^{\mathrm{\prime }}(\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}+[F{]}_{0,\alpha })$; summing, $\mathrm{\parallel }v{\mathrm{\parallel }}_{2,\alpha ;B_{1\mathrm{/}2}(0)}^{\ast }\le C_{n,\alpha }(\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty };B_r(a)}+\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty };B_1(0)}+[F{]}_{0,\alpha ;B_1(0)})$.

Undoing the scaling. The scaling identity of [F1] applied to the sub-ball of radius $1\mathrm{/}2$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{2,\alpha ;B_{r\mathrm{/}2}(a)}^{\ast }=\mathrm{\parallel }v{\mathrm{\parallel }}_{2,\alpha ;B_{1\mathrm{/}2}(0)}^{\ast }$, and step 1.1 converts $\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}+[F{]}_{0,\alpha }$ into $r^2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty };B_r(a)}+r^{2+\alpha }[f{]}_{0,\alpha ;B_r(a)}$; hence step 6.1 gives exactly the displayed estimate with a constant depending only on $n$ and $\alpha$. In particular $\mathrm{\parallel }u{\mathrm{\parallel }}_{2,\alpha ;B_{r\mathrm{/}2}(a)}^{\ast }<\mathrm{\infty }$, so $u\in C^{2,\alpha }(B_{r\mathrm{/}2}(a))$.

The quantitative estimate for the potential and the identity $-\mathrm{\Delta }w=G$ come from [F3], and the estimate for the harmonic remainder comes from [F5]. Although [F3] also gives a cancellation formula for the singular Hessian, the proof uses its stated $C^{2,\alpha }$ bound and does not differentiate $F$; neither the weak maximum principle nor a ball Dirichlet theorem is needed.