Continuous Dirichlet problem on a ball

Work under [F8]. Fix $x_0\in B_R(a)$. By [F1] and [F7], $y\mapsto P_{R,a}(x_0,y)$ is continuous on the compact sphere, hence bounded, and $\parallel g{\parallel }_{\mathrm{\infty }}<+\mathrm{\infty }$; the sphere has finite surface measure by [F5], so $\mathrm{\mid }{P}_{R,a}(x_0,\cdot )g\mathrm{\mid }\le \parallel g{\parallel }_{\mathrm{\infty }}{\sup }_y{P}_{R,a}(x_0,y)$ is integrable and $U_g(x_0)$ is absolutely convergent. Moreover $\parallel U_g{\parallel }_{\mathrm{\infty }}\le \parallel g{\parallel }_{\mathrm{\infty }}$ on $B_R(a)$ by [F1] and [F5].

Smoothness of the kernel and of the parametrised integrals. The map $(x,y)\mapsto \mathrm{\mid }x-y{\mathrm{\mid }}^2={\sum }_i(x_i-y_i{)}^2$ is a polynomial, hence $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^n\times {\mathbb{R}}^n$, and it is strictly positive on the set where $x\mathrm{\ne }y$; composing with $t\mapsto t^{-n\mathrm{/}2}$, which is $C^{\mathrm{\infty }}$ on $(0,\mathrm{\infty })$ by [F6], and multiplying by the polynomial $R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2$ shows that $P$ is $C^{\mathrm{\infty }}$ on its domain by [F6]. Consequently for every multi-index $\alpha$ the function $(x,y)\mapsto D_x^{\alpha }P(x,y)$ is continuous there, and for every nonempty compact $K\subset B_R(a)$ the distance $d_K:=\min \{\mathrm{\mid }x-y\mathrm{\mid }:x\in K,y\in \mathrm{\partial }{B}_R(a)\}$ is positive and $M_{\alpha ,K}:={\sup }_{K\times \mathrm{\partial }{B}_R(a)}\mathrm{\mid }{D}_x^{\alpha }P\mathrm{\mid }<+\mathrm{\infty }$, by [F7] and [F6] applied to the continuous function $(x,y)\mapsto \mathrm{\mid }x-y\mathrm{\mid }$ on the compact set $K\times \mathrm{\partial }{B}_R(a)$.

The kernel is harmonic in the interior variable. Fix $y\in \mathrm{\partial }{B}_R(a)$ and $x\in B_R(a)$, put $m(x):=R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2$ and $\rho (x):=\mathrm{\mid }x-y\mathrm{\mid }>0$. Direct differentiation gives ${\mathrm{\partial }}_{i}m=-2(x_i-a_i)$, $\mathrm{\Delta }m=-2n$, ${\mathrm{\partial }}_i\rho =(x_i-y_i)\mathrm{/}\rho$, and for a $C^2$ radial profile $q$ the formulas ${\mathrm{\partial }}_i(q\circ \rho )=q^{\mathrm{\prime }}(\rho )(x_i-y_i)\mathrm{/}\rho$ and $\mathrm{\Delta }(q\circ \rho )=q^{\mathrm{\prime }\mathrm{\prime }}(\rho )+(n-1)q^{\mathrm{\prime }}(\rho )\mathrm{/}\rho$; with $q(\rho )={\rho }^{-n}$ this gives $\mathrm{\nabla }({\rho }^{-n})=-n{\rho }^{-n-2}(x-y)$ and $\mathrm{\Delta }({\rho }^{-n})=2n{\rho }^{-n-2}$ by [F6]. Hence $\mathrm{\Delta }(m{\rho }^{-n})=\mathrm{\Delta }m\cdot {\rho }^{-n}+2\mathrm{\nabla }m\cdot \mathrm{\nabla }({\rho }^{-n})+m\mathrm{\Delta }({\rho }^{-n})=-2n{\rho }^{-n}+4n(x-a)\cdot (x-y){\rho }^{-n-2}+2n(R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2){\rho }^{-n-2}$, and the identity $\mathrm{\mid }x-y{\mathrm{\mid }}^2=\mathrm{\mid }x-a{\mathrm{\mid }}^2-2(x-a)\cdot (y-a)+R^2$ shows that the last two terms equal $2n{\rho }^{-n}$, so $\mathrm{\Delta }(m{\rho }^{-n})=0$. Since $P_{R,a}(x,y)=m(x)\rho (x{)}^{-n}\mathrm{/}(R{\omega }_{n-1})$, we get ${\mathrm{\Delta }}_x{P}_{R,a}(x,y)=0$ for all $x\in B_R(a)$, $y\in \mathrm{\partial }{B}_R(a)$.

