User blog:LittlePeng9/Fast growing hierarchy of analytic functions

Extending the hyperoperators to all real or complex values is a known open problem. Some time ago I had a somewhat converse idea on how to approach this - instead of starting with a fast growing function(s) on natural numbers and extending it to more inputs, we can directly construct a sequence of fast growing functions defined for all complex values. I do not claim this is of as much interest as extending hyperoperators, but might nevertheless be interesting for some.

Suppose we have chosen a system of fundamental sequences for countable limit ordinals up to some bound. We now construct a sequence \(f_\alpha:\mathbb C\rightarrow\mathbb C\) which satisfies the following propertes:


 * 1) \(f_\alpha\) is an ,
 * 2) \(f_\alpha(z)\in(0,\infty)\) for \(z\in\mathbb R\setminus\{0\}\) and \(f_\alpha(0)=0\),
 * 3) \(f_\alpha\) is increasing on \([0,\infty)\),
 * \(|f_\alpha(z)|\leq|z|\) for \(|z|<1\).

The definition of the hierarchy is as follows:


 * 1) \(f_0(z)=z^2\),
 * 2) \(f_{\alpha+1}(z)=\sum_{n=1}^\infty f_\alpha^n\left(\frac{z}{2^n}\right)\), where \(f_\alpha^n\) means the usual function iteration,
 * 3) \(f_\alpha(z)=\sum_{n=1}^\infty f_{\alpha[n]}\left(\frac{z}{2^n}\right)\) for \(\alpha\) limit ordinal.

We prove that all these functions have properties 1-4 by transfinite induction on \(\alpha\). Everything is clear for \(\alpha=0\). Suppose now \(\beta=\alpha+1\) is a successor ordinal and \(f_\alpha\) satisfy the above properties. By simple induction, \(f_\alpha^n\) satisfy these properties as well. We show that the series \(\sum_{n=1}^\infty f_\alpha^n\left(\frac{z}{2^n}\right)\) converges absolutely and uniformly on disks - fix any \(R>0\) and arbitrary \(|z|\leq R\). Then, as long as \(2^n>R\), we have \(\left|f\left(\frac{z}{2^n}\right)\right|\leq\left|\frac{z}{2^n}\right|\), so by we can conclude the series converges uniformly, hence it converges to an analytic function, so \(f_{\alpha+1}(z)\) satisfies property 1. Properties 2 and 3 are straightforward and to see the property 4 we note, for \(|z|\leq 1\), \[|f_{\alpha+1}(z)leq\sum_{n=1}^\infty\left|f_\alpha^n\left(\frac{z}{2^n}\right)\right|\leq\sum_{n=1}^\infty\left|\left(\frac{z}{2^n}\right|=|z|.\]

Dealing with the limit case looks nearly identical.

Functions in these hierarchy are increasingly fast growing - indeed, for \(z>0\), \(f_{\alpha+1}(z)\geq f_\alpha^n\left(\frac{z}{2^n}\right)\), and we can show, though I won't bother to do it, that \(f_\alpha(z)\geq 2^{n+1}z\) for large \(z\), so we can also say \[f_{\alpha+1}(z)\geq f_\alpha^{n+1}\left(\frac{z}{2^{n+1}}\right)\geq f_\alpha^n(z)\). Similarly, we can see that \(f_\alpha(z)\) grows faster than any \(f_{\alpha[n]}\left(\frac{z}{2^n}\right)\), from which we can deduce (though not so immediately) \(f_\alpha(z)>f_\beta(z)\) for any \(\beta<\alpha\) and \(z\) large enough.