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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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