Assumption on $u$ harmonic in disc making $u$ unique with particular (discontinuous) boundary data From an old qualifier:

Let $$J_1 = \{e^{i\theta}: 0 < \theta < \frac\pi2\}, \,\,J_2 =
 \{e^{i\theta}:\frac\pi2<\theta<  \pi\},\,\, J_3 =
 \{e^{i\theta}:\pi<\theta < 2\pi\}$$ be three arcs and $D=\{z\in
 \mathbb{C}: |z|<1\}.$ Find a harmonic function $u$ on $D$ with
  $u|_{J_1}=1, u|_{J_2}=2, u|_{J_3}=4$. Is $u$ unique? If not, give some
  conditions such that it is unique. Make your assumption as sharp as
  possible.

Ideas: First, I believe that $$u(r,\theta) = \int_0^{\pi/2} \frac{1-r^2}{1-2r\cos\theta + r^2}+ 2\int_{\pi/2}^{\pi}\frac{1-r^2}{1-2r\cos\theta + r^2} + 4\int_{\pi}^{2\pi}\frac{1-r^2}{1-2r\cos\theta + r^2}$$
is such a solution. Perhaps there is a way to write it more explicitly? (I'm not sure if the question wants this).
Regarding uniqueness, there is a very helpful Stack response here which makes the claims 


*

*$u$ is not unique.

*$u$ is the unique such function such that
$\limsup_{z\to\zeta}|u(z)|\leq 4$ for all $\zeta \in \partial D$


Before I read the answer I had thought $u$ was unique, and even Wikipedia's article would seem to suggest that the function is unique. In discussing boundary data which is $L^1$, they say: 

By the maximum principle, u is the only such harmonic function on D.

This seems like a glaring error; perhaps they mean if the boundary data is continuous. (Unless someone can correct me.)
Would $\limsup_{z\to\zeta}u(z)\leq 4$ for all $\zeta \in \partial D$ be the assumption you would give to make $u$ unique? Are there any others? Can anyone sketch a proof of claim 2?
 A: Your solution is correct, and it can also be written using a Fourier series by starting with the Fourier representation of the Poisson kernel:
$$
   \frac{1-r^{2}}{1-2r\cos\theta+r^{2}}=\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}.
$$
Your solution is explicit, too. Your solution is definitely bounded, and has non-tangential radial limits which converge uniformly for $\theta$ in any compact subset of $B=(0,\pi/2)\cup(\pi/2,\pi)\cup(\pi,2\pi)$. That can be seen from the Poisson representation. There is a normal derivative of $u$ on the part of the boundary corresponding to $B$, which can also be seen from the Poisson representation after integrating by parts in $\theta$.
If you have a harmonic function $\psi(r,\theta)$ which is uniformly bounded on the open unit disk, then, as it turns out, $\psi$ must have radial limits for a.e. $\theta$; so the bounded convergence theorem allows you to represent $\psi$ using the Poisson integral representation on any circle of radius $r < 1$, and then take a limit using Lebesgue's bounded convergence theorem in order to obtain the representation of $\psi$ as the Poisson integral of its a.e. radial boundary function. So, boundedness definitely gives you uniqueness. The Poisson kernel itself shows you why you need boundedness. This is because
$$
                  v(r,\theta) = \frac{1-r^{2}}{1-2r\cos\theta+r^{2}}
$$
is a harmonic function with 0 boundary limits everywhere on the boundary, except where $\theta=0$. For $\theta=0$, $v(r,0)=\frac{1+r}{1-r}$ does not have boundary limit as $r\uparrow 1$.
A: I will describe two different approaches to proving the uniqueness of harmonic functions with given boundary values.
Functional Analysis approach
Theorem 1. Suppose that $u$ is a harmonic function on $D$ and there exists $p>1$ such that 
$$\sup_{0<r<1} \int_0^{2\pi} |u(r,\theta)|^p\,d\theta<\infty\tag{1}$$
Then there exists an $L^p$ function $f$ on $\partial D$ such that $u$ is the Poisson integral of $f$. 
Also, the radial limits of $u$ are equal to $f$ a.e. 
As a consequence of Theorem 1, two harmonic functions that satisfy (1) and share boundary values a.e. must be equal.
Proof of Theorem 1. Let $u_r$ be the function $\theta\mapsto u(r,\theta)$ defined on the interval  $ (0,2\pi)$. By the Banach-Alaoglu theorem, closed balls in $L^p$ are weakly compact. Therefore, there is a  sequence $r_k\to 1$ such that $u_{r_k}$ converge weakly to some $f\in L^p(0,2\pi)$. 
Fix a point with polar coordinates $(\rho,\phi)$  in $D$. For all sufficiently large $k$ we have $r_k>\rho$, hence
by the Poisson integral formula (used on the disk of radius $r_k$),
$$u(\rho,\phi) = 
\frac{1}{2\pi} \int_0^{2\pi} \frac{r_k^2-\rho^2}{r_k^2-2r_k\rho \cos(\theta-\phi)+\rho^2}u_{r_k}(\theta) \,d\theta
\tag{2}$$
We can pass to the limit $k\to \infty$ under the integral sign in (2), because $u_{r_k}$ converge weakly in $L^p$, and
the other factor converges strongly in the dual space $L^q$ (indeed, it converges uniformly). Thus,
$$u(\rho,\phi) = 
\frac{1}{2\pi} \int_0^{2\pi} \frac{1-\rho^2}{1-2\rho \cos(\theta-\phi)+\rho^2}f(\theta) \,d\theta
\tag{3}$$
proving the first claim of the theorem. 
Once the representation (3) is available, the claim about radial limits follows in a standard way: 
as $\rho\to 1$, most of the integral comes from a small interval around $\phi$, which implies convergence whenever
$\phi$ is a Lebesgue point of $f$. $\quad \Box$ 

