Problem. Let $U \subset \mathbb{C}$ be open and $[0,1] \subset U$. Assume $f(z)$ is holomorphic on $U$. Is it possible to find a constant $C$ (that depends on $f$) such that, for all $0 \leq a < b \leq 1$, $$ \int_a^b |f'(z)| dz \leq {C \over b-a} \int_a^b |f(z)| dz? $$
Motivation. If $f(z)$ is a polynomial of degree $k$ then the conclusion holds as follows. Denote $P_k$ the space of polynomials of degree $k$. The map $Dp = p'$ is linear and hence continuous on $P_k$ because $P_k$ is finite-dimensional. Regarding $P_k$ as a normed space with ${L^1([0,1])}$ norm, we thus conclude that $$ \lVert p' \rVert_{L^1(0,1)} \leq C \lVert p \rVert_{L^1(0,1)} \text{ for all } p \in P_k, $$ where $C$ is the norm of the linear operator $D$. But, setting $q(z) = p((z-a)/(b-a)) \in P_k$, we have the scalings $$ \lVert p' \rVert_{L^1(a,b)} = \lVert q' \rVert_{L^1(0,1)} \leq C \lVert q \rVert_{L^1(0,1)} = {C \over b-a} \lVert p \rVert_{L^1(a,b)}. $$
A related problem. The above inequality is known as a "reverse Sobolev inequality" for obvious reasons. A related problem is that of "reverse Hölder inequality", which reads $$ \sup_{z \in [a,b]} |f(z)| \leq {C \over b-a} \int_a^b |f(z)| \, dz. $$ For the space $P_k$, by a similar argument as under "Motivation" above, there is indeed a constant $C = C_k$ that works for all polynomials $f(z) = p(z)$ of degree $k$. Furthermore, if $f(z)$ is analytic on $U$ and not identically zero, and $[0,1] \subset U$, then there can only be finitely many roots $W = \{w_1,\ldots,w_k\}$ of $f(z)$ within some $\epsilon$ neighborhood of $[0,1]$ because analytic functions cannot have an accumulation of zeros on their domain. Take any polynomial whose roots are precisely $W$ (with the right multiplicities) and then $f(z)/p(z)$ is bounded below and above on $[0,1]$, so the estimate for polynomials leverages itself to the space of holomorphic functions.
I'm having trouble making this argument work with the reverse Sobolev inequality. In addition to interpolating the zeros of $f(z)$, it seems like you'd also want to interpolate the zeros of $f'(z)$, but that looks hard to do simultaneously...
