Harmonic functions are real analytic

Work under [F7] and suppose first that $u$ is real. Since $\mathrm{\Omega }$ is open and $a\in \mathrm{\Omega }$, choose $r>0$ with $\overline{B_{2r}(a)}\subset \mathrm{\Omega }$. Then $u$ is continuous on the compact set $\overline{B_{2r}(a)}$, so $M:={\sup }_{B_{2r}(a)}\mathrm{\mid }u\mathrm{\mid }$ is a finite nonnegative number.

Poisson representation and derivative bounds from the kernel. For $n\ge 3$, $u$ equals the Poisson integral of its trace $g:=u{\mathrm{\mid }}_{\mathrm{\partial }{B}_{2r}(a)}$ on $B_{2r}(a)$ by [F1], since both functions are $C^2\cap C(\overline{B_{2r}(a)})$ harmonic with trace $g$. For $n=2$, [F9] makes $u$ smooth on a neighbourhood of $\overline{B_{2r}(a)}$, so $g$ is smooth; [F10] then gives the same Poisson representation and uniqueness. In both cases the kernel is $P_{2r,a}$. Differentiating the representation through the integral by [F2] (for $x$ in the compact ball $\overline{B_r(a)}$ the sphere is separated from $x$, and all kernel derivatives are bounded there), we get $D^{\alpha }u(x)={\int }_{\mathrm{\partial }{B}_{2r}(a)}{D}_x^{\alpha }{P}_{2r,a}(x,y)u(y)d{S}_y$ for every multi-index $\alpha$ and every $x\in B_r(a)$.

Kernel derivative bound. Write $x=a+2r\zeta$, $y=a+2r\eta$, with $\mathrm{\mid }\zeta \mathrm{\mid }\le 1\mathrm{/}2$ and $\mathrm{\mid }\eta \mathrm{\mid }=1$. Then $P_{2r,a}(x,y)=(2r{)}^{1-n}g(\zeta ,\eta )\mathrm{/}{\omega }_{n-1}$, where $g=(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)\mathrm{\mid }\zeta -\eta {\mathrm{\mid }}^{-n}$. Fix ${\zeta }_0$ with $\mathrm{\mid }{\zeta }_0\mathrm{\mid }\le 1\mathrm{/}2$, put $d={\zeta }_0-\eta$ and $A=\mathrm{\mid }d{\mathrm{\mid }}^2\ge 1\mathrm{/}4$, and write $\zeta ={\zeta }_0+u$. Then $\mathrm{\mid }\zeta -\eta {\mathrm{\mid }}^{-n}=A^{-n\mathrm{/}2}(1+P_0(u){)}^{-n\mathrm{/}2},P_0(u)=A^{-1}(2d\cdot u+\mathrm{\mid }u{\mathrm{\mid }}^2)\mathrm{.}$ For $\mathrm{\mid }u\mathrm{\mid }\le 1\mathrm{/}4$, $\mathrm{\mid }{P}_0(u)\mathrm{\mid }\le 16\mathrm{\mid }u\mathrm{\mid }$, so the binomial series for $(1+P_0{)}^{-n\mathrm{/}2}$ converges near $u=0$, for example when $\mathrm{\mid }u\mathrm{\mid }<1\mathrm{/}32$. Its coefficients satisfy $\mathrm{\mid }\left(\genfrac{}{}{0px}{}{-n\mathrm{/}2}{m}\right)\mathrm{\mid }={\prod }_{j=1}^m(1+(n\mathrm{/}2-1)\mathrm{/}j)\le (n\mathrm{/}2{)}^m$. The coefficients of the linear and quadratic terms of $P_0$ are bounded by $12$ and $4$, respectively, and it has at most $2n$ monomials. For total degree $k\ge 1$, only powers $m\le k$ contribute; counting at most $(2n{)}^m$ products in $P_0^m$, then multiplying by the degree-two polynomial $1-\mathrm{\mid }{\zeta }_0+u{\mathrm{\mid }}^2$ and by $A^{-n\mathrm{/}2}\le 2^n$, bounds each Taylor coefficient of $g$ of total degree $k$ by $C_2^k$ for a constant $C_2(n)$. Since $D^{\alpha }g({\zeta }_0,\eta )=\alpha !$ times its $u^{\alpha }$ coefficient and $\alpha !\le \mathrm{\mid }\alpha \mathrm{\mid }!$, this gives $\mathrm{\mid }{D}_{\zeta }^{\alpha }g\mathrm{\mid }\le C_2^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\alpha \mathrm{\mid }!$ for $\mathrm{\mid }\alpha \mathrm{\mid }\ge 1$, uniformly in ${\zeta }_0,\eta$. Thus for $k=\mathrm{\mid }\alpha \mathrm{\mid }\ge 1$, $\mathrm{\mid }{D}_x^{\alpha }{P}_{2r,a}(x,y)\mathrm{\mid }\le (2r{)}^{1-n-k}{\omega }_{n-1}^{-1}{C}_2^{k}k!$.

