Holomorphic function that decays faster than any exponential in a half plane? I'm getting some trouble with the following question. I will use the common notation $z=x+iy$.
It is well-known that $f(z)=e^{-z}$ tends to zero when $x$ tends to $+\infty$, since $\vert f(z) \vert =e^{-x}$. Of course, the same happens for the family of functions $f_{\lambda}(z)=e^{-\lambda z}$ where $\lambda >0$ is a positive parameter. In fact, if $\lambda$ is bigger, the decay is faster.
My question is if it is possible to find a (non identically zero) $\textbf{holomorphic}$ function (in the right half plane) that decays faster than any function $f_{\lambda}$. In mathematical terms, the question is if we can find an $\textbf{holomorphic}$ function $g \neq 0$ (in the right half plane) such that for any sequence $x_n+i y_n$ such that $x_n$ goes to $+\infty$, we have that the limit $$g(x_n+iy_n) \cdot e^{\lambda (x_n + i  y_n)}$$ tends to zero for any $\lambda >0$. Informally, when we go to the right in the complex plane (ignoring if $y$ changes or not) we must decay faster than any exponential.
Remark: If $g$ is not required to be holomorphic the answer is trivially "yes". You just make $g(z)$ a function of its real parts (just depending on $x$) in a way that in the interval $[n,n+1]$ $g$ decays as $e^{-n}$.
 A: I thought the answer should be yes, but, at least if we assume in addition that $f$ has no zero, it's no. (It seems clear to me that allowing zeroes can't really matter, but I haven't worked out the details. In any case, this shows that the obvious way to construct a counterexample, setting $f=e^g$ where $g$ is such that whatever, cannot work. Also for example $1/\Gamma(z)$ can't work, clarifying the status of the example given in another answer.)
Note first I'm going to talk about the upper half-plane $\Pi^+$ instead of the right half-plane, because I'm more familiar with a few technicalities in that context.
First,


We may assume that $|f(z)|<1$ for all $z$.


Because if $f\in H(\Pi^+)$ and $f(x+iy)\to0$ super-exponentially as $y\to\infty$ then in particular there exists $y_0>0$ so that $|f(x+iy)|<1$ for $y>y_0$; now if $g(z)=f(z+iy_0)$ then $|g|<1$ in $\Pi^+$ and also $g(x+iy)\to0$ super-exponentially.
And now


Surprise If $f\in H(\Pi^+)$, $|f|<1$ and $f$ has no zero then there exists $\lambda>0$ such that $|f(iy)|\ge e^{-\lambda y}$ for all $y>1$.


Proof: We have $f=e^{-(u+iv)}$ where $u$ and $v$ are real-valued harmonic functions and $u>0$.
Now it's very well  known that a positive harmonic function in the unit disk  is the Poisson integral of a measure on the boundary. A conformal mapping suffices to  derive  the perhaps less well known corresponding fact for the upper half plane:


Lemma If $u>0$ is harmonic in $\Pi^+$ then there exist a constant $A\ge0$ and a (regular Borel) measure  $\mu$ on $\Bbb R$ such that $\int\frac{d\mu(t)}{t^2+1}<\infty$ and $u(x+iy)=Ay+\int\frac{y\,d\mu(t)}{(x-t)^2+y^2}$.


If $y>1$ then $\frac{y}{y^2+t^2}<\frac{y}{1+t^2}$; hence $$u(iy)\le\lambda y\quad(y>1),$$where $$\lambda=A+\int\frac{d\mu(t)}{t^2+1}.$$
Comment on removing the assumption that $f$ has no zero: If $|f|<1$ but we do  not assume $f$ has no zero then $f=gB$, where $|g|<1$, $g$ has no zero, and $B$ is a Blaschke product. Since $|g(iy)|\ge e^{-\lambda y}$ for $y>1$ we only need to show that $B(iy)$ cannot vanish super-exponentially.  This seem pretty clear to me, but the details  might be a little fussy; first we have to figure out exactly what a Blaschke product looks like in $\Pi^+$.
