Qfyilyi

Let consider that $f(z)=sumlimits_{nge 0}a_nz^n$ in $mathcal{D}(O,R)$ with $0<Rle +infty$. If $mid fmid$ has a local minimum in $0$ then $fequiv 0$.

Is it linked to the maximum modulus principle ?

Thanks in advance !

The identity map $f : z mapsto z$ is such that $vert fvert$ has a local minimum at $0$. However $f$ is not the always vanishing map.

Hence the statement of the initial question is wrong.

The assertion appears to be false.

Consider

$f(z) = z^n, 1 le n in Bbb N; tag 1$

writing

$z = re^{itheta}, tag 2$

we have

$z^n = r^n e^{i n theta}, tag 3$

whence

$vert f(z) vert = vert z^n vert = r^n; tag 4$

clearly then,

$vert f(0) vert = 0 tag 5$

so $0$ is an absolute minimum of $vert f(z) vert$; however,

$f(z) ne 0, ; text{for} ; z ne 0. tag 6$

The correct version of the statement: if $|f|$ has a local minimum at zero, then either $f(0)=0$ or $f$ is constant.

Proof? $f$ is continuous; if it's not zero at zero, apply the maximum modulus principle to $frac1f$ on some suitably small neighborhood where it's bounded away from zero.