From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Another look at a Putnam problem
Date: 20 Jan 2000 23:03:15 GMT
Newsgroups: sci.math
Summary: [missing]

During the discussion of the 1999 Putnam exam questions, many interesting
responses were generated to problem B4: given a C^3 function on the real
line with all derivatives positive and  f'''<f  we were asked to show
f' < 2 f . The most interesting feature (to me) is that the proposed
inequality is not optimal. Rob Johnson provided a solution and added these
comments:

>Note that the only property of f''' we used was that f''' < f; we used
>neither f''' > 0 nor f'' > 0, yet we got a constant better than 2.

(He got 1.651 for the constant.)

My analysis also suggested better results should be true: that perhaps
 { f>0, f'>0, f'''<f } ==> { f' <= f }

At the time, I suggested a method of approach: to show that the functions
under consideration admit a representation

 f(t) = integral( P(u) exp( u t) du,  u \in [0,1] )

for some positive kernel  P  on the unit interval (taking a Stieltjes 
integral or allowing a measure with point masses or something). All these
functions meet the stated conditions, and any such function satisfies the
desired conclusion.

This is very intriguing but cannot really be true, since it would suggest
that much more is actually true: that all higher derivatives are also
positive and each is less than the ones before it.


Well, I never really got very far with this idea, except to generalize
it a bit: For what values of  (k,l,m,n)  can we show that
(*) { f>0, f'>0, f^(k) < f^(l) } ==> { f^(m) <= f^(n) } ?

In order to construct interesting examples (and, perhaps, to prove something)
I found it useful to refer to  h = f'/f, a well-defined positive smooth
function on the line. The ordering (and positivity) of the derivatives of  f
is then equivalent to the ordering (and positivity) of these functions:
  1
  h
  h' + h^2
  h'' + 3 h h' + h^3 
  h''' + 4 h h'' + 3 h'^2 + 6 h' h^2 + h^4
  ...

It's easy to generate some negative examples to conjecture (*) when 
m > k. For example, when (k,l)=(1,0) we may take  
h = 1/2 + (1/pi) arctan(4 x)  and discover that  h'+h^2  isn't even less
than  1,  much less being less than  h. So it seems more productive to
restrict attention to (*) in cases with  m <= k. (I've also looked only
at  k > l  and  m > n,  in keeping with the Putnam problem.)

The conjecture (*) is true in the case  (k,l)=(2,1), (m,n)=(1,0), and
thus when (k,l)=(2,1) and (m,n)=(2,0). I can also prove it when (k,l)=(2,0)
and (m,n)=(1,0), but the other case (k,l)=(2,0) and (m,n)=(2,1) is not
true; try  h=1/sqrt(1+exp(-2x)).

The next interesting cases are those with  k=3; this includes the Putnam
problem. Can anyone prove any of the various cases, or provide counterexamples?

One thing is certainly true: the proposed integral respresentation 
_must fail_ because there are smooth functions having  f'''<f but not
satisfying the additional inequalities on  f^(iv)  which would be a
consequence of such a representation. Behold:

	h = 2/3 - 1/4*sin(x/3) - 1/13*sin(2*x/3)

This makes  1 > h > h'+h^2 > h"+3hh'+h^3  but the next inequality does
not hold; that is, we may construct examples with  f>0, f'>0, and  
f''' < f'' < f' < f  but  f^(iv) not even less than  f'', much less f''' .
In particular, this  f  is certainly not expressible as an 
integral \int P(u) exp(ut)  as above. I don't know what does characterize
the functions representable in this way (a characterization would have
to include more assumptions, e.g. that _all_ derivatives are positive).

dave