Start by taking the function \( \beta_\lambda \), and define a sequence of functions \( \tau_\lambda^n(s) \) such that,
\(
\beta_\lambda(s) + \tau_\lambda^{n+1}(s) = \log(\beta_\lambda(s+1) + \tau_\lambda^n(s+1))\\
\)
We aren't too concerned about where \( \tau_n \) is holomorphic at the moment. We know that,
\(
\tau_\lambda^0(s) = 0\\
\tau_\lambda^1(s) = \sum_{k=1}^\infty \frac{(-1)^k}{k} e^{-k\lambda s}\\
\)
Assume that,
\(
\tau_\lambda^n(s) = \sum_{k=1}^\infty c_{nk}(\lambda) e^{-k\lambda s}\\
\)
And we are reduced to the equation,
\(
\tau_\lambda^{n+1}(s) = \sum_{k=1}^\infty c_{(n+1)k}(\lambda) e^{-k\lambda s} = \log(\beta_\lambda(s+1) + \sum_{k=1}^\infty c_{nk}(\lambda)e^{-k\lambda} e^{-k\lambda s}) - \beta_\lambda(s)\\
\)
Now we can calculate \( c_{(n+1)k}(\lambda) \) inductively on \( k \). Assuming we know \( c_{(n+1)j}(\lambda) \) for \( 1 \le j \le k \); then we can find \( c_{(n+1)(k+1)}(\lambda) \) by,
\(
\lim_{\Re(s) \to \infty} \frac{\tau_\lambda^{n+1}(s) - \sum_{j=1}^k c_{(n+1)j}(\lambda) e^{-j\lambda s}}{e^{(k+1)\lambda s}} = c_{(n+1)(k+1)}(\lambda)\\
\)
Of which this limit necessarily exists; and constructs a sequence \( c_{(n+1)k}(\lambda) \). And now our problem is reduced to solving \( \lim_{n\to\infty} c_{nk}(\lambda) \).
We are also given the quick relation,
\(
\tau^{n+1}_\lambda(s) = \log(\beta_\lambda(s+1) + \tau^n_\lambda(s+1)) - \beta_\lambda(s)\\
= -\log(1+e^{-\lambda s}) + \log(1 + e^{-\beta_\lambda(s)}(e^{-\lambda s} + 1) \tau^n(s+1))\\
=\sum_{k=1}^\infty \frac{(-1)^k}{k}e^{-k\lambda s} + \log(1+e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau^{(n)}(s+1))\\
\)
Then since we know that \( \tau^{n+1}_\lambda(s) \to 0 \) we know that,
\(
e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau_\lambda^{n}(s+1) \to 0\\
\)
Because, \( e^{-\beta_\lambda(s)}(e^{-\lambda s} +1) = \frac{1}{\beta(s+1)} \), this implies that,
\(
\frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} \to 0\\
\)
This gives us hope we can devise a normality condition on \( \tau^n_\lambda(s) \) as it controls how fast \( \beta(s) \) can dip to zero; which is the exact problem we had with \( \phi \). This can be more aptly written,
\(
\frac{1}{\beta(s)} = o(e^{\lambda s})\\
\)
But we can say something stronger. We know that \( \tau^n_\lambda(s) = \mathcal{O}(e^{-\lambda s}) \) and this means that,
\(
\frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} = \mathcal{O}(e^{-\lambda s})\\
\)
As this is the term in the logarithm, and in order for \( \tau^{n+1}_\lambda(s) \) to tend to zero exponentially the term in the logarithm has to as well. Therefore we can say that,
\(
\frac{1}{\beta(s)} = \mathcal{O}(h(s))\\
\)
Where for any \( \delta > 0 \) we have \( e^{-\delta s} h(s) \to 0 \) as \( \Re(s) \to \infty \). Which tells us that,
\(
|\beta(s)| \ge \frac{A}{h(s)}
\)
Which informs us asymptotically how \( \beta(s) \) approaches zero at infinity... slower than any exponential. This should help us with the limit; in the end, and making sure we stay away from zero when defining our function \( F_n(s) = \log^{\circ n} \beta(s+n) \).
As to the second part, when we actually go about and take this limit we want \( \lambda \) to be a function of \( s \). And in this sense, we want \( \beta_{\lambda(s)}(s) : \{s \in \mathbb{C} \,:\, |\arg(s)| < \theta\} \to \mathbb{C} \) where \( (s,\lambda(s)) \in \mathbb{L} \) and \( \lambda : \mathbb{R}^+ \to \mathbb{R}^+ \). This would mean for every imaginary part of \( s \) there is a large enough \( N \) in which \( F_n(s) \) exists for \( n > N \). This is where we'll need a clever Riemann mapping on \( \lambda \) over some simply connected domain \( \mathbb{D} \subset \mathbb{L} \); which I'm not sure how to do; especially as this requires a two variable ideation.
Forget I said anything about periodic functions, I don't think that's necessary here. I'm going to keep plugging and playing with this, but the more I fiddle the more this has none of the problems \( \phi \) has.
Regards, James.
EDIT:
Some even more evidence this is doable. Let \( \lambda(s) \) be an implicitly defined function such that,
\(
\log \beta_{\lambda(s+1)}(s+1) - \log \beta_{\lambda(s)}(s+1) = \log(1+ e^{-\lambda(s) s})\\
\log \beta_{\lambda(s+1)}(s+1) - \beta_{\lambda(s)}(s) + \log(1+e^{-\lambda(s)s}) = \log(1+ e^{-\lambda(s) s})\\
\log \beta_{\lambda(s+1)}(s+1) = \beta_{\lambda(s)}(s)\\
\)
Which would require an implicit solution \( \lambda(s) \); hinting that the logarithm trick might work for correctly chosen \( \lambda \); as it's idempotent in this case.
