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The following 2 problems are past exit exam problems for my major. I see that they're worded differently but are asking me to do the same thing. Not sure how they differ much I'd appreciate if anyone filled me in on that.

Can I prove the following 2 problems in the manner done from this post from user17762?: Why: A holomorphic function with constant magnitude must be constant.

1) Suppose $u(x,y)$ is a real valued function which is harmonic on the whole plane such that $|u(x,y)| \le 17$ for every $z=x+iy$ in $\mathbb{C}$. Show that $u$ must be constant

2) Suppose $u: \mathbb{R^2} \to \mathbb{R}$ is harmonic on the whole plane and that $u(x,y)<0$ for all $(x,y)$ in $\mathbb{R^2}$. Show that $u$ must be constant.

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User69127
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1 Answers1

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For these problems, an approach using the Cauchy-Riemann equations isn't the most convenient. Liouville's theorem is far more straightforward.

An entire (real-valued) harmonic function is the real part of an entire holomorphic function, let's call that $f$.

Both conditions, $\lvert u(x,y)\rvert \leqslant 17$, and $u(x,y) < 0$ imply that

$$\operatorname{Re} f \leqslant M$$

on all of $\mathbb{C}$ for some $M < +\infty$. And that implies that

$$\left\lvert e^{f(z)}\right\rvert \leqslant e^M$$

on all of $\mathbb{C}$, so by Liouville's theorem $e^f$ is constant. Taking the logarithm (which works, since $\mathbb{C}$ is simply connected) we see that $f$ is constant. But then of course $u = \operatorname{Re} f$ is constant too.

Daniel Fischer
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  • Wow this is great I'd rather use this approach than the cauchy approach if I'm asked a similar problem on the exam. Will this also work if $f$ is entire and $Re(f(z))=Im(f(z))$ for every $z$ in $\mathbb{C}$? – User69127 Feb 08 '14 at 21:17
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    If $u(z) = \operatorname{Im} f(z)$, it works pretty much the same, you use $e^{i f(z)}$ then ($e^{-f(z)}$ or $e^{-i f(z)}$ if the real or imaginary part is bounded below instead of above). If you have something like $\operatorname{Re} f(z) = \operatorname{Im} f(z)$ - or more generally that the image of $f$ is contained in a straight line - the Cauchy-Riemann approach is simpler, you can then always multiply with a constant (and add another constant maybe) to have a function with constant real part, or a real-valued function (whichever you prefer). Then CR immediately produces $f'\equiv 0$. – Daniel Fischer Feb 08 '14 at 21:20
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    But, an even easier way, if you have it, is the theorem that a non-constant holomorphic function has an open image, so if the image is contained in a set with empty interior (a straight line, a circle, any piecewise differentiable curve), you know it's a constant function. – Daniel Fischer Feb 08 '14 at 21:24
  • ^^^that's great! – User69127 Feb 08 '14 at 21:25
  • Daniel, I forgot to ask - why did you take both sides of the inequality to a power with base $e$? – User69127 Mar 24 '14 at 07:40
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    To get something we can directly apply Liouville's theorem to. If you have Picard's theorem, or the weaker theorem that an entire function whose range is not dense in the plane (which is a corollary of Liouville's theorem, composition with a Möbius transformation yields a bounded function), one can invoke that directly. But since I don't know if these are already known, I make a small transformation to get a function Liouville's theorem says is constant. – Daniel Fischer Mar 24 '14 at 10:19
  • Yes I have Liouville's theorem. Can't I just say that if $f(z)$ happens to be entire, then the conclusion immediately follows from Liouville's theorem? Therefore $u=Re f$ is constant too? – User69127 Mar 28 '14 at 06:54
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    Sorry, I don't understand. At any point where you have a theorem that tells you $u$ or $f$ is constant, you can stop. If you have the theorem that an entire harmonic function bounded above or below is constant, you can appeal to that directly. If not, you have to infer it from something else. Considering an entire holomorphic function with $\operatorname{Re} f = u$ is a standard way. If you know that a holomorphic function whose real part is bounded above or below is constant, you can stop there. If not, you need a transformation that gets it into a shape where you have a theorem telling you – Daniel Fischer Mar 28 '14 at 11:26
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    that it's constant. Liouville's theorem that a bounded entire holomorphic function is constant is a natural candidate. So you need a transformation that gives you a bounded function. $z\mapsto e^{f(z)}$ or $z \mapsto \frac{1}{f(z) - 18}$ for example do that. From the constancy of the transformed function, you can infer that $f$ is constant, and hence $u$. – Daniel Fischer Mar 28 '14 at 11:29
  • That clears it up! I see! – User69127 Mar 28 '14 at 21:45
  • Daniel, forgive me again for reverting to this question again. I've been asked to prove functions are constant in different manners. I really like the way you have taught me by applying a function. I'm going to ask just one last question on this topic hope you don't mind. If $Re(f(z))=Im(f(z))$ and want to prove $f$ is constant using $e^{if(z)}$. Then $|e^{if(z)}|=e^{Re(if)}=e^{-Im(f(z))}$ Do you do it in this manner? How do I get $f$ to be constant from here though? – User69127 Apr 09 '14 at 06:32
  • In the situation $\operatorname{Re} f(z) = \operatorname{Im} f(z)$, you can a) appeal to the open mapping principle, a non-constant holomorphic function (on a domain) is an open mapping, so if the image is contained in a line (which has empty interior) the function must be constant; b) use the Cauchy-Riemann equations, which then yield that the real and imaginary parts must be constant; c) multiply by a suitable constant to achieve a constant real part, here $(1+i)(u+iv) = (u-v) + i(u+v)$ makes the real part of $(1+i)f$ constantly $0$. – Daniel Fischer Apr 09 '14 at 07:56