Hey, everyone.
I'm going to run through a proof of uniform convergence of \( \log^{\circ n} (\beta_\lambda(s+n)) \); to get a feel for what it might look like. I might make some errors, but this is looking like it might work.
Let's start by considering \( \lambda \in \mathbb{C} \) and \( s \in \mathbb{C} \) as variables; assume that \( \Re(\lambda) > \delta \) and \( |\lambda(j-s) - (2k+1)\pi i| > \delta \) for some \( \delta > 0 \); and \( j,k \in \mathbb{Z} \) with \( j \ge 1 \). We'll call this set \( \mathbb{L} \). As such, \( (s,\lambda) \in \mathbb{L} \). We have a holomorphic function for \( (s,\lambda) \in \mathbb{L} \) given by,
\(
\beta_\lambda(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{\lambda(j-s)} + 1}\,\bullet z\\
\)
This satisfies the functional equation,
\(
\beta_\lambda(s+1) = \frac{e^{\beta_\lambda(s)}}{e^{-\lambda s} + 1}\\
\)
And the way we've constructed \( \mathbb{L} \) is such that for any compact subset \( \mathcal{U} \subset \mathbb{L} \) we must have,
\(
\sum_{j=1}^\infty ||\frac{1}{e^{\lambda(j-s)} + 1}||_{(s,\lambda) \in \mathcal{U}} < \infty\\
\)
This is all that's needed to prove holomorphy of \( \beta_\lambda(s) \) by the same proof which constructed \( \phi \). In which,
\(
\log \beta_{\lambda}(s+1) = \beta_\lambda(s) + \mathcal{O}(e^{-\delta \Re(s)})\\
\)
And just as well by Taylor's theorem,
\(
\log(\beta_\lambda(s+1) + \mathcal{O}(e^{-\delta \Re(s)})) = \beta_\lambda(s) + \mathcal{O}(e^{-\delta \Re(s)})\\
\)
Which holds as \( \Re(s) \to \infty \) by construction. So in defining a sequence of convergents \( \tau_\lambda^{n}(s) \) where,
\(
\tau_\lambda^n(s) = \log(\beta_\lambda(s+1) + \tau_\lambda^{n-1}(s+1)) - \beta(s)\\
\)
As \( \Re(s)\to \infty \) for \( (s,\lambda) \in \mathbb{L} \), we get by induction \( \tau_\lambda^n(s) = \mathcal{O}(e^{-\delta \Re(s)}) \to 0 \) as \( \Re(s) \to \infty \). We may expect a swift enough convergence in this manner. Then,
\(
\beta_\lambda(s+1) + \tau_\lambda^n(s+1) = e^{\beta(s) + \tau_\lambda^{n+1}(s)}\\
\)
To which,
\(
|\tau_\lambda^n(s) - \tau_\lambda^m(s)| \le Ae^{-\delta \Re(s)}\\
\)
So at infinity we have a normality condition, in which \( \tau_\lambda^n(s) \) is bounded for large enough \( n > N \). Where convergence should follow for \( (s,\lambda) \in \mathbb{L} \) when \( \Re(s) > X \). And here is where we need a Riemann mapping on \( \lambda \)...
Please, I need help at this point of the proof. But I think I have it.
I'm going to run through a proof of uniform convergence of \( \log^{\circ n} (\beta_\lambda(s+n)) \); to get a feel for what it might look like. I might make some errors, but this is looking like it might work.
Let's start by considering \( \lambda \in \mathbb{C} \) and \( s \in \mathbb{C} \) as variables; assume that \( \Re(\lambda) > \delta \) and \( |\lambda(j-s) - (2k+1)\pi i| > \delta \) for some \( \delta > 0 \); and \( j,k \in \mathbb{Z} \) with \( j \ge 1 \). We'll call this set \( \mathbb{L} \). As such, \( (s,\lambda) \in \mathbb{L} \). We have a holomorphic function for \( (s,\lambda) \in \mathbb{L} \) given by,
\(
\beta_\lambda(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{\lambda(j-s)} + 1}\,\bullet z\\
\)
This satisfies the functional equation,
\(
\beta_\lambda(s+1) = \frac{e^{\beta_\lambda(s)}}{e^{-\lambda s} + 1}\\
\)
And the way we've constructed \( \mathbb{L} \) is such that for any compact subset \( \mathcal{U} \subset \mathbb{L} \) we must have,
\(
\sum_{j=1}^\infty ||\frac{1}{e^{\lambda(j-s)} + 1}||_{(s,\lambda) \in \mathcal{U}} < \infty\\
\)
This is all that's needed to prove holomorphy of \( \beta_\lambda(s) \) by the same proof which constructed \( \phi \). In which,
\(
\log \beta_{\lambda}(s+1) = \beta_\lambda(s) + \mathcal{O}(e^{-\delta \Re(s)})\\
\)
And just as well by Taylor's theorem,
\(
\log(\beta_\lambda(s+1) + \mathcal{O}(e^{-\delta \Re(s)})) = \beta_\lambda(s) + \mathcal{O}(e^{-\delta \Re(s)})\\
\)
Which holds as \( \Re(s) \to \infty \) by construction. So in defining a sequence of convergents \( \tau_\lambda^{n}(s) \) where,
\(
\tau_\lambda^n(s) = \log(\beta_\lambda(s+1) + \tau_\lambda^{n-1}(s+1)) - \beta(s)\\
\)
As \( \Re(s)\to \infty \) for \( (s,\lambda) \in \mathbb{L} \), we get by induction \( \tau_\lambda^n(s) = \mathcal{O}(e^{-\delta \Re(s)}) \to 0 \) as \( \Re(s) \to \infty \). We may expect a swift enough convergence in this manner. Then,
\(
\beta_\lambda(s+1) + \tau_\lambda^n(s+1) = e^{\beta(s) + \tau_\lambda^{n+1}(s)}\\
\)
To which,
\(
|\tau_\lambda^n(s) - \tau_\lambda^m(s)| \le Ae^{-\delta \Re(s)}\\
\)
So at infinity we have a normality condition, in which \( \tau_\lambda^n(s) \) is bounded for large enough \( n > N \). Where convergence should follow for \( (s,\lambda) \in \mathbb{L} \) when \( \Re(s) > X \). And here is where we need a Riemann mapping on \( \lambda \)...
Please, I need help at this point of the proof. But I think I have it.