Ball Poisson series: interior convergence, boundary data and the Poisson kernel

Finite projection, density and Parseval. For any $g\in L^2(S^2)$ put $a_{lm}=⟨g,Y_{lm}⟩$ and $Q_{N}g={\sum }_{l\le N,m}{a}_{lm}{Y}_{lm}$, with finite-dimensional span $V_N$. Because the $Y_{lm}$ are real, these coefficients are $\int g{Y}_{lm}d\mathrm{\Omega }$ also for complex $g$. Finite inner-product algebra [F1, F2] gives $⟨g-Q_{N}g,Y_{lm}⟩=0$ for $l\le N$, $\mathrm{\parallel }{Q}_{N}g{\mathrm{\parallel }}_2^2={\sum }_{l\le N,m}\mathrm{\mid }{a}_{lm}{\mathrm{\mid }}^2$, and the Pythagorean identity $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=\mathrm{\parallel }g-Q_{N}g{\mathrm{\parallel }}_2^2+{\sum }_{l\le N,m}\mathrm{\mid }{a}_{lm}{\mathrm{\mid }}^2$. Thus the finite coefficient sums are bounded by $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$ (Bessel). For every $h\in V_N$, the same orthogonality yields $\mathrm{\parallel }g-h{\mathrm{\parallel }}_2^2=\mathrm{\parallel }g-Q_{N}g{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{Q}_{N}g-h{\mathrm{\parallel }}_2^2$, hence $\mathrm{\parallel }g-Q_{N}g{\mathrm{\parallel }}_2\le \mathrm{\parallel }g-h{\mathrm{\parallel }}_2$. Completeness in [F1] means that the finite linear span is dense. For each fixed $\epsilon >0$, choose one finite combination $h$ with $\mathrm{\parallel }g-h{\mathrm{\parallel }}_2<\epsilon$ and let $N_0$ contain all its degrees (take $N_0=0$ for empty support). Then $h\in V_N$ for all $N\ge N_0$, so $Q_{N}g\to g$ in norm. In the complex case the same density follows by separately approximating the real and imaginary parts by finite real combinations. This epsilon argument selects no sequence of approximants and introduces no new Choice. Taking limits in the finite Pythagorean identity proves ${\sum }_{l,m}\mathrm{\mid }{a}_{lm}{\mathrm{\mid }}^2=\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$ (Parseval); subtracting its finite sum gives $\mathrm{\parallel }g-Q_{N}g{\mathrm{\parallel }}_2^2={\sum }_{l>N,m}\mathrm{\mid }{a}_{lm}{\mathrm{\mid }}^2\to 0$. Apply this to $g=f$, so $Q_{N}f=S_N$.

For the general $L^2$ uniqueness, let $v$ be a smooth harmonic function on the unit ball with ${\sup }_{r<1}\mathrm{\parallel }v(r\cdot ){\mathrm{\parallel }}_2<\mathrm{\infty }$ and $L^2$ boundary limit $f$. For $0<s<1$ the published Dirichlet theorem [F4] applied on $B_s(0)$ to the continuous data $v(s\cdot )$ gives, for $0<r<s$, $v(rn)={\int }_{S^2}\frac{1-(r\mathrm{/}s{)}^2}{4\pi \mathrm{\mid }(r\mathrm{/}s)n-n^{\mathrm{\prime }}{\mathrm{\mid }}^3}v(s{n}^{\mathrm{\prime }})d\mathrm{\Omega }(n^{\mathrm{\prime }})$, the scaled form of the radius-$s$ Poisson kernel; expanding this kernel with the addition theorem and integrating against $v(s\cdot )$ [F1, F6] gives $v(rn)={\sum }_{l,m}(r\mathrm{/}s{)}^l{v}_{lm}(s)Y_{lm}(n)$ with $v_{lm}(s)={\int }_{S^2}v(s{n}^{\mathrm{\prime }})Y_{lm}(n^{\mathrm{\prime }})d\mathrm{\Omega }(n^{\mathrm{\prime }})$.

