Let $f(z)$ be an entire function (holomorphic function on $\mathbb{C}$) satifying the following condition: $$|f(z)|\leq \max (e^{\text{im}(z)},1 ),\ \forall z\in\mathbb{C}$$ $$\lim_{\mathbb{R}\ni t\rightarrow+\infty} f(t)=0$$
My question is: can we prove that the following limit exists? $$\lim_{\mathbb{R}\ni t\rightarrow-\infty} f(t)$$
Maybe we can even prove $f(z)=0$. But I do not know how to prove it.
My trying: if $f(z)$ has finitely many zeros, then by Hadamard factorization theorem, $f(x)=e^{az+b}P(z)$ where $P(z)$ is a polynomial. such a function can not be bounded by $\max(e^{\text{im}(z)},1)$, unless $P(z)\in\mathbb{C}$ and $ia\in\mathbb{R}$. Then we know $f(z)$ has to be $0$. But if $f(z)=0$ has infinitely many solution, then I do not know how to proceed.
I will sketch the construction of such an $f$ which satisfies all the required conditions but for which the limit at $-\infty$ doesn't exist. Pick a sequence $a_n \to 0, n \to \infty$ very fast eg $a_n=(2n)^{-2n}, n \ge 1$ and another sequence that is bounded by $1$ but oscillates eg $a_{-2n+1}=1$ and $a_{-2n}=1/2$ when $n \ge 1$. Consider the function $$g(z)=\sum_{n \in \mathbb Z, n \ne 0}a_n\frac{1}{(z-n)^2}$$
We claim that $g$ is meromorphic with double poles at non zero integers and $|g(z)| \le K$ on $\mathbb C -\cup_{n \ne 0}B(n,1/4)$.
The series converges uniformly on any compacts that stay away from the nonzero integers so $g$ is meromorphic.
For the second part, noting that $|a_n| \le 1$, let $z=x+iy$ and then the at most three terms where $|x-n| \le 1$ can be bound by $16$ since $|z-n| \ge 1/4$ for all $n$, while for the rest we can bound $|1/(z-n)^2|$ by $1/(x-n)^2$ and that sum is clearly uniformly bounded by a multiple of $\zeta(2)$
We claim that $h(z)=g(z)\frac{\sin^2 \pi z}{\pi^2}$ is entire and satisfies $h(n)=a_n, h(r) \to 0, r \to \infty$ while $h(-n)=a_{-n}$ so there is no limit of $h(r)$ as $r \to -\infty$. Also $h(z) \le Ce^{7|\Im z|}$ for some $C>0$
Clearly $h$ entire and $h(n)=a_n$ for all nonzero integers since $\frac{\sin^2 \pi z}{\pi^2}=(z-n)^2+O((z-n)^3$ when $|z-n|$ small enough while $|\frac{\sin^2 \pi z}{\pi^2}| \le C_1e^{2\pi |\Im z|}$ means that $|h(z)| \le C_1Ke^{2\pi |\Im z|} \le C_2e^{7|\Im z|}$ first on the domain $\mathbb C -\cup_{n \ne 0}B(n,1/4)$ and then with a possibly higher $C$ also on the discs $B(n,1/4)$ hence on the full plane
(if we fix $n$ we notice that for all $m \ne n$ and $|z-n| \le 1/4$ we have $|z-m| \ge 1/4$ so the constant bound above for $g$ applies for $g-a_n\frac{1}{(z-n)^2}$ hence the required bound applies for $(g-a_n\frac{1}{(z-n)^2})\frac{\sin^2 \pi z}{\pi^2}$ while for the special term $\frac{\sin^2 \pi z}{\pi^2}a_n\frac{1}{(z-n)^2}$ we use that $|\frac{\sin^2 \pi z}{(z-n)^2}|=|\frac{\sin^2 \pi (z-n)}{(z-n)^2}| \le C_4$ by continuity where $C_4=\max_{|x| \le 1/4} |\frac{\sin^2 \pi x}{x^2}|$)
It remains to prove that $h(r) \to 0, r \to \infty$ and again taking out the at most one possible value of $n$ for which $|r-n| \le 1/4$ where we need to use the $\sin^2 \pi z$ factor and we get a $C_5a_n \to 0, r \to \infty$ bound as above, we can ignore the bounded $\sin^2 \pi r$ term for the rest of the $n$ and focus on the sum $\sum_{n \ne 0, |n-r| \ge 1/4}a_n\frac{1}{(r-n)^2}$ and analyze it closer.
The part where $r-n \ge r/2$ is $O(\int_{-\infty}^{r/2}\frac{dx}{(r-x)^2})=O(1/r)$ by the integral test, while the part where $n-r \ge r/2$ is similarly $O(1/r)$ by the integral test. For $r/2 \le n \le 3r/2, |r-n| \ge 1/4$ we bound each $1/(r-n)^2$ by $1/16$ and we get a bound of $16\sum_{r/2 \le n \le 3r/2}a_n=O(r^{-r+1}) \to 0$ by our choice of $a_n$ so we proved that $h(r) \to 0, r \to \infty$
But now let's take $f(z)=Ah(z/700)e^{-iz/10}$. Clearly $f(r) \to 0, r \to \infty$ and $|f(-700n)|=A$ or $A/2$ depending on parity, so no limit as $r \to -\infty$
But if $\Im z >0$ we have $|f(z)| \le ACe^{\Im z/100}e^{\Im z/10} \le AC e^{\Im z/5}$ while if $\Im z \le 0$ we have $|f(z)| \le ACe^{-\Im z/100}e^{\Im z/10} \le AC$ so choosing $AC \le 1$ so $f$ satisfies the required condition and we are done!