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




Re: Unbounded second derivative
Posted:
Dec 3, 2012 11:31 AM


On Mon, 03 Dec 2012 11:17:24 +0000, José Carlos Santos <jcsantos@fc.up.pt> wrote:
>Hi all, > >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.
Yes, it's probably very simple. Let's see. Assume that f'' is bounded. The MVT shows that
f'(x)  f'(y) <= c xy.
In particular f' is uniformly continuous. And that's enough to get a contradiction:
Prop. If f(x) > 0 as x > infinity and f' is uniformly continuous then f'(x) > 0.
Pf. Assume not. Then there exist x_n > infinity and eps > 0 with f'(x_n) > eps. WLOG
(*) f'(x_n) > eps.
Now since f' is uniformly continuous there exists delta > 0 such that if
I_n = (x_n  delta, x_n + delta)
then
f'(t) > eps/2 for all t in I_n.
Now if we say I_n = (a_n, b_n) it follows by the MVT that f(b_n)  f(a_n) > eps*delta.
Hence
max(f(a_n), f(b_n)) > eps*delta/2,
which implies that f(x) does not tend to 0 at infinity. QED.
Yes, it's very simple.
> >Best regards, > >Jose Carlos Santos



