# harmonic 1-form with bounded energy on a strip in $\mathbb{R}^2$

Let $S=[-a,a]\times[b,+\infty] \mod \{(-a,t) =(a,t) \mid t \in [b,+\infty]\}$ be a strip in $\mathbb{R}^2$ with identified sides. Let $w$ be a real harmonic 1-form on $S$, which has a primitive $f$ on $S$, i.e. $df=w$.
Suppose that $f$ has bounded energy $E(f)$ on $S$, i.e.
$E(f)= \int \limits_{S} \|df\|^2 < \infty$,
can we conclude that $\mathop {\lim } \limits_{t \to \infty} f(c,t)$ is independent of $c \in [-a,a]$ ?

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It is enough to consider the case $a=\pi$, $b=0$. Since $\Delta df=d\Delta f=0\;$ then $\Delta f=c$ in $S$. So function $g(x,y)=f(x,y)-cy^2/2\;$ is harmonic and $g(x,0)=f(x,0)\;$, $\partial_yg(x,0)=\partial_yf(x,0)\;$. Also we can assume that $f(\cdot,0)\in C^\infty(\mathbb T)$, otherwise it is enough to consider the strip $S\cap\lbrace y\ge1\rbrace$. Let $$f(x,0)=\frac{a_0}2+\sum_{k=1}^\infty a_k \cos kx+b_k\sin kx.$$ Then $$g(x,y)=\frac{a_0}2+\sum_{k=1}^\infty (a_k \cos kx+b_k\sin kx)e^{-ky}$$ is a solution of the Cauchy problem $\Delta g=0$ in $S$, $g(x,0)=f(x,0)\;$, $\partial_yg(x,0)=\partial_yf(x,0)\;$. As far as I remember it is unique in the class of bounded functions, though I can't pinpoint the reference. The integrability condition for $g$, namely $E(g)=\pi\sum_{k=1}^\infty k(a_k^2+b_k^2)<\infty\;$, is fulfilled because $g(\cdot,0)\in C^\infty(\mathbb T)$. Now we can conclude what the condition $\int\limits_{S} \|df\|^2 < \infty$ for $f=g+cy^2/2$ requires that $c=0$. So $f=g$ and the positive answer follows from the representation for $g$ above.

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Thank you for your answer ! Just one question. What is $\Delta$ for you in the first line of your answer ? Is it the Laplace operator or not ? – Bjoern Muetzel Aug 28 '11 at 20:59
@Bjoern yes, it is the Laplace operator. – Andrew Aug 29 '11 at 6:44
@ Andrew : If $w=df$ is harmonic, then $\Delta f =0$. This probably does not change your arguments anyway. – Bjoern Muetzel Aug 29 '11 at 14:50
@Bjoern it simplifies them a bit, no need to consider the function g. – Andrew Aug 29 '11 at 17:27
@ Andrew : I'm not so sure I can follow you here. $f(x,y)$ and $g(x,y)$ are two-dimensional functions, so the coefficients in the Fourier series can depend on $y$, i.e. $a_k(y)$ and $b_k(y)$. If $g(x,y)=f(x,y)−c\cdot \frac{y^2}{2}$, then only the $a_0(y)$ from $f(x,y)$ should change to $a_0(y)−c\cdot \frac{y^2}{2}$ in $g(x,y)$ by the integral formula for the Fourier coefficients. Or is the second $g(x,y)$ in your first text another function than $g(x,y)=f(x,y)−c\cdot \frac{y^2}{2}$ ? – Bjoern Muetzel Aug 30 '11 at 15:39

I don't think you need harmonicity of the form $w$ for your claim. It is actually true for any function $f$ with bounded energy. The point is that by Royden's decomposition the space of all such functions is the direct sum of the closure (in the Dirichlet norm) of compactly supported functions and of the space of harmonic functions of bounded energy. Then the claim follows from the fact that in your case the latter space consists of constants only.

EDIT: Unfortunately, Royden's decomposition, in spite of its importance, hasn't got a good coverage in the literature (although its discrete version is now quite popular in the theory of networks). You may look either at the original paper by Royden Harmonic functions on open Riemann surfaces, Trans. Amer. Math. Soc. 73 (1952), 40–94 or at the references from the paper by Ancona, Lyons and Peres Crossing estimates and convergence of Dirichlet functions along random walk and diffusion paths, Ann. Probab. 27 (1999), 970–989.