Higher derivatives under the integral. Induct on the length of an ordered word of coordinate derivatives. The empty word gives the defining integral for $U_g$. Suppose a word gives $V(x)=\int Q(x,y)g(y)d{S}_y$, where $Q$ is the same ordered derivative of $P$. Fix $x_0$ and a closed ball $K$ with $x_0\in \mathrm{int}K$ and $K\subset B_R(a)$. By step 1.2, both $Q$ and ${\mathrm{\partial }}_{i}Q$ are continuous and bounded on $K\times \mathrm{\partial }{B}_R(a)$. For $x=x_0+t{e}_i$ on a sufficiently small open interval, each slice $Q(x,\cdot )g$ is Borel and integrable, and its $t$-derivative is Borel and bounded by ${\sup }_{K\times \mathrm{\partial }{B}_R(a)}\mathrm{\mid }{\mathrm{\partial }}_{i}Q\mathrm{\mid }\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$, an integrable constant by [F5]. Thus all hypotheses of [F4] hold, with empty exceptional set, and ${\mathrm{\partial }}_{i}V(x_0)=\int {\mathrm{\partial }}_{i}Q(x_0,y)g(y)d{S}_y$. The integral expressions for both $V$ and this derivative are continuous near $x_0$ by dominated convergence [F5], using the respective bounded continuous kernels on $K$. This proves existence and continuity for every ordered derivative, hence $U_g\in C^{\mathrm{\infty }}$ under [F6]; choosing the canonical word for a multi-index gives $D^{\alpha }{U}_g(x)=\int D_x^{\alpha }P(x,y)g(y)d{S}_y$. No interchange of derivative order is required.

$U_g$ is smooth and harmonic. Step 2.1 with $\alpha =0$ gives $C^{\mathrm{\infty }}$, and for $\mathrm{\mid }\alpha \mathrm{\mid }=2$ it gives $\mathrm{\Delta }{U}_g(x)={\sum }_i\int {\mathrm{\partial }}_i{\mathrm{\partial }}_i{P}_{R,a}(x,y)g(y)d{S}_y=\int {\mathrm{\Delta }}_x{P}_{R,a}(x,y)g(y)d{S}_y$ by [F5] and the Laplacian definition of [F6]; step 1.3 makes every value of ${\mathrm{\Delta }}_x{P}_{R,a}$ vanish, so $\mathrm{\Delta }{U}_g=0$ on $B_R(a)$ and $U_g$ is smooth harmonic.

Boundary trace and continuity on the closed ball. Interior continuity holds by step 2.1 with $\alpha =0$. Define $\tilde{U}:=U_g$ on $B_R(a)$ and $\tilde{U}:=g$ on $\mathrm{\partial }{B}_R(a)$. At a boundary point $p$, [F2] gives $U_g(x)\to g(p)=\tilde{U}(p)$ along every interior approach, and $g$ is continuous on the sphere by hypothesis; hence $\tilde{U}$ is continuous at every point of $\overline{B_R(a)}$ and $U_g$ extends continuously to the closed ball with trace $g$.

Uniqueness. Let $v\in C^2(B_R(a))\cap C(\overline{B_R(a)})$ be harmonic on $B_R(a)$ with $v=g$ on $\mathrm{\partial }{B}_R(a)$, and put $w:=v-\tilde{U}$, which is continuous on the closure, $C^2$ inside and harmonic inside by step 3.1. Apply [F3] to $\mathrm{R}\mathrm{e}w$ and to $-\mathrm{R}\mathrm{e}w$, and to $\mathrm{I}\mathrm{m}w$ and $-\mathrm{I}\mathrm{m}w$: on the boundary all four functions vanish, so their maxima over $\overline{B_R(a)}$ are zero. Hence $w\equiv 0$ and $v=U_g$.

Steps 1.1, 3.1 and 3.2 show that $U_g$ is absolutely convergent, smooth harmonic and continuously extendible with trace $g$, and step 4.1 shows that every such classical solution equals $U_g$; this is exactly the assertion. The boundary convergence was obtained from the cap/complement estimate [F2], which depends only on the kernel formula, its positivity and its unit mass, so the later uniform-radial corollary is not presupposed.