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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.

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  • $\begingroup$ As far as I can see, the only difference to your earlier question is that you now require the inequality to hold uniformly in $y\in K$. Since $K$ is compact, this still follows from the argument that I gave earlier. $\endgroup$ Commented Feb 5, 2023 at 18:16
  • $\begingroup$ The earlier version is here: mathoverflow.net/questions/439821/… $\endgroup$ Commented Feb 5, 2023 at 18:16
  • $\begingroup$ Thanks for this. Your previous argument did work, but I'm not sure I get it here. Say that you've got some ball $B$ and for $(a,y) \in B$, you have $f(a+h,y) = c(y)h + O(h^2)$. Your argument relied on $c(y)h$ dominating over $O(h^2)$ but here the most important case is where $c(y)=0$ somewhere in $B$ (usually at its center). Then I don't know how to compare $c(y)h$ and $O(h^2)$. Am I misunderstanding something? $\endgroup$ Commented Feb 5, 2023 at 18:53
  • $\begingroup$ The argument works the same way if $f=ch^n + O(h^{n+1})$, with $c\not= 0$, which holds everywhere (with a uniform bound on $n$ in $K$). $\endgroup$ Commented Feb 5, 2023 at 21:41
  • $\begingroup$ Right but that is specifically not the case I'm interested in. I've got a case where $f(x,y)$ has zeros of various depths over its domain. For example, the case I'm worried about now is that $f(x,y)$ has a zero of high order in $x$ (like, order 4), but if you perturb $y$, you instead get 4 distinct roots in $x$. $\endgroup$ Commented Feb 5, 2023 at 21:44

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