Schauder Theory and the Continuity Method

Proof. Let $v(x) = \int_B\partial_i\Phi(x - y)f(y)\,dy$, which converges since $|\partial_i\Phi(z)| \leq C|z|^{1-n}$ is locally integrable. Fix a smooth cut-off $\eta(t)$ with $\eta = 0$ for $t \leq 1$, $\eta = 1$ for $t \geq 2$, $0 \leq \eta \leq 1$, $|\eta'| \leq 2$, and let $\eta_\varepsilon(z) = \eta(|z|/\varepsilon)$. The function $w_\varepsilon(x) = \int_B\Phi(x - y)\eta_\varepsilon(x - y)f(y)\,dy$ has a smooth kernel, so $\partial_iw_\varepsilon = \int\partial_i(\Phi\eta_\varepsilon)(x - y)f(y)\,dy$. The kernels of $v$ and $\partial_iw_\varepsilon$ agree where $|x - y| \geq 2\varepsilon$, so

(the $\log$ only for $n = 2$), and similarly $|w - w_\varepsilon| \leq C|f|_0\varepsilon^2|\log\varepsilon|$. Both tend to $0$ uniformly in $x$, so $w_\varepsilon \to w$ and $\partial_iw_\varepsilon \to v$ uniformly; hence $w \in C^1$ with $\partial_iw = v$ (2B.5 Uniform Convergence and Arzelà–Ascoli).

Proof. Let $v(x)$ be the right side; the first integral converges because its integrand is at most $C[f]_\alpha|x - y|^{\alpha - n}$. Let $u_\varepsilon(x) = \int_B\partial_i\Phi(x - y)\eta_\varepsilon(x - y)f(y)\,dy$, a smooth function of $x$, which converges uniformly to $\partial_iw$ by the previous proof. For $x \in B$ and $2\varepsilon < \operatorname{dist}(x, \partial B)$, differentiate under the integral and split $f(y) = (f(y) - f(x)) + f(x)$:

In the last integral, $\partial_{x_j}$ of a function of $x - y$ is $-\partial_{y_j}$ of it, so by the divergence theorem it equals $-\int_{\partial B}(\partial_i\Phi\,\eta_\varepsilon)(x - y)\nu_j(y)\,dS(y)$, and $\eta_\varepsilon = 1$ on $\partial B$ by the choice of $\varepsilon$. So $\partial_ju_\varepsilon(x)$ differs from $v(x)$ only through the first integral, over $|x - y| \leq 2\varepsilon$:

This tends to $0$ uniformly on compact subsets of $B$, so $\partial_iw \in C^1(B)$ with $\partial_j\partial_iw = v$.

For the equation, sum over $i = j$. Since $\sum_i\partial_{ii}\Phi = \Delta\Phi = 0$ away from $0$, the first integral contributes nothing. In the second, $\sum_i\partial_i\Phi(x - y)\nu_i(y) = -\nabla_y[\Phi(x - y)]\cdot\nu$, and the flux of $\nabla_y\Phi(\cdot - x)$ out of any domain containing $x$ is $-1$ (6A.2 Harmonic Functions). So $\Delta w(x) = -f(x)\cdot(-1)\cdot(-1) = -f(x)$.

Proof. Throughout, $C$ depends only on $n$ and $\alpha$, and we use the formula of Lemma 7.2 on $B = B_2$. Write $J(x) = \int_{\partial B_2}\partial_i\Phi(x - y)\nu_j(y)\,dS(y)$ and $V(x) = \int_{B_2}\partial_{ij}\Phi(x - y)(f(y) - f(x))\,dy$, so $\partial_{ij}w = V - fJ$.

Sup bound. For $x \in B_1$ and $y \in \partial B_2$, $|x - y| \geq R$, so $|J(x)| \leq CR^{1-n}\cdot R^{n-1} = C$. And $|V(x)| \leq [f]_\alpha\int_{|z|\leq3R}C|z|^{\alpha-n}dz = C[f]_\alpha R^\alpha$. So $|D^2w|_{0;B_1} \leq C(|f|_0 + R^\alpha[f]_\alpha)$.

Hölder bound, boundary term. Let $x, \bar x \in B_1$, $\delta = |x - \bar x| \leq 2R$. Then $f(x)J(x) - f(\bar x)J(\bar x) = f(x)(J(x) - J(\bar x)) + (f(x) - f(\bar x))J(\bar x)$. By the mean value theorem, $|J(x) - J(\bar x)| \leq \delta\sup|D^2\Phi|\cdot|\partial B_2| \leq C\frac\delta R \leq C\big(\frac\delta R\big)^\alpha$ (as $\frac\delta R \leq 2$). So this part is at most $C(|f|_0R^{-\alpha} + [f]_\alpha)\delta^\alpha$.

