totally monotonic
#1
As Jay serveral times mentioned there is a condition that all odd derivatives are positive. This condition occured also in Szekeres paper [1] however in a slightly different context:
Definition. We call \( f(x) \) totally monotonic at \( x_0 \) if it has derivatives of any order and \( (-1)^{k+1}f^{(k)}(x_0)>0 \) for every \( k>0 \).

Then he shows that if the inverse of a function \( f \) (\( f \) real analytic for \( x\ge 0 \), \( f(x)>x \), \( f'(x)>0 \) for \( x>0 \) and \( f(x)=x+ax^2+\dots \), \( a>0 \)) is totally monotonic then the regular Abel function is also totally monotonic and is uniquely determined by this property.

\( x\mapsto e^x-1 \) meets the criteria and its inverse is \( x\mapsto \log(x+1) \) and is totally monotonic. Hence the regular Abel function is also totally monotonic.

In our case however the situation is a bit different. The function \( e^x \) has no fixed point. \( \text{slog}_e \) is an Abel function for it but is not totally monotonic, but the inverse of slog is (/seems to be) totally monotonic.

I would bet there is no proof for the uniqueness claim by total monotonicity, though it sound quite plausible.

[1] G. Szekeres, Fractional iteration of exponentially growing functions, 1961.
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Messages In This Thread
totally monotonic - by bo198214 - 08/23/2007, 08:43 PM
RE: totally monotonic - by Catullus - 07/11/2022, 06:54 AM
RE: totally monotonic - by bo198214 - 07/13/2022, 07:15 PM
RE: totally monotonic - by JmsNxn - 07/13/2022, 11:13 PM
RE: totally monotonic - by Gottfried - 07/14/2022, 08:06 AM



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