intersection of holomorphic curve with hyperplane

Question

Let $f : \mathbb{C} \rightarrow \mathbb{C}^n$, $n>1$ be an entire function. Assume for simplicity that $f(0)=0$. Let $B$ be the closed ball of centre $O$ and radius $R$.

Is there an upper bound for the number of points of intersection $f(\mathbb{C})\cap B \cap H$ where $H$ is a hyperplane in $\mathbb{C}^n$ (not containing $f(\mathbb{C})$), independent of $H$ (depending only on $f$ and $R$)?

I know that there is no general analogue of Bezout's theorem for analytic sets, but I am asking for something much weaker - I do not care about the dependence on $R$...

Answer 1

EDITED. Suppose first that $f(C)$ is not contained in any hyperplane. Let us describe hyperplanes by equations $L(w)=a_1w_1+\ldots+a_nw_n-b=0$, where we can normalize so that $\sum |a_j|^2+b^2=1$. Now our assumption that $f(C)$ does not contained in any hyperplane, implies that you can find points $z_0,\ldots,z_n$ so that $f(z_j)$ do not lie in any hyperplane. This means that $\max\{ |L(f(z_j))|: 0\leq 0\leq n\}\geq c>0$, where $c$ depends only on $f$. Let $r_0=\max_j|z_j|$. Now apply Jensen's formula to the disks $|z-z_j|\leq R:= 6r+r_0$. They all contain the disk $|z|<2r$. Let $n_j(t,L,f)$ be the counting function of the intersections with the hyprplane $L$ in the disk $|z-z_j|\leq t$, $$N_j(R,L,f)=\int_0^R(n_j(t,L,f)-n_j(0,L,f))\frac{dt}{t}+n_j(0,L,f)\log R.$$ Notice the inequality $n(r,L,f)\leq N(er,L,f)$ for $r>e$. Then By Jensen's formula applied to the subharmonic function $\log|L\circ f|$, we obtain $$N_j(R,L,f)\leq\int_0^{2\pi}\log| L\circ f(Re^{it}+z_j)| dt/(2\pi)-\log| L\circ f(z_j)|.$$ The first term has an upper bound depending only on $f$ and $r$ and the second term has an upper bound $-\log c$ for some $j$. So take minimum in $j$ and you obtain the estimate independent of $L$.

Now if $f(C)$ is contained in some hyperplane, we reduce to the previous case. Let $H$ be the affine space of the smallest dimension that contains $f(C)$. Then change the coordinates so that $H$ is described by equations $w_{m+1}=w_{m+2}=\ldots=w_n=0$. In these new coordinates $f$ can be considered as a map to $C^m$ whose image is not contained in any hyperplane. The hyperplanes for which the number of preimages is discrete are exactly those which do not contain $H$.

Answer 2

I think the general answer would be unbounded unless you restrict $f$ to be some special class of entire functions. But the reason for this is trivial; namely you can approximate any continuous complex-valued function over $B$ using $f$. Certain variations of Weierstrass pproximation theorem allows you to do this. So if you allow $f$ to be "as bad as you like", then the behavior could be out of control.

My guess is if you consider a more refined class of $f$ using Weierstrass factorization theorem, then the problem may be treatable following the answer above. But even then it depends on the specific class of $f$ involved and you may need tools like Hadamard factorization theorem.