Factorial derivative bound on the inner ball. For $k:=\mathrm{\mid }\alpha \mathrm{\mid }\ge 1$, combining steps 2.1 and 3.1 with [F5] gives, for $x\in B_r(a)$, $\mathrm{\mid }{D}^{\alpha }u(x)\mathrm{\mid }\le C_2^{k}k!(2r{)}^{1-n-k}{\omega }_{n-1}^{-1}{\int }_{\mathrm{\partial }{B}_{2r}(a)}\mathrm{\mid }u\mathrm{\mid }dS\le M{C}_2^{k}k!(2r{)}^{-k}\mathrm{.}$ For $k=0$, the bound $\mathrm{\mid }u(x)\mathrm{\mid }\le M$ follows directly from the definition of $M$.

Taylor remainder. Let $k\ge 0$ and let $h$ satisfy $\mathrm{\mid }h\mathrm{\mid }<r$. The ball $B_r(a)$ is convex and open, contains $a$ and $a+h$, and $u$ is $C^{k+1}$ on it by step 2.1, so [F3] gives some $\theta \in (0,1)$ with $u(a+h)-T_{k}u(a;h)={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=k+1}{D}^{\alpha }u(a+\theta h)h^{\alpha }\mathrm{/}\alpha !$. Since $a+\theta h\in B_r(a)$, step 4.1 bounds each term by $M(2r{)}^{-(k+1)}{C}_2^{k+1}(k+1)!\mathrm{\mid }{h}^{\alpha }\mathrm{\mid }\mathrm{/}\alpha !$, and ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=k+1}\mathrm{\mid }{h}^{\alpha }\mathrm{\mid }\mathrm{/}\alpha !\le {\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=k+1}\mathrm{\mid }h{\mathrm{\mid }}^{k+1}\mathrm{/}\alpha !=n^{k+1}\mathrm{\mid }h{\mathrm{\mid }}^{k+1}\mathrm{/}(k+1)!$ by [F3]; the two $(k+1)!$ factors cancel and the remainder is at most $M(n{C}_2\mathrm{\mid }h\mathrm{\mid }\mathrm{/}(2r){)}^{k+1}$.

The factorial bound. If $\overline{B_{2r}(a)}\subset \mathrm{\Omega }$ and $M={\sup }_{B_{2r}(a)}\mathrm{\mid }u\mathrm{\mid }$, step 4.1 gives the claimed estimate for $\mathrm{\mid }\alpha \mathrm{\mid }\ge 1$ with $C_n:=C_2\mathrm{/}2$; for $\mathrm{\mid }\alpha \mathrm{\mid }=0$ it is $\mathrm{\mid }u(a)\mathrm{\mid }\le M$. The constant depends only on $n$.

Convergence and analyticity. Choose $0<\rho <2r\mathrm{/}(n{C}_2)$. For $\mathrm{\mid }h\mathrm{\mid }<\rho$ the Taylor remainder bound in step 5.1 tends to zero, so the Taylor polynomials converge to $u(a+h)$. The degree-zero term is at most $M$, while for each $k\ge 1$ step 4.1 and the multinomial bound in [F3] give ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=k}\mathrm{\mid }{D}^{\alpha }u(a)h^{\alpha }\mathrm{\mid }\mathrm{/}\alpha !\le M(n{C}_2\mathrm{\mid }h\mathrm{\mid }\mathrm{/}(2r){)}^k$. The geometric series converges, so the Taylor series converges absolutely and equals $u(a+h)$; the ball $\mathrm{\mid }h\mathrm{\mid }<\rho$ contains the polydisc $\mathrm{\mid }{h}_i\mathrm{\mid }<\rho \mathrm{/}\sqrt{n}$, hence $u$ is real analytic at $a$ in the sense of [F4].

Complex $u$: apply steps 1.1 through 6.1 to $\mathrm{R}\mathrm{e}u$ and $\mathrm{I}\mathrm{m}u$, which are real harmonic; the Taylor coefficients of $u$ are the sums of the corresponding coefficients, and the two real series give an absolutely convergent complex series. For $\mathrm{\mid }\alpha \mathrm{\mid }=0$, $\mathrm{\mid }u(a)\mathrm{\mid }\le M$ directly. For $k:=\mathrm{\mid }\alpha \mathrm{\mid }\ge 1$, the real estimates give $\mathrm{\mid }{D}^{\alpha }u(a)\mathrm{\mid }\le 2M{C}_n^{k}k!r^{-k}\le M(2C_n{)}^{k}k!r^{-k}$. Thus the stated estimate, including order zero, holds with constant $2C_n$ in place of $C_n$. Since $a\in \mathrm{\Omega }$ was arbitrary, every harmonic function on $\mathrm{\Omega }$ is real analytic.