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How can I describe the set of all functions which satisfy the constraint. My first thought was using Cauchy-Riemann, but I am not certain if the inequality is going to be preserved when I take the partial derivative of $u(x,y)$ and $v(x,y)$

Johann Martens
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3 Answers3

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An entire function whose real part is bounded must be constant. – and the same is true for entire functions bounded imaginary part.

Now look at $g(z) = f(z) - z$.

Martin R
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  • But we are not given that $\text{ Im }f(z)-y$ is bounded; we are given that it is bounded above. – zhw. Sep 27 '20 at 02:22
  • @zhw.: If you look at the referenced Q&A then you'll see that it suffices that the real (or imaginary) part is bounded above. The question is not clear about that, but the answers are. – Martin R Sep 27 '20 at 05:13
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Martin R.’s answer is very good, but I wanted to provide a different argument, which works for a slightly larger class of functions.

Let $g(x,y)=-v(x,y)+y+c$. Then $g$ is a harmonic real nonnegative function on the plane. We'll show this implies $g$ is constant, thus $v(x,y)=y+C$ and thus $f$ is a translation.

Take $x,y \in \mathbb{C}$. Take a large $R > 1+2|x-y|$. As $g$ is harmonic, $g(x) = \frac{1}{\pi R^2}\int_{D(x,R)}{g} \geq \frac{1}{\pi R^2}\int_{D(y,R-|x-y|)}{g} = \frac{(R-|x-y|)^2}{R^2}g(y)$. Let $R$ go to infinity, thus $g(x) \geq g(y)$ and the conclusion follows.

Aphelli
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  • There is an easier proof that if $u$ is nonnegative and harmonic on $\mathbb C,$ then it is constant: Let $v$ be a harmonic conjugate of $u.$ Then $\exp(-(u+iv))$ is bounded and entire, hence is constant by Liouville. That implies $u$ is constant.

    That works because we're in $\mathbb C=\mathbb R^2.$ In higher dimensions, something like your proof seems to be the way to go. Is that your own proof? It's nice–simpler than the one I know.

     – zhw. Sep 27 '20 at 02:32
  • @zhw: What you’re suggesting is exactly Martin R’s answer. This isn’t really my own proof, I probably remembered this one from some lectures on PDEs. – Aphelli Sep 27 '20 at 07:49
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Let $f(z)=u(z)+iv(z), z=x+iy.$ Then $v(z)\le y+c.$ Thus

$$\left|\frac{\exp(-i(u(z)+iv(z))}{\exp(-i(x+iy))}\right| = \frac{e^v(z)}{e^{y}} = e^{v(z)-y}\le e^c.$$

By Liouville, the function inside the absolute values is constant–in fact a nonzero constant, which we can write as $e^b.$ We thus have $\exp(-if(z))= \exp(b-iz).$ From that we get $f(z) = z+\text { constant}.$

zhw.
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