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OK. Could you give me a reference for that ? I'm not so familiar with this decomposition. – Bjoern Muetzel Aug 30 '11 at 15:22
@ R W : Thanks. I might cite these articles. – Bjoern Muetzel Sep 5 '11 at 21:11

This solution relies only on the fact that the energy $E(f)$ of $f$ on $S$ is bounded.

It is enough to consider the case $a=\pi$ and $b=0$. On $S$ we can expand $f$ into a Fourier series. We have :
$f(x,y):= \sum\limits_{n \in \mathbb{Z}} c_n(y) \exp(i n x),$ where $c_n(y) = \int \limits_{-\pi}^{\pi} f(x,y) \exp(-i n x) \,dx.$

As $f$ is real, we have $c_{-n}(y) = \overline{c_n(y)}$.

We have for the derivatives :

$\frac{\partial}{\partial x} f(x,y) = \sum\limits_{n \in \mathbb{Z} \backslash \{0\}} i n c_n(y) \exp(i n x) \, \,$ and $\, \, \frac{\partial }{\partial y} f(x,y) = \sum\limits_{n \in \mathbb{Z} } \frac{\partial}{\partial y} c_n(y) \exp(i n x).$

Now we obtain for $E_S(f)$ : $E_S(f) = \int \limits_{b}^{+\infty} {\int \limits_{-\pi}^{\pi} \left(\sum\limits_{n \in \mathbb{Z} \backslash \{0\}} i n c_n(y) \exp(i n x)\right)^2 + \left(\sum\limits_{n \in \mathbb{Z} } \frac{\partial}{\partial y} c_n(y) \exp(i n x)\right)^2 \,dx} \,dy$

Due to the orthogonality of the functions $\left(\exp(inx)\right)_{n\in \mathbb{Z}}$, this simplifies to $E_S(f) =2 \pi \cdot \int \limits_{b}^{+\infty} \sum\limits_{n \in \mathbb{N} \backslash \{0\}} \left( n^2 c_n(y)\cdot c_{-n}(y) + \frac{\partial}{\partial y} c_n(y) \frac{\partial}{\partial y} c_{-n}(y) \right) + \frac{\partial}{\partial y} c_0(y) \frac{\partial}{\partial y} c_{0}(y) \,dy$

As $c_n(y)\cdot c_{-n}(y) = \| c_n(y) \|^2$, we have : $E_S(f) =2 \pi \cdot \int \limits_{b}^{+\infty} \sum\limits_{n \in \mathbb{N} \backslash \{0\}} \left( n^2 \| c_n(y) \|^2 + \| \frac{\partial}{\partial y} c_n(y) \|^2 \right) + \| \frac{\partial}{\partial y} c_0(y) \|^2 \,dy < +\infty.$

ATTENTION : There is a mistake here. The following implication is not true in general.

This implies that $\mathop{\lim} \limits_{y \to +\infty} \sum\limits_{n \in \mathbb{N} \backslash \{0\}} n^2 \| c_n(y) \|^2 =0 \, \,$ and $\, \, \mathop{\lim} \limits_{y \to +\infty} \| \frac{\partial}{\partial y} c_0(y) \|^2 = 0. \,\,\,(*)$

Furthermore $f(x,y) -c_0(y) = \sum\limits_{n \in \mathbb{Z} \backslash \{ 0 \} } (n \cdot c_n(y)) \frac{\exp(i n x)}{n}$. Applying the Cauchy-Schwarz inequality we obtain :

$\|f(x,y) -c_0(y) \|^2 \leq \sum\limits_{n \in \mathbb{Z} \backslash \{ 0 \}} \| n \cdot c_n(y) \|^2 \cdot \sum\limits_{n \in \mathbb{Z} \backslash \{ 0 \}}\| \frac{\overline{\exp(i n x)}}{n} \|^2 = 2 \cdot \sum\limits_{n \in \mathbb{N} \backslash \{ 0 \}} n^2 \| c_n(y) \|^2 \cdot \sum\limits_{n \in \mathbb{N} \backslash \{ 0 \}} \frac{1}{n^2}.$

As $\sum \limits_{n \in \mathbb{N} \backslash \{ 0 \}} \frac{1}{n^2}$ is finite, we obtain together with $(*)$ that

$\mathop{\lim} \limits_{y \to +\infty} f(x,y) = \mathop{\lim} \limits_{y \to +\infty} c_0(y)$

and the result is independent of $x$. It follows furthermore from the second equation in $(*)$ that

$\mathop{\lim} \limits_{y \to +\infty} f(x,y) =c$.

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