A bounded harmonic function near an isolated puncture extends harmonically

Fix $z_0$ with $0<\mathrm{\mid }{z}_0-a\mathrm{\mid }<R\mathrm{/}2$, and let $\rho :=\mathrm{\mid }{z}_0-a\mathrm{\mid }\mathrm{/}4$. Then the disc $D(z_0,2\rho )$ lies in $0<\mathrm{\mid }z-a\mathrm{\mid }<R$. By [L1], choose a holomorphic function $F$ on $D(z_0,2\rho )$ with $\mathrm{Re}F=u$ there.

The function $H:=\exp (F)$ is holomorphic on $D(z_0,2\rho )$ by [L6], and its modulus is $\mathrm{\mid }H\mathrm{\mid }=e^u\le e^M$ there by the bound on $u$. Applying [L2] on the circle of radius $\rho$ about $z_0$ gives $\mathrm{\mid }{H}^{\mathrm{\prime }}(z_0)\mathrm{\mid }\le e^M\mathrm{/}\rho =4e^M\mathrm{/}\mathrm{\mid }{z}_0-a\mathrm{\mid }$. Since $H^{\mathrm{\prime }}=F^{\mathrm{\prime }}\exp (F)$ and $\mathrm{\mid }\exp (-F(z_0))\mathrm{\mid }=e^{-u(z_0)}\le e^M$, one gets $\mathrm{\mid }{F}^{\mathrm{\prime }}(z_0)\mathrm{\mid }\le \frac{4e^{2M}}{\mathrm{\mid }{z}_0-a\mathrm{\mid }}\mathrm{.}$

On every local potential disc from step 1.1 one has $F^{\mathrm{\prime }}=u_x-i{u}_y$, so $g:=u_x-i{u}_y$ is holomorphic on the punctured disc. Because the point $z_0$ of step 1.1 was arbitrary in $0<\mathrm{\mid }{z}_0-a\mathrm{\mid }<R\mathrm{/}2$, step 2.1 yields $\mathrm{\mid }(z-a)g(z)\mathrm{\mid }\le 4e^{2M}$ throughout $0<\mathrm{\mid }z-a\mathrm{\mid }<R\mathrm{/}2$. Therefore $h(z):=(z-a)g(z)$ is holomorphic on $0<\mathrm{\mid }z-a\mathrm{\mid }<R$ and bounded on the punctured neighbourhood $0<\mathrm{\mid }z-a\mathrm{\mid }<R\mathrm{/}2$, so [L3] extends it holomorphically across $a$; write $c:=h(a)$.

Because $h-c$ is holomorphic on $D(a,R)$ and vanishes at $a$, [L4] gives a holomorphic function $k$ on $D(a,R)$ with $h(z)-c=(z-a)k(z)$. Hence $g(z)=c\mathrm{/}(z-a)+k(z)$ on the punctured disc. Restricting to the real ray $z=a+t$ with $t>0$, the identity $u_x(a+t)=\mathrm{Re}g(a+t)=\mathrm{Re}(c)\mathrm{/}t+\mathrm{Re}k(a+t)$ and [L9] imply $u(a+t_0)-u(a+t)=\mathrm{Re}(c)\log \text{\hspace{0.17em}⁣}\frac{t_0}{t}+{\int }_t^{t_0}\mathrm{Re}k(a+s)ds\mathrm{.}$ Because $u$ is bounded and $k$ is continuous near $a$, the logarithmic term cannot diverge; hence $\mathrm{Re}(c)=0$.

For $0<r<R$, parameterize the circle by ${\gamma }_r(t)=a+r{e}^{it}$. Since $u\circ {\gamma }_r$ is $C^1$ and periodic, $0=u({\gamma }_r(2\pi ))-u({\gamma }_r(0))={\int }_0^{2\pi }\frac{d}{dt}u({\gamma }_r(t))dt=\mathrm{Re}{\int }_{{\gamma }_r}g(z)dz\mathrm{.}$ Writing $g=c\mathrm{/}(z-a)+k(z)$, the holomorphic function $k$ has a primitive on $D(a,R)$ by [L5], so its circle integral is $0$; therefore $0=\mathrm{Re}(2\pi ic)$, which means $\mathrm{Im}(c)=0$. Combined with step 4.1, this gives $c=0$.

Step 5.1 shows $h(a)=0$, so [L4] gives a holomorphic extension $g$ on $D(a,R)$ with $h(z)=(z-a)g(z)$. Since the disc is star-shaped, [L5] gives a primitive $G$ of $g$ on $D(a,R)$. On the punctured disc, $H:=u-\mathrm{Re}G$ has continuous first partials with $H_x=H_y=0$, so [L7] makes $H$ holomorphic there; being real-valued, $H$ is constant by [L7]. Therefore some real constant $b$ makes $u=\mathrm{Re}G+b$ on the punctured disc. Since $G$ is holomorphic on the full disc, [L8] makes $U:=\mathrm{Re}G+b$ harmonic on $D(a,R)$, and this $U$ extends $u$.