Unique continuation for harmonic functions

Work under [F4] and let $F:=\{x\in \mathrm{\Omega }:D^{\alpha }u(x)=0\text{ for every multi-index }\alpha \}$, the set of points where all derivatives vanish. Since $V$ is open and $u=0$ identically on $V$, every derivative of $u$ vanishes on $V$ (a derivative of the zero function), so $V\subseteq F$ and $F\mathrm{\ne }\mathrm{\varnothing }$.

$F$ contains $V$, and $F$ is closed in $\mathrm{\Omega }$: by [F1] each function $D^{\alpha }u$ is continuous on $\mathrm{\Omega }$, and $F={\bigcap }_{\alpha }\{x\in \mathrm{\Omega }:D^{\alpha }u(x)=0\}$ is an intersection of closed subsets of $\mathrm{\Omega }$.

$F$ is open in $\mathrm{\Omega }$: let $x\in F$. By [F1] choose $\rho >0$ such that $u(x+h)={\sum }_{\alpha }{D}^{\alpha }u(x)h^{\alpha }\mathrm{/}\alpha !$ with absolute convergence for $\mathrm{\mid }h\mathrm{\mid }<\rho$. Since $x\in F$, every coefficient $D^{\alpha }u(x)$ vanishes, so the series is identically zero and $u=0$ on the ball $B_{\rho }(x)\subseteq \mathrm{\Omega }$; that ball is open, so every derivative of $u$ vanishes on it and $B_{\rho }(x)\subseteq F$. Hence $F$ is open in $\mathrm{\Omega }$.

Therefore $F$ is clopen in $\mathrm{\Omega }$ and nonempty, while $\mathrm{\Omega }$ is connected; by [F2] the only clopen subsets of $\mathrm{\Omega }$ are $\mathrm{\varnothing }$ and $\mathrm{\Omega }$, so $F=\mathrm{\Omega }$. Hence all derivatives of $u$ vanish everywhere and, in particular, $u(x)=D^{0}u(x)=0$ for every $x\in \mathrm{\Omega }$ by [F3]; that is, $u$ vanishes identically on $\mathrm{\Omega }$. The argument applies to real and complex $u$ alike because the Taylor representation and continuity are available in both cases.