Reverse Sobolev inequality for family of holomorphic functions Denote by $P_m$ the space of polynomials of degree $m$ in a single complex variable $x$. This is a "Reverse Sobolev inequality":
Theorem. Let $U \subset \mathbb{C}^2$ be an open domain and $m>0$. Let $K \subset \mathbb{C}$ be compact and assume $[0,1] \times K \subset U$. Assume $f(x,y)$ is holomorphic on $U$ and,
\begin{align}
x \to f(x,y) \in P_m \text{ for each } y \in K. \tag{1}
\end{align}
There exists a constant $C<\infty$ such that, for all $0 \leq a < b \leq 1$ and all $y \in K$,
$$
\int_a^b |f_x(x,y)| dx \leq {C \over b-a} \int_a^b |f(x,y)| dx
$$
Proof.
Denote the derivative operator $D : P_m \to P_m$. We endow $P_m$ with the norm $\lVert p\rVert_{L^1([0,1])}$. By finite dimensionality of $P_m$, $D$ must be continuous, since it is linear, and thus there is a constant $C$ such that
$$
\lVert q'\rVert_{L^1(0,1)} \leq C \lVert q\rVert_{L^1(0,1)} \text{ for } q \in P_m.
$$
Putting $q(\xi) = p((b-a)\xi+a)$, we find the scalings
$$
\lVert p'\rVert_{L^1(a,b)} = \lVert q'\rVert_{L^1(0,1)} \leq C \lVert q\rVert_{L^1(0,1)} = {C \over b-a} \lVert p\rVert_{L^1(a,b)}
$$
We then apply this estimate to the polynomial $p(x) = f(x,y)$ for each $y$ separately and the conclusion holds because $C$ only depends on $m$, and not on $y$. □
Problem. Can the hypothesis (1) be deleted?
In the proof, we didn't use the compactness of $K$, but I think it will be necessary for the following. I want to delete the hypothesis (1), i.e. $f(x,y)$ is not assumed to be a polynomial in $x$, but is a general holomorphic function on $U$. In that case, something like the Weierstrass preparation theorem gives a "quasi-polynomial" $q(x,y)$ (i.e. a polynomial in the variable $x$) such that $c_1 |q| \leq |f| \leq c_2 |q|$ in a small ball around any desired $(x,y)$, but unfortunately we do not have $|f_x| \leq c_3 |q_x|$. I was thinking maybe an inequality of the type $|f_x| \leq c_3 (|q|+|q_x|)$ might hold but I can't prove it.
I asked a simpler version of this question a few days ago, but that version was accidentally too simple and didn't capture the main difficulty of my problem above.
 A: I must be getting old, it seems like the answer comes straight from the Weierstrass preparation theorem.
Recall that a Weierstrass polynomial $W(x,y)$ has the form
$$
W(x,y) = \sum_{k=0}^n a_k(y) x^k,
$$
where each $a_k(y)$ is holomorphic in $y$ near the origin, and $a_n(y) = 1$. Also recall that a holomorphic function $u$ is said to be a unit near the origin if $1/u$ is also holomorphic near the origin.
Theorem (Weierstrass preparation). Let $f(x,y)$ be holomorphic near the origin. There is a unique factorization $f(x,y) = W(x,y)u(x,y)$, where $W$ is a Weierstrass polynomial and $u$ is a unit.
From this theorem, we immediately compute
$$
f_x = W_xu + Wu_x.
$$
Thus we obtain
$$
|f_x| \leq C(|W_x|+|W|),
$$
where $C$ is a constant that is larger than both $|u|$ and $|u_x|$ in a neighborhood of the origin. When you integrate with respect to $x$, the right-hand-side becomes an $W^{1,1}$ norm of the polynomial $W$ and then you use polynomial trickery as I did in my original question to find $\|W\|_{W^{1,1}(a,b)} \leq {C' \over b-a}\|W\|_{L^1(a,b)}$. Finally, since $u$ is a unit, it must be bounded below by some $c>0$ in a neighborhood of the origin, and thus $\|W\|_{L^1(a,b)} \leq {1 \over c} \|f\|_{L^1(a,b)}$, i.e.
$$\|f_x\|_{L^1(a,b)} \leq {C C' \over c (b-a)}\|f\|_{L^1(a,b)},$$
at least in a neighborhood of the origin.
Compactness of $[0,1] \times K$ leads to a global result.
