# Understanding Maximum Principle

Understanding Maximum Principle

One of the point of that theorem is: If $f$ is analytic on the open connected set $\Omega$ and $|f|$ assumes a local maximum at some point in $\Omega$, then $f$ is constant on $\Omega$

What does ''local maximum'' here mean, if you take any closed bounded set in $\Omega$, then, how can $f$ not attain a maximum ?

• A little bit outside your context, but I think it illustrates your last question better than a more correct example: if you have the function $f:\Bbb R \to \Bbb R$ given by $f(x) = x$, and the open, bounded set $U = (0, 1) \subseteq \Bbb R$, then $f$ has no maxima or minima at all on $U$. – Arthur Apr 24 '15 at 13:07
• @Arthur that's why I asked what ''local means'', can you not just take any closed subset of $(0,1)$ ? – ketum Apr 24 '15 at 13:09
• Local (almost) always means open neighbourhoods. – Arthur Apr 24 '15 at 13:09

First of all, $f$ is complex valued, so it doesn't really make sense to talk about a (local) maximum of $f$. It's $|f|$ that the theorem says something about.
And yes, $|f|$ is continuous, so if $f$ is analytic on the open (bounded) set $\Omega$ and continuous on $\bar\Omega$, the maximum modulus principle implies that that maximum of $|f|$ must be attained on the boundary $\partial\Omega$.
If $f(z)$ assumes a local maximum, then it satisfies $|f(z_0)|\geq|f(z)|$ for some open $U\subseteq \Omega$. This then means that $f(z)$ is a constant function.
The theorem isn't saying that $f(z)$ won't attain a maximum on some closed subset of $\Omega$, just that if it assumes a local maximum anywhere inside $\Omega$ (as stated above), then the function is necessarily constant.