@Suvrit Related material can be found in the paper S. Abramovich, J. Baric, M. Matic, J. Pecaric, "On van de Lune-Alzer's Inequality" J. of Math. Inequalities, 1, (2007) 563-587. <jmi.ele-math.com/01-47/On-Van-de-Lune-Alzer-s-inequality>

@Suvrit Much of the work of van de Lune about his conjecture is described in the CWI Report PNA - R0502, May 2005: J. van de Lune, H. J. J. te Riele, "On some conjectural inequalities and their consequences"

@Agno The real points $t$ where $\chi(t)=1$ are the Gram points. At a Gram point $\zeta(1/2+i g_n,2)=\zeta(1/2-i g_n,2)$ your function $f(t)$ would be $0$ at $g_n$ unless $\zeta(1/2+i g_n,2)<0$. This is equivalent to $\zeta(1/2+i g_n)<1$. This happen for $g_3 = 31.71$, $g_8=48.71$, $g_{12}=60.35$, $g_{18}=76.17$, $g_{23}=88.38$, $g_{26}=95.41$ $\dots$

@Agno Observe that $\zeta(s)=\chi(s)\zeta(1-s)$ gives $\chi(\frac12+it)=\zeta(\frac12+it)/\zeta(\frac12-it)$. Then your function $f(t)=|\sqrt{\zeta(\frac12-it)}-\sqrt{\zeta(\frac12+it)}|=|\sqrt{\zeta(\frac12-it)}|\cdot|1-\sqrt{\chi(\frac12+it)}|$. (If you take adequate roots). I think that the graph is not correct $f(t)$ vanish at the Gram point t=31.717979954764 or you are taking differents roots of $\zeta(\frac12+it)=\zeta(\frac12-it)=$ that at this point is a negative real number. At this point $\chi(t)=1$.

$\chi(s)$ is a meromorphic function. I do not understand the assertion that $|\chi(s)-1|$ is entire. The modulus of a non constant meromorphic function is not entire. If you think only on the critical line $|\chi(0.5+it)-1|= 2 |\sin\vartheta(t)|$ has real zeros, and the modulus is not differentiable as a function of a real variable.

Yes, but as I said before, the first example in the question refers to factorization in polynomials of arbitrary degree. This is other question, for which I suspect the only condition must be on the value at $z=0$ and the order of the possible zero there. In this sense the exponential admits convergent factorizations possibly not unique. But this requires some proofs.

See the OEIS A170910 and the references there.

Why are you so sure that the exponential has no factorization on polynomials convergent on a certain disk $|z|<r$ ? Specially as if in your first example you admit polynomial of arbitrary degree.

Solving $\exp(-\pi/x)=0.78177$ gives us x=12.7606. The remaining portion is $12.76 <x <+\infty$. In this range the other argument I think will be relatively easy to implement. For these large values of $x$ we have $1-\vartheta_4(e^{-\pi x})=8.48 \cdot 10^{-17}$.

We have $f(e^{-\pi/x}) = g(x) \vartheta_4^2(e^{-\pi x})$. The factor $\vartheta_4^2(e^{-\pi x})$ is almost constant $=1$. Therefore the character of $f(e^{-\pi/x}) $ will be that of $g(x)$. Of course this need careful bounds. Those of $\vartheta_4^2(e^{-\pi x})$ are easily obtained from the expression given for the $\vartheta$ functions at the start of my answer. Those of $g(x)$ must be elementary. For example the error in $\vartheta_4^2(e^{-\pi x})\approx1$ is of the order of $C e^{-\pi x}$. This will work for $x\gg1$.

There exists a power series $q = a_2 k^2+\cdots$, after changing $k$ into $4k$ $a_n$ in this expansion is A005797 in the OEIS. Then your $f(k)$ is $(1-q)\vartheta_2^2/\sqrt{q}$. After substituting $q$ by this series. This may be a way to get your property. I do not say this is easy. Certainly in this way dissapear your terrible looking hyperbolic sin.