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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 but I do not know where to start to prove that there is no harmonic function that is positive everywhere.

Any help/ 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. – WimC Dec 30 '12 at 8:29

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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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this is elementary! – K.Ghosh Dec 30 '12 at 8:55
You mean if $u(x,y)>0$? – Pantelis Damianou 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
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Right, $h(z)$ is constant, and hence $f(z)$ is constant. – Isaac Solomon Dec 31 '12 at 1:08
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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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Do you mean that since $f(z)$ assumes the whole complex plane, $Re(f)$ which is harmonic should be negative for some z? – Yeonjoo Yoo Dec 30 '12 at 21:29
yes,Re(f) which is harmonic must also assume negative values – K.Ghosh Dec 31 '12 at 3:03

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