Fonction smaller than iterated logs, but not constant

[fortin946]

I am looking for a continuous and increasing function f(x) which tends to infinity as x tends to infinity.
This function must have the property that it is eventually smaller than log_k(x)
(the k-th iterated logarithm) for all k>=1

I have no hint how to find such a function!
One of my problems is that log_k(x) tends to 0 when k tends to infinity...
then how is it possible ton find f(x) increasing?!

 

[Stephen Tashi]

My thoughts:

Suppose we have found an increasing function h(x) on the interval (0,x_k] such that h(x) is less than than the k th iterated log function log[k](x). Can we extend the definition of h(x) to a larger interval?

There should be some point x_(k+1) > x_k so that log[k+1] (x_(k+1)) = log[k] (x_k) + 1.0
(Eventually log[k+1] (x) must climb up well past the height where log[k] is now.)

You could extend the definition of h(x) by making it a line from h(x_k) to log[k+1](x_(k+1)).

Starting with k = 1, after N steps, you would have defined a function h(x) that is less than log[1+N] (x) on the interval from x_(1+N) to infinity.

To make the above idea precise you have to worry about a lot of details. For example, does repeating the above process extend the domain of definition to arbitrarily large values of x instead of reaching some sort of limit on the x-axis.

 

[fortin946]

Thanks for the idea! Do we call this function log*(x) ?

 

[zetafunction]

you can always think of an step function that is increasing but piecewise constant and that grows slower than the logarithm and iterated logarithm

 

[disregardthat]

Did you want a function f such that f at some point is smaller than log^k for each k (for which log^k(x) makes sense anyway)? This is different from being smaller than f being less than log^k at some point for each k at some point (the point depending on k that is). There are no function satisfying the first condition, but the latter is already dealt with here.

 

[g_edgar]

One of my problems is that log_k(x) tends to 0 when k tends to infinity...

No, it doesn't