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

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David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: Unbounded second derivative
Posted: Dec 3, 2012 11:31 AM
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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 |x-y|.

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





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