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.

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