\(
\beta_\lambda(s) + \tau_\lambda^{n+1}(s) = \log(\beta_\lambda(s+1) + \tau_\lambda^n(s+1))\\
\)
We aren't too concerned about where \( \tau_n \) is holomorphic at the moment. We know that,
\(
\tau_\lambda^0(s) = 0\\
\tau_\lambda^1(s) = \sum_{k=1}^\infty \frac{(-1)^k}{k} e^{-k\lambda s}\\
\)
Assume that,
\(
\tau_\lambda^n(s) = \sum_{k=1}^\infty c_{nk}(\lambda) e^{-k\lambda s}\\
\)
And we are reduced to the equation,
\(
\tau_\lambda^{n+1}(s) = \sum_{k=1}^\infty c_{(n+1)k}(\lambda) e^{-k\lambda s} = \log(\beta_\lambda(s+1) + \sum_{k=1}^\infty c_{nk}(\lambda)e^{-k\lambda} e^{-k\lambda s}) - \beta_\lambda(s)\\
\)
Now we can calculate \( c_{(n+1)k}(\lambda) \) inductively on \( k \). Assuming we know \( c_{(n+1)j}(\lambda) \) for \( 1 \le j \le k \); then we can find \( c_{(n+1)(k+1)}(\lambda) \) by,
\(
\lim_{\Re(s) \to \infty} \frac{\tau_\lambda^{n+1}(s) - \sum_{j=1}^k c_{(n+1)j}(\lambda) e^{-j\lambda s}}{e^{(k+1)\lambda s}} = c_{(n+1)(k+1)}(\lambda)\\
\)
Of which this limit necessarily exists; and constructs a sequence \( c_{(n+1)k}(\lambda) \). And now our problem is reduced to solving \( \lim_{n\to\infty} c_{nk}(\lambda) \).
We are also given the quick relation,
\(
\tau^{n+1}_\lambda(s) = \log(\beta_\lambda(s+1) + \tau^n_\lambda(s+1)) - \beta_\lambda(s)\\
= -\log(1+e^{-\lambda s}) + \log(1 + e^{-\beta_\lambda(s)}(e^{-\lambda s} + 1) \tau^n(s+1))\\
=\sum_{k=1}^\infty \frac{(-1)^k}{k}e^{-k\lambda s} + \log(1+e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau^{(n)}(s+1))\\
\)
Then since we know that \( \tau^{n+1}_\lambda(s) \to 0 \) we know that,
\(
e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau_\lambda^{n}(s+1) \to 0\\
\)
Because, \( e^{-\beta_\lambda(s)}(e^{-\lambda s} +1) = \frac{1}{\beta(s+1)} \), this implies that,
\(
\frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} \to 0\\
\)
This gives us hope we can devise a normality condition on \( \tau^n_\lambda(s) \) as it controls how fast \( \beta(s) \) can dip to zero; which is the exact problem we had with \( \phi \). This can be more aptly written,
\(
\frac{1}{\beta(s)} = o(e^{\lambda s})\\
\)
But we can say something stronger. We know that \( \tau^n_\lambda(s) = \mathcal{O}(e^{-\lambda s}) \) and this means that,
\(
\frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} = \mathcal{O}(e^{-\lambda s})\\
\)
As this is the term in the logarithm, and in order for \( \tau^{n+1}_\lambda(s) \) to tend to zero exponentially the term in the logarithm has to as well. Therefore we can say that,
\(
\frac{1}{\beta(s)} = \mathcal{O}(h(s))\\
\)
Where for any \( \delta > 0 \) we have \( e^{-\delta s} h(s) \to 0 \) as \( \Re(s) \to \infty \). Which tells us that,
\(
|\beta(s)| \ge \frac{A}{h(s)}
\)
Which informs us asymptotically how \( \beta(s) \) approaches zero at infinity... slower than any exponential. This should help us with the limit; in the end, and making sure we stay away from zero when defining our function \( F_n(s) = \log^{\circ n} \beta(s+n) \).
As to the second part, when we actually go about and take this limit we want \( \lambda \) to be a function of \( s \). And in this sense, we want \( \beta_{\lambda(s)}(s) : \{s \in \mathbb{C} \,:\, |\arg(s)| < \theta\} \to \mathbb{C} \) where \( (s,\lambda(s)) \in \mathbb{L} \) and \( \lambda : \mathbb{R}^+ \to \mathbb{R}^+ \). This would mean for every imaginary part of \( s \) there is a large enough \( N \) in which \( F_n(s) \) exists for \( n > N \). This is where we'll need a clever Riemann mapping on \( \lambda \) over some simply connected domain \( \mathbb{D} \subset \mathbb{L} \); which I'm not sure how to do; especially as this requires a two variable ideation.
Forget I said anything about periodic functions, I don't think that's necessary here. I'm going to keep plugging and playing with this, but the more I fiddle the more this has none of the problems \( \phi \) has.
Regards, James.
EDIT:
Some even more evidence this is doable. Let \( \lambda(s) \) be an implicitly defined function such that,
\(
\log \beta_{\lambda(s+1)}(s+1) - \log \beta_{\lambda(s)}(s+1) = \log(1+ e^{-\lambda(s) s})\\
\log \beta_{\lambda(s+1)}(s+1) - \beta_{\lambda(s)}(s) + \log(1+e^{-\lambda(s)s}) = \log(1+ e^{-\lambda(s) s})\\
\log \beta_{\lambda(s+1)}(s+1) = \beta_{\lambda(s)}(s)\\
\)
Which would require an implicit solution \( \lambda(s) \); hinting that the logarithm trick might work for correctly chosen \( \lambda \); as it's idempotent in this case.