Hölder bound, volume term. Let $\xi = \frac{x + \bar x}{2}$ and split $B_2$ into $B(\xi, \delta)\cap B_2$ (near) and the rest (far).

Near part: $B(\xi, \delta) \subset B(x, \frac32\delta)$, so $\big|\int_{\text{near}}\partial_{ij}\Phi(x - y)(f(y) - f(x))\,dy\big| \leq C[f]_\alpha\int_{|z|\leq\frac32\delta}|z|^{\alpha-n}dz = C[f]_\alpha\delta^\alpha$, and the same for $\bar x$.

Far part: write the difference of integrands as

For the first piece: for $y$ outside $B(\xi, \delta)$ and $\hat x$ on the segment $[x, \bar x]$, $|\hat x - y| \geq |\xi - y| - \frac\delta2 \geq \frac12|\xi - y|$, so the mean value theorem and $|D^3\Phi(z)| \leq C|z|^{-n-1}$ give $|\partial_{ij}\Phi(x - y) - \partial_{ij}\Phi(\bar x - y)| \leq C\delta|\xi - y|^{-n-1}$. Also $|f(y) - f(\bar x)| \leq [f]_\alpha(|y - \xi| + \frac\delta2)^\alpha \leq C[f]_\alpha|y - \xi|^\alpha$. So the first piece contributes at most

which is finite exactly because $\alpha < 1$. For the second piece, $|f(\bar x) - f(x)| \leq [f]_\alpha\delta^\alpha$, and by the divergence theorem

with $\nu$ the outward normals of the two spheres (the region's boundary is $\partial B_2$ outward and $\partial B(\xi, \delta)$ inward, and $\partial_{x_j} = -\partial_{y_j}$ supplies the signs). On $\partial B(\xi, \delta)$, $|x - y| \geq \frac\delta2$, so the first boundary integral is at most $C\delta^{1-n}\cdot\delta^{n-1} = C$; the second is at most $C$ as before. This cancellation, the kernel integrating to something bounded over an annulus, is the one described in the previous section.

Adding everything, $|\partial_{ij}w(x) - \partial_{ij}w(\bar x)| \leq C(|f|_{0;B_2}R^{-\alpha} + [f]_{\alpha;B_2})\delta^\alpha$, which is the claim after multiplying by $R^\alpha$.

Proof. Let $w$ be the Newtonian potential of $-f$ over $B_{2R}$, so $\Delta w = f$ in $B_{2R}$, and $h = u - w$, harmonic in $B_{2R}$. By Theorem 7.3, with the balls $B_R \subset B_{2R}$, $w$ satisfies the estimate. For $h$, use the derivative estimates for harmonic functions (6A.2 Harmonic Functions): on $B_{3R/2}$, $|D^kh| \leq C_kR^{-k}|h|_{0;B_{2R}}$ for $k = 2, 3$. Then for $x, y \in B_R$: if $|x - y| \leq \frac R2$ the segment lies in $B_{3R/2}$ and $|D^2h(x) - D^2h(y)| \leq C\frac{|h|_0}{R^3}|x - y| \leq C\frac{|h|_0}{R^{2+\alpha}}|x - y|^\alpha$; if $|x - y| > \frac R2$, then $|D^2h(x) - D^2h(y)| \leq 2C\frac{|h|_0}{R^2} \leq C'\frac{|h|_0}{R^{2+\alpha}}|x - y|^\alpha$. So $R^2|D^2h|_{0;B_R} + R^{2+\alpha}[D^2h]_{\alpha;B_R} \leq C|h|_{0;B_{2R}}$. Finally $|h|_0 \leq |u|_0 + |w|_0$ and $|w|_{0;B_{2R}} \leq |f|_0\sup_x\int_{B_{2R}}|\Phi(x - y)|\,dy \leq CR^2|f|_0$ (with a $\log R$ correction when $n = 2$, which is removed by scaling to $R = 1$ first). Add the estimates for $w$ and $h$.

