Entire functions with controlled growth in an angular region and bounded elsewhere

Question

I am trying to construct a holomorphic function $U \to L^2(\mathbb R)$ for some open $U \subseteq \mathbb C$, which is not locally bounded by an $L^2$ function. (I believe this should exist.)

To do so, I was hoping that there exist entire functions $\mathbb C \to \mathbb C$ that are simultaneously:

1. bounded (say by $1$) outside an arbitrarily small angular region $D_\alpha = \{0 \leq \arg z \leq \alpha\}$
2. of arbitrarily large absolute value at a point of modulus $1$ in that region, say of size $y_\alpha$
3. bounded by $M y_\alpha$ in arbitrarily large disks $B(0, R_\alpha)$, where $M > 0$ does not depend on $\alpha$

Do they exist?

Here the "arbitrary large/small" can be dependent on each other, i.e. I'm looking for such a sequence of functions with $(\alpha, y_\alpha, R_\alpha) \to(0, \infty, \infty)$.

The first two conditions can be satisfied, e.g. by Mittag-Leffler functions as mentioned here: https://mathoverflow.net/questions/29734

The additional difficulty is the controlled growth in large disks. I would be happy too if these functions are only defined on $B(0,R_\alpha)$.

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To give more context: if $f_n$ is such a sequence with $\alpha_n$ of size $1/n^2$ (say) so that the angular regions $D_n$ can be chosen pairwise disjoint, and take a subsequence for which the $y_{\alpha_n}$ are at distance $\geq 1$ from each other, then

$$g(y, s) = \sum_{n=1}^\infty \rho(y-y_{\alpha_n})\frac{f_n(s R_{\alpha_n})}{y}$$

with $\rho$ a bump function such that $\rho(y-y_{\alpha_n})$ have disjoint support, is a holomorphic $L^2(\mathbb R_{>1})$-valued function on $B(0,1)$ which is not bounded by an $L^2$-function in any neighborhood of $0$.

Answer 1

It turns out that this is not possible, by Montel's theorem:

Call those functions $f_\alpha$, and w.l.o.g. assume that they attain $y_\alpha$ at the point $s=1$, so that the angular regions $D_\alpha$ all contain the positive real ray. Then $f_\alpha/y_\alpha$ is uniformly bounded on (say) $B(0,2)$ hence is a normal family. But the uniform limit of any subsequence is $0$ outside $[0,2]$ and equals $1$ at $s=1$, hence cannot be continuous (let alone holomorphic).

In fact, from $L^{\infty}$ norm is bounded by $L^2$ norm for holomorphic functions it follows that it is true, that holomorphic $U \to L^2$ are locally bounded by an $L^2$ function.