Hey, Sheldon.
So I've thought of a different way of phrasing your conjecture, which may be easier to prove. It's based off where I see the flaw in my proof.
If we write, for \( 0 \le y \le 2\pi \)
\(
f(y,t) = e^{-iy}\phi(t+iy)\\
\)
Then,
\(
\lim_{t \to -\infty} \frac{f(y,t)}{e^{t-1}} = 1\\
\)
and,
\(
\frac{d}{dy}|f(t,y)| = 0\,\,\text{at least when}\,\, y = 0,\pi,2\pi\\
\)
To see this,
\(
\frac{d}{dy} |f(t+1,y)| = \frac{d}{dy}|e^{t+e^{iy}f(t,y)}| = 0\\
\frac{d}{dy} \Re(e^{iy}f(t,y)) = \frac{d}{dy} \big(\cos(y)\Re f(t,y) - \sin(y) \Im f(t,y)\big) = 0\\
\sin(y)\Re f(t,y) + \cos(y)\frac{d}{dy}\Re f(t,y) - \cos(y)\Im f(t,y) - \sin (y) \frac{d}{dy}\Im f(t,y) = 0\\
\)
Now when \( y = 0, \pi , 2\pi \), we know \( \sin(y) = 0 \) and \( \Im f(t,y) = 0 \) (\( f \) is real valued here.) So we only care about,
\(
\frac{d}{dy}\Re f(t,y)= \frac{d}{dy}|f(t,y)|\cos(\arg f(t,y)) = 0
\)
Assuming \( \frac{d}{dy}|f(t,y)| = 0 \) implies \( \frac{d}{dy}|f(t+1,y)| = 0 \) when \( y = 0, \pi , 2\pi \). Since \( \frac{d}{dy}|f(t,y)| = 0 \) as \( \Re(t) \to -\infty \). We're done.
Now, I INCORRECTLY assumed these are the only places where we have zeroes in \( y \). What actually happens, and what you are noticing, is that,
\(
\frac{d}{dy}|_{y=y_k} |f(t,y)| = 0\\
\)
has solutions, but as \( t \) grows, these solutions \( y_k \to 0 \). Which means, we get a whole bunch of minima clustering near the real axis. And where there are minima we get solutions to \( \phi(s) + s = 0 \). From your conjecture we can derive this (as it hits \( \phi(s) + s = 0 \) then \( \phi(s+1) =1 \), which implies there's a minima somewhere about there), and from this we can derive your conjecture; if we hit minima we should approach \( \phi(s) + s =0 \).
Now, the conjecture I implore, which equates to this; is that for all these \( y_k \) they cluster towards the real line. This would imply,
\(
|f(t,y)| \ge |f(t,\pi)|\,\,\text{while}\,\, \delta < y < 2\pi - \delta\\
\)
For \( \delta \to 0 \) as \( t \to \infty \). I.e: there are no more minima/maxima, in this strip for large enough \( t \) other than the minima at \( \pi \).
This would allow us to show,
\(
e \uparrow \uparrow s : \{ 0 < \Im(s) < 2\pi\} \to \mathbb{C}\\
\)
is analytic; but at the real line, it cannot be. Nor at multiples of \( 2 \pi ik + \mathbb{R} \).
This actually meshes very well with the things I can say; it makes a lot of sense frankly.
So I've thought of a different way of phrasing your conjecture, which may be easier to prove. It's based off where I see the flaw in my proof.
If we write, for \( 0 \le y \le 2\pi \)
\(
f(y,t) = e^{-iy}\phi(t+iy)\\
\)
Then,
\(
\lim_{t \to -\infty} \frac{f(y,t)}{e^{t-1}} = 1\\
\)
and,
\(
\frac{d}{dy}|f(t,y)| = 0\,\,\text{at least when}\,\, y = 0,\pi,2\pi\\
\)
To see this,
\(
\frac{d}{dy} |f(t+1,y)| = \frac{d}{dy}|e^{t+e^{iy}f(t,y)}| = 0\\
\frac{d}{dy} \Re(e^{iy}f(t,y)) = \frac{d}{dy} \big(\cos(y)\Re f(t,y) - \sin(y) \Im f(t,y)\big) = 0\\
\sin(y)\Re f(t,y) + \cos(y)\frac{d}{dy}\Re f(t,y) - \cos(y)\Im f(t,y) - \sin (y) \frac{d}{dy}\Im f(t,y) = 0\\
\)
Now when \( y = 0, \pi , 2\pi \), we know \( \sin(y) = 0 \) and \( \Im f(t,y) = 0 \) (\( f \) is real valued here.) So we only care about,
\(
\frac{d}{dy}\Re f(t,y)= \frac{d}{dy}|f(t,y)|\cos(\arg f(t,y)) = 0
\)
Assuming \( \frac{d}{dy}|f(t,y)| = 0 \) implies \( \frac{d}{dy}|f(t+1,y)| = 0 \) when \( y = 0, \pi , 2\pi \). Since \( \frac{d}{dy}|f(t,y)| = 0 \) as \( \Re(t) \to -\infty \). We're done.
Now, I INCORRECTLY assumed these are the only places where we have zeroes in \( y \). What actually happens, and what you are noticing, is that,
\(
\frac{d}{dy}|_{y=y_k} |f(t,y)| = 0\\
\)
has solutions, but as \( t \) grows, these solutions \( y_k \to 0 \). Which means, we get a whole bunch of minima clustering near the real axis. And where there are minima we get solutions to \( \phi(s) + s = 0 \). From your conjecture we can derive this (as it hits \( \phi(s) + s = 0 \) then \( \phi(s+1) =1 \), which implies there's a minima somewhere about there), and from this we can derive your conjecture; if we hit minima we should approach \( \phi(s) + s =0 \).
Now, the conjecture I implore, which equates to this; is that for all these \( y_k \) they cluster towards the real line. This would imply,
\(
|f(t,y)| \ge |f(t,\pi)|\,\,\text{while}\,\, \delta < y < 2\pi - \delta\\
\)
For \( \delta \to 0 \) as \( t \to \infty \). I.e: there are no more minima/maxima, in this strip for large enough \( t \) other than the minima at \( \pi \).
This would allow us to show,
\(
e \uparrow \uparrow s : \{ 0 < \Im(s) < 2\pi\} \to \mathbb{C}\\
\)
is analytic; but at the real line, it cannot be. Nor at multiples of \( 2 \pi ik + \mathbb{R} \).
This actually meshes very well with the things I can say; it makes a lot of sense frankly.