An inf-sup estimate for holomorphic functions

Question

Is the following true?

Conjecture? Let $U \subset \mathbb{C}^n$ be open and $\eta : U \to \mathbb{C}$ be holomorphic. Denote by $B(z,r)$ the usual ball of radius $r$. There is a constant $\kappa<\infty$ such that, for all $B(z,2r) \subset U$, if $\eta$ is nonzero on $B(z,2r)$ then $$\sup_{z \in B(z,r)} |\eta(z)| \leq \kappa \inf_{z \in B(z,r)} |\eta(z)|.$$

By way of the Weierstrass preparation theorem, it might suffice to show a related estimate for polynomials of a fixed degree $m$. It may also be necessary to assume that $\eta$ is in fact holomorphic on some $V$ with $U \subset \subset V$.

Answer 1

This is not true: take $n=1$, $r=1$, $\eta(z)=e^{az},\; a>0,$ then $$\max_{z\in B(0,r)}|\eta(z)|=e^a,$$ while $$\min_{z\in B(0,1)}|\eta(z)|=e^{-a}.$$ Since $a>0$ is arbitrary, no $\kappa$ with required property exists.

Answer 2

Inspired by Alexandre Emerenko, the counter example is $U = \mathbb{C}$ and $\eta(z) = e^z$. Because $\eta$ is nowhere zero, the condition on $B(z,2r)$ is moot. Thus, one should have $\kappa \geq \sup_{z \in B(0,r)} |\eta(z)| / \inf_{z \in B(0,r)} |\eta(z)| = e^{2r}$, which is unbounded as $r \to \infty$.

What ended up working for me is Leo Moos's suggestion of using the Harnack inequality. The modified statement is in part as follows:

Theorem. Let $V \subset \mathbb{C}^m$ be an open domain, $\eta : V \to \mathbb{C}$ be holomorphic. Let $\Gamma = \{z \in V \; : \; \Re \eta(z) = 0 \}$, note that $\Gamma$ is not a complex manifold, but assume it is a real manifold of codimension $1$ in $\mathbb{R}^{2m} \sim \mathbb{C}^m$. Let $K$ be a compact subset of $V$ such that $\Re \eta(z) \geq 0$ on $K$. Then, there is $\epsilon>0$ such that, for $z \in K$, $$\Re \eta(z) \geq \epsilon d(z,\Gamma),$$ where $d(z,\Gamma) = \inf_{w \in \Gamma} \|z-w\|_2$, as usual.

Proof sketch. The Harnack inequality applied to $\Re \eta$, regarded as a function of $\mathbb{R}^{2m}$, gives that if $B(z_0,R)$ is free of zeros, and if $\|z-z_0\|_2 = r < R$, then $$ {1-(r/R) \over [1+(r/R)]^{2m-1}}\Re \eta(z_0) \leq \Re \eta(z). $$ We can make this "best possible", as follows. Let $w \in \Gamma$ be nearest to $z$, and denote by $v$ the unit normal of $\Gamma$ at $w$. Note that the ball centered at $z$ of radius $\delta = \|w-z\|_2$ is tangent to $\Gamma$ at $w$, i.e. the normal of the ball is also $\pm v$.

At $w$, there are two "largest possible tangent balls", one for each side of $\Gamma$, which we denote $B(z_0,R_0)$ and $B(z_1,R_1)$, such that $\Re \eta > 0$ in each ball. Without loss of generality, the ball $B(z_0,R_0)$ is on the "same side" of $\Gamma$ as $B(z,\delta)$ and so $B(z,\delta) \subset B(z_0,R_0)$. We then use $R = R_0 = R_0(w)$, and $r = R-\delta$ in the Harnack inequality to find that $$ {d(z,\Gamma)/R(w) \over [2-(d(z,\Gamma)/R(w))]^{2m-1}}\Re \eta(z_0) \leq \Re \eta(z). $$ Here, $R(w)$ can be estimated as a function of the curvature of the manifold at the point $w \in \Gamma$, and in particular, if the curvature of $\Gamma$ is bounded above, then $R(w)$ is bounded below. Because I am restricting to some compact set $K$, we can effectively find a lower bound on $R(w)$ even if $\Gamma$ were to have unbounded curvature at infinity or some other pathology. QED