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




Re: Unbounded second derivative
Posted:
Dec 4, 2012 7:00 AM


On 03122012 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



