Start by taking the function , and define a sequence of functions such that,

We aren't too concerned about where is holomorphic at the moment. We know that,

Assume that,

And we are reduced to the equation,

Now we can calculate inductively on . Assuming we know for ; then we can find by,

Of which this limit necessarily exists; and constructs a sequence . And now our problem is reduced to solving .

We are also given the quick relation,

Then since we know that we know that,

Because, , this implies that,

This gives us hope we can devise a normality condition on as it controls how fast can dip to zero; which is the exact problem we had with . This can be more aptly written,

But we can say something stronger. We know that and this means that,

As this is the term in the logarithm, and in order for to tend to zero exponentially the term in the logarithm has to as well. Therefore we can say that,

Where for any we have as . Which tells us that,

Which informs us asymptotically how 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 .

As to the second part, when we actually go about and take this limit we want to be a function of . And in this sense, we want where and . This would mean for every imaginary part of there is a large enough in which exists for . This is where we'll need a clever Riemann mapping on over some simply connected domain ; 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 has.

Regards, James.

EDIT:

Some even more evidence this is doable. Let be an implicitly defined function such that,

Which would require an implicit solution ; hinting that the logarithm trick might work for correctly chosen ; as it's idempotent in this case.

We aren't too concerned about where is holomorphic at the moment. We know that,

Assume that,

And we are reduced to the equation,

Now we can calculate inductively on . Assuming we know for ; then we can find by,

Of which this limit necessarily exists; and constructs a sequence . And now our problem is reduced to solving .

We are also given the quick relation,

Then since we know that we know that,

Because, , this implies that,

This gives us hope we can devise a normality condition on as it controls how fast can dip to zero; which is the exact problem we had with . This can be more aptly written,

But we can say something stronger. We know that and this means that,

As this is the term in the logarithm, and in order for to tend to zero exponentially the term in the logarithm has to as well. Therefore we can say that,

Where for any we have as . Which tells us that,

Which informs us asymptotically how 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 .

As to the second part, when we actually go about and take this limit we want to be a function of . And in this sense, we want where and . This would mean for every imaginary part of there is a large enough in which exists for . This is where we'll need a clever Riemann mapping on over some simply connected domain ; 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 has.

Regards, James.

EDIT:

Some even more evidence this is doable. Let be an implicitly defined function such that,

Which would require an implicit solution ; hinting that the logarithm trick might work for correctly chosen ; as it's idempotent in this case.