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I was thinking about this question - If $f$ is analytic in $D$ and $z_{0}$ is a point in $D$ such that $|f(z)| \leq |f(z_{0})|$ holds for all $z \in D$ ,then show that $f$ is constant in $D$.

My attempt -

For a fixed $f$ , $|f(z_{0})|$ will be a constant say $c$ , now $|f(z)|< c $ $\forall z$ implies that $f$ is constant by Liouville's Theorem, is this approach correct.?

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    Liouville's theorem says something about entire functions. If $D = \mathbb{C}$, that works, but not in general. – 2017-02-13
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    Yes,Liouville's theorem tells when we are dealing with entire functions on the whole complex plane.so how to approach this ? – 2017-02-13
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    What properties of holomorphic functions do you know? Maximum modulus principle (most likely not yet ;-), open mapping theorem? – 2017-02-13
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    No,not yet! , i think any other idea might help? – 2017-02-13
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    There are lots of ways. What things have you to work with? What stuff do you know about holomorphic functions? – 2017-02-13
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    C-R equations , reflexion principles,.. – 2017-02-13
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53562/discussion-between-baymax-and-daniel-fischer). – 2017-02-13

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As mentioned in the comments, Liouville's Theorem only applies in the case of $D=\mathbb{C}$.

If there are no restrictions on $D$, then it's easy to see that this "theorem" is false: Let $f(z)=z$ and let $D=\{0,1\}$. Then the inequality holds, but $f$ is non-constant. Now let's look at a more interesting counterexample: $f(z)=z^2$ and $D=\{w:\Re(z)\in[-1,1]\land\Im(z)=0\}$.

Both of these sets have the property that they don't contain any balls of radius $>0$. Can you make a counterexample that does contain a ball of radius $>0$?

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    Yes,but in the question there is no specific domain given , but still very nice observation and counterexamples! – 2017-02-14
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    @BAYMAX if no specific domain is given than the theorem then, as these examples show, it is not true. – 2017-02-14
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    yes , i will try to figure it out and will tell.. – 2017-02-14
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    What happens if D is simple,open,connected...?@Stella Biderman – 2017-02-16