Each $r^l{Y}_{lm}(n)$ is the homogeneous harmonic polynomial of degree $l$ representing $Y_{lm}$, evaluated at $x=rn$ [F1], so the partial sums $P_N(x)={\sum }_{l\le N,m}{f}_{lm}{r}^l{Y}_{lm}(n)$ are harmonic polynomials; by the addition theorem [F1] and finite counting-measure Cauchy-Schwarz [F2], first on $N<l\le M$ and then letting $M\to \mathrm{\infty }$ using step 1.1, the absolute tail on $\mathrm{\mid }x\mathrm{\mid }\le r_0<1$ is bounded by $\mathrm{\parallel }f-S_N{\mathrm{\parallel }}_2({\sum }_{l>N}{r}_0^{2l}\frac{2l+1}{4\pi }{)}^{1\mathrm{/}2}\le C(r_0)\mathrm{\parallel }f-S_N{\mathrm{\parallel }}_2$, which tends to $0$ by step 1.1; hence the series converges uniformly on every closed smaller ball and defines a continuous function $u$ there.

Uniform convergence on the fixed sphere from step 2.1 implies $L^2$ convergence there, so Cauchy-Schwarz [F2] identifies the coefficients of $u(r\cdot )$ as $r^l{f}_{lm}$. Parseval from step 1.1 applied to $u(r\cdot )$ and $u(r\cdot )-f$ gives $\mathrm{\parallel }u(r\cdot ){\mathrm{\parallel }}_2^2={\sum }_{l,m}{r}^{2l}\mathrm{\mid }{f}_{lm}{\mathrm{\mid }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }u(r\cdot )-f{\mathrm{\parallel }}_2^2={\sum }_{l,m}(1-r^l{)}^2\mathrm{\mid }{f}_{lm}{\mathrm{\mid }}^2$; since $0\le (1-r^l{)}^2\le 4$ and $(1-r^l{)}^2\to 0$ for each $(l,m)$ as $r\uparrow 1$, dominated convergence [F6] on the countable coefficient family with the summable bound $4\sum \mathrm{\mid }{f}_{lm}{\mathrm{\mid }}^2$ gives $u(r\cdot )\to f$ in $L^2(S^2)$ as $r\uparrow 1$.

For $x\in B_{r_1}$ with $r_1<r_0<1$ and $0<\rho <r_0-r_1$, the harmonic estimate [F3] applied to the harmonic differences $P_N-P_M$ gives $\mathrm{\mid }{D}^{\alpha }(P_N-P_M)(x)\mathrm{\mid }\le C_{3,\alpha }{\rho }^{-3-\mathrm{\mid }\alpha \mathrm{\mid }}{\int }_{B_{\rho }(x)}\mathrm{\mid }{P}_N-P_M\mathrm{\mid }$, and the $L^2$ bound inherited from the uniform tail of step 2.1 makes this tend to $0$ uniformly on $B_{r_1}$; hence every derivative series converges uniformly on each closed smaller ball, and the FTC [F6] identifies the derivatives of the limit: $u\in C^{\mathrm{\infty }}(B_1)$ and $\mathrm{\Delta }u={\lim }_N\mathrm{\Delta }{P}_N=0$, so $u$ is harmonic on the unit ball.

Inserting $f_{lm}={\int }_{S^2}f(n^{\mathrm{\prime }})Y_{lm}(n^{\mathrm{\prime }})d\mathrm{\Omega }(n^{\mathrm{\prime }})$ and the addition theorem [F1] into the partial sums gives $P_N(rn)={\int }_{S^2}[{\sum }_{l\le N}\frac{2l+1}{4\pi }{r}^l{P}_l(n\cdot n^{\mathrm{\prime }})]f(n^{\mathrm{\prime }})d\mathrm{\Omega }(n^{\mathrm{\prime }})$; for $r\le r_0<1$ the zonal series converges absolutely and uniformly by [F5], and letting $N\to \mathrm{\infty }$ with the linearity and dominated-convergence rules [F6] identifies $u(rn)={\int }_{S^2}{K}_r(n,n^{\mathrm{\prime }})f(n^{\mathrm{\prime }})d\mathrm{\Omega }(n^{\mathrm{\prime }})$ with $K_r(n,n^{\mathrm{\prime }})=\frac{1}{4\pi }{\sum }_l(2l+1)r^l{P}_l(n\cdot n^{\mathrm{\prime }})$.

