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Topic: Unbounded second derivative
Replies: 4   Last Post: Dec 4, 2012 7:00 AM

 Messages: [ Previous | Next ]
 Rotwang Posts: 1,685 From: Swansea Registered: 7/26/06
Re: Unbounded second derivative
Posted: Dec 3, 2012 11:57 AM

On 03/12/2012 16:08, José Carlos Santos wrote:
> On 03-12-2012 11:17, José Carlos Santos 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?

--
I have made a thing that superficially resembles music:

http://soundcloud.com/eroneity/we-berated-our-own-crapiness

Date Subject Author
12/3/12 Jose Carlos Santos
12/3/12 Jose Carlos Santos
12/3/12 Rotwang
12/4/12 Jose Carlos Santos
12/3/12 David C. Ullrich