
> 
> Let $f : U \to \Bbb C$ be a holomorphic function in a neighborhood of 0 (or a
>  polynomial). Is it true that for any integer $k \geq 1$,
>  $$ |f^{(k)}(0)| \leq \|D^k\_0 |f|\| \left( = \max\_{|u\_1|=\dotsb = |u\_k|=1} \| D^k\_0 |f| (u\_1,\dotsc,u\_k)\| \right)? $$ where $D^k\_0 |f|$ is the
>  $k$th derivative at zero of the map $|f| : \Bbb C \simeq \Bbb R^2 \to \Bbb R$.
> 
> 
> 


For $k=1$ there is an equality. For $k=2$, the inequality holds true (I computed it, but I have no easy way to prove it).


It holds that $\| D^k\_0 \log |f|\| = \left|\frac{\partial^k}{\partial z^k} \log f(z)\right|$ (because $\log |f| = \Re(\log f)$) but I can't obtain anything out of that.


I tried also to exploit the formula $f^{(k)}(0) = \frac{1}{2\pi i}\oint z^{-k-1} f(z) dz$ but without success.


Any idea?



For a holomorphic function $f$, is $|f^{(k)}(0)| \leq \| D_0^k |f| \|$?
It is true for $k=1,2$ but not for $k\ge 3$.


Write $f=g^2$ near $0$ and let $g(z)=\sum\_{m\ge 0}a\_m z^m$. Then $|f|(z)=\overline{g(z)}g(z)$, so we want (up to $k!$ on both sides) that 
$$
\left|\sum\_{0\le m\le k/2}a\_ma\_{k-m}\right|\le \max\_u\left|\Re\sum\_{0\le m\le k/2}\bar a\_ma\_{k-m}P\_{m}(u)\right|
$$
where $P\_{m}(u)$ is the average of all products of $u\_1,\dots,u\_k$ with exactly $m$ conjugation bars over $u$'s and if $k=2m$, then the last term in the sum should be taken with the coefficient $1/2$.


Now notice that if we multiply $a\_m$ and $a\_{k-m}$ by any $\zeta\_m\in\mathbb T$, then we can rotate the terms on the LHS in any way we want without changing the ones on the RHS. Also we can make $\bar a\_ma\_{k-m}$ any numbers we want except $a\_m\bar a\_m$ should be positive when $k$ is even and $m=k/2$.
Thus, the desired inequality is equivalent to
$$
\sum\_{0\le m\le k/2}|b\_m|\le \max\left|\Re\sum\_{0\le m\le k/2}b\_mP\_m(u)\right|
$$
However, the only way to achieve this is to align all terms in all $b\_mP\_m(u)$ with either $+1$ or $-1$ simultaneously.


If $k=1$, then $P\_0(u)=u\_1$, so it is not a problem.
If $k=2$, then $P\_0(u)=u\_1u\_2$ and $P\_1(u)=\frac 12(u\_1\bar u\_2+u\_2\bar u\_1)$ and we know that $b\_1$ is already aligned with $+1$, so we can align $P\_1$ with $+1$ by choosing $u\_1=u\_2=\zeta\in\mathbb T$ and we still have enough freedom to rotate $u\_1u\_2$ to anything we want.


However, when $k>2$, we have arbitrary $b\_0$ and $b\_1$ and aligning all terms in $P\_1$ between themselves requires that all $u\_k=\pm\zeta$ for some $\zeta\in \mathbb T$. Thus, our sum starts with $\pm (b\_0\zeta^k+b\_1\zeta^{k-2})$ and for generic $b\_0,b\_1$ we just don't have enough freedom to bring even these two terms to the real line simultaneously. 


