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It isn't too hard to show that if $f:\mathbb{C}\to\mathbb{C}$ holomorphic everywhere (entire) and $\Re (f)$ is bounded, then $f$ is constant: it suffices to consider $\exp f$, which is entire, and by $|\exp f|=\exp\Re (f)$ bounded and therefore constant. That question has been asked on this site before.

I am now asked to show that this holds if $f$ is only defined on $U\subseteq\mathbb{C}$, $U$ path-connected and non empty, and the bound is attained in the interior of $U$. Because of the domain I can't use Liouville anymore, so I am not sure how to proceed.

Is it true that there exists a differentiable bijection $\phi: U\to\mathbb{C}$ for instance? That would resolve the issue... Otherwise some help would be welcome.

user118224
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    Every proper simply connected open subset of the complex space is biholomorphic to the unit disk. This is Riemann's mapping theorem. This fails if the space is not simply connected, of course. – Pedro Feb 28 '16 at 01:59
  • @PedroTamaroff So does that mean the $\phi$ I mentioned definitely exists? – user118224 Feb 28 '16 at 02:01
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    @Pedro : the riemann's mapping theorem is that it is biholomorphic to the unit disk, which is biholomorphic to the half plane $Re(z) > 0$ https://en.wikipedia.org/wiki/Unit_disk#The_open_unit_disk.2C_the_plane.2C_and_the_upper_half-plane – reuns Feb 28 '16 at 02:03
  • with $f(z)$ entire, $Re(f)$ is bounded on any finite subset of $\mathbb{C}$, so we need that $U$ contains $\infty$, and because it has been shown there exists an entire function bounded on every path from $0$ to $\infty$ except one (and on the neighborhood of that path), I don't see how you'll prove that $f$ is constant except if $\overline{U} = \mathbb{C}$ – reuns Feb 28 '16 at 02:12
  • No, the $\phi$ doesn't exist. Riemann's mapping theorem only works if your simply connected domain is not all of $\Bbb C$. In fact a holomorphic map from $\Bbb C$ into the unit disc is by definition bounded, hence constant, hence non-injective. Hence Pedro's comment is incorrect. – shalop Feb 28 '16 at 02:12
  • by the way, for that reason, that there exists such an entire function, bounded everywhere except on the neighborhood of one path to $\infty$ should prove that there is no biholomorphic map $\mathbb{C} \to U$ ? – reuns Feb 28 '16 at 02:19
  • did you add that the maximum is attained at $a$ inside U just now ? so yes because $f(a+z) = f(a) + c z^n + \mathcal{0}(|z|^{n+1})$ it shows that $c = 0$ hence $f$ is constant (this is how is proved the maximum modulus principle) – reuns Feb 28 '16 at 02:47

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Suppose $f= u+iv:U\to \mathbb C$ is holomorphic, with $U$ as you describe, with $u$ bounded in $U.$ If there exists $a\in U$ such that $\sup_U u = u(a),$ then $f$ is constant in $U.$

The proof is simple: $|e^{f}|$ achieves a global max at $a,$ hence $e^f$ is constant in $U$ by the maximum modulus principle. Once you know that, it's a nice after-dinner walk to deduce that $f$ is constant in $U.$

zhw.
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I think that the result is not true.

For example, if $U$ is the unit open disk centered at the origin, and $f(z)=\exp(z)$, then we have that $\exp(z)=\exp(x)\cos(y)+i\exp(x)\sin(y)$, for $z=x+iy$. We have that the absolute value of the real part, $|\exp(x)\cos(y)|$ is bounded by $\exp(1)$ for all $z\in U$, but we know that $\exp(z)$ is not constant in $U$.

If the bound is attained in the interior of $U$, the you can use the Maximum modulus principle as Sangchul Lee says.

Larara
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