## Rigidity of upper half plane

Thm: Suppose ${\mathbb{R}^n_+}$ is the upper half plane. ${u\in C^0(\overline{\mathbb{R}^n_+})}$ is non-negative harmonic function in ${\mathbb{R}^n_+}$ and ${u\equiv 0}$ on ${\partial \mathbb{R}^n_+}$. Then ${u=ax_n}$ for some ${a\geq 0}$. Note that there is no requirement on the growth of $u$ at infinity.

Lemma: Let ${u}$, ${v}$ ba positive solution to ${Lw=0}$ in ${B^+_1}$ continuously vanishing on ${\{x_n=0\}}$ with

$\displaystyle u(\frac{1}{2}x_n)=v(\frac{1}{2}x_n)=1$

Then, in ${\bar{B}^+_{1/2}}$,

$\displaystyle \frac{v}{u}\text{ is of class }C^\alpha$

and

$\displaystyle ||\frac{v}{u}||_{L^\infty(B^+_{1/2})},\quad ||\frac{v}{u}||_{C^\alpha(B^+_{1/2})}\leq C(n,\lambda)$

Remark: Refer to Caffarelli’s book, A Geometric Approach to Free Boundary Problems, this lemma is related to boundary harnack inequality or Carleson estimate.

Proof: For ${x=(x_1,\cdots,x_{n})}$, denote ${x'=(x_1,\cdots,x_{n-1})}$, then ${x=(x',x_n)}$. Fix ${\forall \,t>0}$, define ${\tilde{u}(x)=u(2xt)/u(0,t)}$, ${\tilde{v}(x)=2x_n}$ for ${x\in B^+_1}$. Apply the lemma, we can get

$\displaystyle \tilde{u}(x)\leq C\tilde{v}(x)\quad \forall \, x\in B^+_{1/2}$

$\displaystyle C^{-1}\tilde{v}(x)\leq \tilde{u}(x)\quad \forall \, x\in B^+_{1/2}$

where ${C=C(n)\geq 1}$. These are equivalent to

$\displaystyle u(x)\leq Cx_n\frac{u(0,t)}{t} \quad \forall \,x=(x',x_n)\in B^+_{t}\quad (1)$

$\displaystyle C^{-1}x_n\frac{u(0,t)}{t}\leq u(x)\quad \forall \, x=(x',x_n)\in B^+_{t}\quad (2)$

Suppose

$\displaystyle \lim\limits_{\overline{t\rightarrow+\infty}}\frac{u(0,t)}{t}=a$

(i) If ${a=0}$, then ${\forall \epsilon>0}$, ${\exists\, t>0}$ large enough such that ${(1)}$ implies

$\displaystyle u(x)\leq \epsilon Cx_n\quad \forall \,x=(x',x_n)\in B^+_1\subset B^+_{t}\quad$

This means ${u\equiv 0}$ in ${B^+_1}$. Hence ${u\equiv 0}$ on ${\mathbb{R}^n_+}$.

(ii) If ${a>0}$, then ${\forall\, \epsilon>0}$, ${\exists\, t>r}$ large enough such that ${(2)}$ implies that

$\displaystyle u(x)\geq (a-\epsilon)C^{-1}x_n \quad \forall \,x\in B^+_r\subset B^+_{t}$

Since ${r,\epsilon>0}$ are arbitrary, then

$\displaystyle u(x)\geq aC^{-1}x_n\text{ in }\mathbb{R}^n_+$

Define

$\displaystyle v_1=u(x)-aC^{-1}x_n$

Then ${v_1}$ is non-negative and harmonic in ${\mathbb{R}^n_+}$, however

$\displaystyle \lim\limits_{\overline{t\rightarrow+\infty}}\frac{v_1(0,t)}{t}=a(1-C^{-1})$

Applying the above lemma and repeating the above procedure, we get

$\displaystyle v_1\geq a(1-C^{-1})C^{-1}x_n$

So we construct

$\displaystyle v_2=v_1-a(1-C^{-1})C^{-1}x_n$

Then ${v_2}$ is non-negative and harmonic in ${\mathbb{R}^n_+}$ with ${\lim\limits_{\overline{t\rightarrow+\infty}}\frac{v_2(0,t)}{t}=a(1-C^{-1})^{2}}$. Inductively, we construct

$\displaystyle v_k=v_{k-1}-a(1-C^{-1})^{k-1}C^{-1}x_n=u-\sum_{i=0}^{k-1}a(1-C^{-1})^{i}C^{-1}x_n$

Then ${v_\infty=u-ax_n}$ is non-negative and harmonic in ${\mathbb{R}^n_+}$ with

$\displaystyle \lim\limits_{\overline{t\rightarrow+\infty}}\frac{v_\infty(0,t)}{t}=0$

From (i), we know $v_\infty\equiv 0$, that is $u=ax_n$.

Remark: This problem was presented to me by Tianling Jin. As he said, there should have a very elementary proof, which do not require such advanced theorem.