4

This is one of the past qualifying exam problems that I was working on.

I know that, when we let $z=x+iy$, ${|z|}^2=x^2+y^2$ is not harmonic. I do not know where to start to prove that there is no harmonic function that is positive everywhere.

Any help or ideas idea will be really appreciated.

Thank you in advance.

Martin R
  • 113,040
Yeonjoo Yoo
  • 975
  • 2
    Constant functions are harmonic... An harmonic function in two variables is the real part of an entire holomorphic function. So try to construct some bounded entire function. – WimC Dec 30 '12 at 08:29

3 Answers3

8

Let $f(z)$ be entire with real part everywhere positive. Consider

$$h(z) = e^{-f(z)}$$

If $\Re f(z) > 0$, we have $-\Re f(z) < 0$, so $h(z)$ is bounded and entire. What can you conclude about $h(z)$, and hence about $f(z)$?

To go into a little more detail, note that

$$|h(z)| = e^{-\Re f(z)} < e^{0} < 1$$

Dan Asimov
  • 1,157
Elchanan Solomon
  • 30,807
  • this is elementary! – Koushik Dec 30 '12 at 08:55
  • You mean if $u(x,y)>0$? – PAD Dec 30 '12 at 11:55
  • Even if we rewrite $f$ as $f(x+iy)=u(x+iy)+iv(x+iy)$ and $u>0$, $h(z)=e^{-u(z)}e^{-iv(z)}=e^{-u(z)}(\cos(v(z))-i\sin(v(z)))$ and $Re(h(z))=e^{-u(z)}(\cos(v(z))$ can be negative depending on $v(z)$. Is it right? Would you give me more explanation? – Yeonjoo Yoo Dec 30 '12 at 21:21
  • @ Isaac Solomon: Thank you for the additional note, so since $h(z)$ is a bounded entire function, it is a constant function by Liouville's Theorem and so we conclude that the only harmonic function that is positive everywhere is a constant function. Is it correct? – Yeonjoo Yoo Dec 30 '12 at 22:06
  • 2
    Right, $h(z)$ is constant, and hence $f(z)$ is constant. – Elchanan Solomon Dec 31 '12 at 01:08
2

Continuing,a non-constant harmonic function is the real part of a non-constant entire function.so the real part must be positive.Little Picard Theorem: If a function is entire and non-constant, then the set of values that f(z) assumes is either the whole complex plane or the plane minus a single point. so we get a contradiction

Koushik
  • 4,472
0

MORE GENERAL AND SIMPLE ANSWER

Liouville Type (Theorem's) Suppose $u:\mathbb R^d\to \mathbb R$ be a nonegaitve harmonic function Then, $u$ is constant.

Proof $u\ge 0$ then $\inf_{\mathbb R^d} u<\infty$. Hence we set $$v= u-\inf_{\mathbb R^d} u$$ and

\begin{split} \begin{cases}\Delta v=0\\ v\ge 0\\ \inf_{\mathbb R^d} v =0\end{cases}\end{split}

Let $\varepsilon>0,$ and $y\in\mathbb R^d$ then there exists $x_\varepsilon$ such that,

$$ v(x_\varepsilon)\le \inf_{\mathbb R^d} v+\varepsilon$$

Let $R>\max(|x_\varepsilon|,|y|)+1$ therefore, $x_\varepsilon,y\in B_R(0)$ and

$$ B_R(y)\subset B_{3R}(x_\varepsilon) $$

By Mean value property (The mean value is true in any dimension for harmonics functions),

\begin{split} v(y) &=& \frac{1}{|B_R(y)|}\int_{B_R(y)} v(z) dz\\ &=& \frac{3^d}{|B_{3R}(x_\varepsilon)|}\int_{B_R(y)} v(z) dz \\&\le&\frac{3^d}{|B_{3R}(x_\varepsilon)|}\int_{B_{3R}(x_\varepsilon)} v(z) dz \\&= & 3^d v(x_\varepsilon) \end{split} This leads to,

$$ v(y)\le 3^d v(x_\varepsilon)\le 3^d(\inf_{\mathbb R^d} v+\varepsilon)~~\forall y\in \mathbb R^d$$

That is $$ \sup_{\mathbb R^d} v < 3^d \varepsilon $$

Since $\inf_{\mathbb R^d} v=0$ ~~ but $\varepsilon>0$ was arbitrarily chosen, letting $\varepsilon \to 0$ we get

$$ 0\le \sup_{\mathbb R^d} v \le 0$$

i.e $v = 0$ or $u= \inf_{\mathbb R^d} u $

Guy Fsone
  • 23,903