# $f$ is bounded iff $\Omega$ is bounded

Let $\Omega \subseteq \Bbb C$ be an open and connected set and let $f:\Omega \to \Bbb C$ be an analytic function .Pick out true ones:

1. $f$ is bounded if $\Omega$ is bounded

2.$f$ is bounded only if $\Omega$ is bounded

3.$f$ is bounded iff $\Omega$ is bounded.

For $3$ I took $\Omega=\Bbb C$ and I took the map $z\mapsto e^{iy}$ where $z=x+iy$. Then $|f(z)|=1$ which is bounded but $\Omega=\Bbb C$ is not.

Am I right? For $1,2$ I am not getting any examples .Any help will behelpful

• Hint: Consider $1/z$ and constant functions. For 3 notice that this is the combination of both 1 and 2. Your counterexample is not holomorphic and thus does not work as an counterexample (I assume that you mean complex analytic functions). – Jendrik Stelzner Jan 22 '16 at 6:02
• Regarding your counterexample you might consider Liouville's theorem. – copper.hat Jan 22 '16 at 6:16
• @JendrikStelzner; I have got your point ;Thanks – Learnmore Jan 22 '16 at 9:06

Unfortunately, no: you are not correct. Liouville's theorem shows that there is no bounded non-constant entire function on $\mathbb{C}$; but to be explicit, $e^{iy}$ is not an analytic function. You can check that the Cauchy-Riemann equations aren't satisfied. It's worth pointing out that in general, a non-constant function that depends only on the real or imaginary part of $z$ will never be analytic.

Regarding the other points, notice that if you take a function with a pole, such as $f(z) = z^{-1}$, you can easily construct unbounded analytic functions on bounded domains. This shows 1 is false.

2 has a trivial counterexample.

• There are many counterexamples to 2), not just the trivial ones. (You don't have to take $\Omega=\mathbb{C}$.) – mrf Jan 22 '16 at 6:50
• @mrf Quite right, I misspoke. Thanks. – user296602 Jan 22 '16 at 6:55

With the help of @Jendrik Slender I have got the following counterexamples:

1.$f(z)=\frac{1}{z};\Omega: |z-1|=3$

2.$f(z)=c;\Omega=\Bbb C\setminus \{0\}\subset \Bbb C$.

3.$f(z)=c;\Omega=\Bbb C\setminus \{0\}\subset \Bbb C$.

Hence none are correct.