The norm-convergent expansion proved in step 1.1, applied to $v(r\cdot )$, gives $v(rn)={\sum }_{l,m}{v}_{lm}(r)Y_{lm}(n)$; comparing with step 1.2 and using orthonormality yields $v_{lm}(r)=(r\mathrm{/}s{)}^l{v}_{lm}(s)$ for all $0<r<s<1$, and letting $s\uparrow 1$ with Cauchy-Schwarz [F2], $\mathrm{\mid }{v}_{lm}(s)-f_{lm}\mathrm{\mid }=\mathrm{\mid }\int (v(s{n}^{\mathrm{\prime }})-f(n^{\mathrm{\prime }}))Y_{lm}(n^{\mathrm{\prime }})d\mathrm{\Omega }\mathrm{\mid }\le \mathrm{\parallel }v(s\cdot )-f{\mathrm{\parallel }}_2\to 0$, so $v_{lm}(r)=r^l{f}_{lm}$ for every degree and every $0<r<1$.

Applying the operator $1+2r{\mathrm{\partial }}_r$ to the generating function of [F5] - legitimate inside $\mathrm{\mid }r\mathrm{\mid }<1$ by termwise differentiation [F5] - gives ${\sum }_l(2l+1)r^l{P}_l(z)=\frac{1-r^2}{(1-2rz+r^2{)}^{3\mathrm{/}2}}$; since $1-2r(n\cdot n^{\mathrm{\prime }})+r^2=\mathrm{\mid }rn-n^{\mathrm{\prime }}{\mathrm{\mid }}^2$, the kernel is $K_r(n,n^{\mathrm{\prime }})=\frac{1-r^2}{4\pi \mathrm{\mid }rn-n^{\mathrm{\prime }}{\mathrm{\mid }}^3}$, positive with numerator $1-r^2>0$ and positive denominator; integrating over $n^{\mathrm{\prime }}$ leaves only the constant $l=0$ harmonic in the zonal expansion, so ${\int }_{S^2}{K}_r(n,n^{\mathrm{\prime }})d\mathrm{\Omega }(n^{\mathrm{\prime }})=1$.

For continuous $f$, uniform continuity of $f$ on the compact sphere and the concentration of the normalised kernel [F4] give ${\sup }_n\mathrm{\mid }u(rn)-f(n)\mathrm{\mid }\le \int K_r(n,n^{\mathrm{\prime }})\mathrm{\mid }f(n^{\mathrm{\prime }})-f(n)\mathrm{\mid }d\mathrm{\Omega }(n^{\mathrm{\prime }})\to 0$ as $r\uparrow 1$: near $n^{\mathrm{\prime }}=n$ the integrand is small by continuity and the kernel mass is one, while the complement carries vanishing kernel mass because on a fixed complement $\mathrm{\mid }n-n^{\mathrm{\prime }}\mathrm{\mid }\ge \delta >0$, for $r>1-\delta \mathrm{/}2$ one has $\mathrm{\mid }rn-n^{\mathrm{\prime }}\mathrm{\mid }\ge \delta \mathrm{/}2$ by the triangle inequality; its kernel mass is bounded by $(1-r^2)(\delta \mathrm{/}2{)}^{-3}$ and tends to zero.

For continuous $f$ the Poisson integral of [F4] on the unit ball is exactly the integral representation of step 4.2, so the published ball Dirichlet theorem [F4] shows that $u$ extends continuously to the closed ball with boundary value $f$ and is the unique $C^2\cap C(\overline{B_1})$ harmonic function with that trace; the formula of the Statement is the kernel of step 4.2.

Hence $v(r\cdot )$ and $u(r\cdot )$ have the same complete orthonormal expansion on every sphere $r<1$, so they agree in $L^2$ there and, being continuous on each sphere, agree pointwise; equality at the centre follows by continuity, so $v=u$ on the unit ball. This proves the interior convergence and harmonicity, the $L^2$ boundary limit, the kernel formula for continuous data with its continuous extension, and $L^2$-boundary uniqueness, with Countable Choice used only through the cited harmonic, $L^2$ and Poisson interfaces.