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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.

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    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.2012-12-30

3 Answers 3

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Let $f(z)$ be entire and 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

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

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    Right, $h(z)$ is constant, and hence $f(z)$ is constant.2012-12-31
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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

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    yes,Re(f) which is harmonic must also assume negative values2012-12-31
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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 $