Date: Dec 4, 2012 7:00 AM
Author: Jose Carlos Santos
Subject: Re: Unbounded second derivative

On 03-12-2012 16:57, Rotwang wrote:

>>> This is probably very simple, but I can't see it. :-) Let f be a twice
>>> derivable function from [0,+oo[ into R such that:
>>>
>>> 1) lim_{x to +oo} f(x) = 0;
>>>
>>> 2) lim_{x to +oo} f'(x) does not exist.
>>>
>>> Prove that the function f'' is unbounded.

>>
>> Forget it! I've done it.

>
> I was half way through a solution when you posted your reply, and it's a
> fun problem so I may as well finish it:
>
> Since lim_{x -> oo} f'(x) != 0, there exists d > 0 such that at least
> one of the sets {x | f'(x) > 2d} or {x | f'(x) < -2d} is unbounded;
> wolog let's suppose the first one is. Let M > 0, and let X be such that
> |f(x)| < d^2/2M whenever x > X. There exists x > X such that f'(x) > 2d.
> If f'(x) >= d for all y in [x, x + d/M] then we would have f(x + d/M) >=
> d^2/2M, so there must exist y such that x < y < x + d/M and f'(y) < d.
> By the mean value theorem there exists z in [x, y] such that
>
> |f''(z)| = |(f'(y) - f'(x))/(y - x)| >= d/(d/M) = M.
>
> Since this is true for any M, f'' is unbounded.
>
> Did you find an easier way?


No. Here it is. Since lim_{x to +oo} f(x) = 0, there can be no a > 0
such that f'(x) >= a for each large enough _x_ and there can be no a < 0
such that f'(x) <= a for each large enough _x_. It follows from this
and from the fact that the limit lim_{x to +oo} f'(x) does not exist
that there are numbers _a_ and _b_ such that the equations f'(x) = a and
f'(x) = b has arbitrarily large solutions and we can assume wlog that
a > b > 0. Now, let a_1 be some number such f'(a_1) = a, let a_2 be the
first number after a_1 such that f'(a_2) = b, let a_3 be the first
number after a_2 such that f'(a_3) = a and so on. Then

|(f'(a_{n + 1}) - f'(a_n))/(a_{n + 1} - a_n)| = (*)

= (a - b)/(a_{n + 1} - a_n)

and if I prove that lim_n(a_{n + 1} - a_n) = 0, it follows that (*)
takes arbitrarily large values. Since (*) = |f''(x)| for some _x_, this
solves the problem.

But on any interval [a_n,a_{n + 1}] of length _d_, _f_ increases by b*d,
at least. Since lim_{x to +oo} f(x) = 0, it follows that the lengths of
the intervals must tend to 0.

Best regards,

Jose Carlos Santos