As Sheldon has thoroughly convinced me of non-holomorphy of my tetration. I thought I'd provide the proof I have that it is on the line . I sat on this proof and didn't develop it much because I was too fixated on the holomorphy part. But, I thought it'd be nice to have a proof of .

Now, the idea is to apply Banach's fixed point theorem, but it's a bit more symbol heavy now. We will go by induction on the degree of the derivative. So let's assume that,

Where,

And is the sup-norm across some interval . As a forewarning, this is going to be very messy...

Now to begin we can bound,

And that next,

Now,

So, we ask you to put on your thinking cap, and excuse me if I write,

And by the induction hypothesis,

Which is because these terms are made up of finite sums and products of and these are said to be summable. Now the proof is a walk in the park.

Where, we've continued the iteration and set and for , and for (but we're tossing this away because we know it's differentiable). Therefore,

Of which, I've played a little fast and loose, but filling in the blanks would just require too much tex code.

EDIT: I'll do it properly as I correct my paper and lower my expectations of the result.

***********************

As to the second part of this post--now that we have out of the way, we ask if we can continue this iteration and get pentation. Now, will certainly be and so it's a well defined bijection of . So, first up to bat is to get another phi function,

This will be (it'll be a bit trickier to prove because we aren't using analytic functions, but just bear with me). And it satisfies the equation,

By now, I think you might know where i'm going with this.

And now I'm going to focus on showing this converges... Wish me luck; after being trampled by this holomorphy I thought I'd stick to where things are nice--no nasty dips to zero and the like...

Now, the idea is to apply Banach's fixed point theorem, but it's a bit more symbol heavy now. We will go by induction on the degree of the derivative. So let's assume that,

Where,

And is the sup-norm across some interval . As a forewarning, this is going to be very messy...

Now to begin we can bound,

And that next,

Now,

So, we ask you to put on your thinking cap, and excuse me if I write,

And by the induction hypothesis,

Which is because these terms are made up of finite sums and products of and these are said to be summable. Now the proof is a walk in the park.

Where, we've continued the iteration and set and for , and for (but we're tossing this away because we know it's differentiable). Therefore,

Of which, I've played a little fast and loose, but filling in the blanks would just require too much tex code.

EDIT: I'll do it properly as I correct my paper and lower my expectations of the result.

***********************

As to the second part of this post--now that we have out of the way, we ask if we can continue this iteration and get pentation. Now, will certainly be and so it's a well defined bijection of . So, first up to bat is to get another phi function,

This will be (it'll be a bit trickier to prove because we aren't using analytic functions, but just bear with me). And it satisfies the equation,

By now, I think you might know where i'm going with this.

And now I'm going to focus on showing this converges... Wish me luck; after being trampled by this holomorphy I thought I'd stick to where things are nice--no nasty dips to zero and the like...