Potential Theory  approach
Let's first state the maximum principle in a convenient form.
Maximum principle: Suppose $u$ is  harmonic in $D$. If for all $\zeta\in\partial D$
we have $\limsup_{z\to\zeta } u(z)\le M$, then $u\le M$ in $D$.
Proof. Let  $u_\epsilon(z)=u(z)+\epsilon |z|^2 $. The function $u_\epsilon$ does not have local maxima by the 2nd derivative test. Let $(z_k)$ be a  sequence such that $u_\epsilon(z_k)\to \sup_D u_\epsilon$. 
Then $z_k$ approach the boundary of $D$. It follows that  $\sup_D u_\epsilon \le M+ \epsilon$. 
Let $\epsilon\to 0$. $\quad \Box$
We can generalize the maximum principle with one weird trick. 
Theorem 2. (Generalized maximum principle) 
Suppose that $u$ is harmonic in $D$ and bounded above. If $\limsup_{z\to \zeta} u(z)\le M$
for all $\zeta\in \partial D$, with finitely many exceptions, then $u\le M$ in $D$.
As a consequence of Theorem 2, two harmonic functions that are bounded in $D$ and share boundary values on a cofinite subset of $\partial D$  must be equal.
Proof of Theorem 2. Let $\zeta_1,\dots,\zeta_n$ be the exceptional points. Define 
$$u_\epsilon(z)=u(z)+\epsilon\sum_{j=1}^n \log|z-\zeta_j|$$
Observe that $u_\epsilon\le u+\epsilon\, n \log \operatorname{diam} D$. 
The function $u_\epsilon$ is harmonic in $D$ and tends to $-\infty$ as $z\to \zeta_j$. Hence, 
for all $\zeta\in\partial D$ we have $$\limsup_{z\to\zeta } u_\epsilon(z)\le M+
\epsilon\, n \log \operatorname{diam} D\tag{4}$$ 
By the maximum principle $u_\epsilon\le M+ \epsilon\, n \log \operatorname{diam} D$ in $D$. 
Let $\epsilon\to 0$. $\quad \Box$

Remarks


*

*The potential-theoretic approach required stronger assumptions on $u$, but with it we did not actually
need $D$ to be a disk; it could be any bounded domain. 

*The sum $\sum_{j=1}^n \log|z-\zeta_j|$ is the logarithmic potential of a finite sum of point masses. 
Using more general potentials for the same purpose, one can prove a stronger form of Theorem 2, in which the exceptional set is allowed to be any polar set. See Ransford's book Potential theory in the complex plane. 