Proof. Let $x \in B_s$ and $e$ a unit vector. Move $x$ towards the centre by $\frac\eta2$ (or to the centre, if it is closer): the new point $y$ satisfies $|x - y| \leq \frac\eta2$ and $B(y, \frac\eta2) \subset B_s$. With $h = \frac\eta2$, Taylor's theorem gives a point $\theta$ on the segment from $y - he$ to $y + he$ with

so $|D^2u(\theta)(e, e)| \leq \frac{4|u|_0}{h^2} = \frac{16|u|_0}{\eta^2}$. And $|x - \theta| \leq \eta$, so $|D^2u(x)(e, e)| \leq |D^2u(\theta)(e, e)| + [D^2u]_\alpha\eta^\alpha$. The norm of a symmetric matrix is $\max_{|e|=1}|M(e, e)|$ (1A.6 Symmetric Matrices and the Spectral Theorem), which gives the first inequality. The second is the same with the first difference $u(y + he) - u(y - he) = 2hDu(\theta)\cdot e$ and $|x - \theta| \leq \eta$.

Proof. Choose $\tau \in (0, 1)$ with $\theta\tau^{-\beta} < 1$, and points $r_0 = \frac12$, $r_{k+1} = r_k + (1 - \tau)\tau^k\cdot\frac12$, which increase to $1$. Applying the hypothesis with $r = r_k$, $s = r_{k+1}$ and iterating $m$ times,

Let $m \to \infty$: the first term tends to $0$ because $\varphi$ is bounded, and both series converge.

Proof. Let $\varphi(r) = [D^2u]_{\alpha;B_r}$ for $\frac12 \leq r \leq 1$, bounded because $u \in C^{2,\alpha}(\overline{B_1})$. Fix $\frac12 \leq r < s \leq 1$ and let $\rho = \sigma(s - r)$, where $\sigma \in (0, \frac12]$ will be chosen small. Throughout, $C$ denotes constants depending only on $n$, $\alpha$, $\lambda$, $\Lambda$, and all sup norms and seminorms without a subscript are over $B_s$.

Freezing. Fix $x_0 \in B_r$; then $B(x_0, \rho) \subset B_s$. Let $L_0 = a^{ij}(x_0)\partial_{ij}$, so that on $B(x_0, \rho)$

The point is that the first term is small on a small ball: $|a^{ij}(x_0) - a^{ij}(x)| \leq \Lambda\rho^\alpha$ there.

Estimating $F$. Using $[gh]_\alpha \leq |g|_0[h]_\alpha + [g]_\alpha|h|_0$ and, on a ball of radius $\rho$, $[g]_\alpha \leq (2\rho)^{1-\alpha}|Dg|_0$ for $g = \partial_iu$ and $g = u$,

(The first term comes from $[(a(x_0) - a)D^2u]_\alpha \leq \Lambda\rho^\alpha[D^2u]_\alpha + \Lambda|D^2u|_0$; the rest from the lower-order terms.) Since $\rho \leq 1$, this simplifies to

The frozen estimate. Apply $(*)$ on $B(x_0, \rho)$ and divide by $\rho^{2+\alpha}$:

From small balls to $B_r$. For $x, y \in B_r$: if $|x - y| < \frac\rho2$, then $y \in B(x, \frac\rho2)$ and the quotient $\frac{|D^2u(x) - D^2u(y)|}{|x - y|^\alpha}$ is bounded by the left side above (with $x_0 = x$); if $|x - y| \geq \frac\rho2$, it is at most $2|D^2u|_0(\frac\rho2)^{-\alpha}$. So

Interpolating away the lower derivatives. By Lemma 7.5 on $B_s$ with $\eta = \varepsilon\rho$ (note $\eta \leq s$): $\rho^{-\alpha}|D^2u|_0 \leq \varepsilon^\alpha\varphi(s) + 16\varepsilon^{-2}\rho^{-2-\alpha}|u|_0$, and $\rho^{-\alpha}|Du|_0 \leq \rho^{-\alpha}\big(\rho|D^2u|_0 + 4\rho^{-1}|u|_0\big)$, which is bounded the same way. Hence

Choosing the constants and absorbing. Choose $\varepsilon$ with $C\varepsilon^\alpha \leq \frac14$, and then $\sigma$ with $C\sigma^\alpha \leq \frac14$, so that $C(\rho^\alpha + \varepsilon^\alpha) \leq \frac12$ (recall $\rho = \sigma(s - r) \leq \sigma$). With $\rho^{-2-\alpha} = \sigma^{-2-\alpha}(s - r)^{-2-\alpha}$,

Lemma 7.6 gives $[D^2u]_{\alpha;B_{1/2}} = \varphi(\frac12) \leq C(|u|_{0;B_1} + |f|_{0;B_1} + [f]_{\alpha;B_1})$. The bounds for $|D^2u|_0$ and $|Du|_0$ on $B_{1/2}$ follow from Lemma 7.5.