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From: Maury Barbato on 11 May 2010 01:56
Hello,
let R be the field of real numbers and f:(c,+inf) -> R^n
a twice-differentiable function, where c is a real number
and n an integer greater than 1.
Let M_0, M_1, M_2 be the least upper bounds of |f(x)|,
|f'(x)| and |f''(x)|, respectively, on (c, +inf).
Can we have (M_1)^2 > 4*(M_0)*(M_2) for some f?

Thank you very much for your attention.
My Best Regards,
Maury Barbato

PS The question is from Rudin, Principles of Mathematical
Analysis. If n = 1, the answer is negative, that is
we have (M_1)^2 <= 4*(M_0)*(M_2) for every f. This is
the proof. If h>0, then Tayolr's theorem shows

f'(x) = [f(x+2h) - f(x)]/(2h) - h*f''(y),

for some y in (x, x+2h). Hence

|f'(x)| <= h*M_2 + (M_0)/h,

and since h is arbitrary, you obtain, choosing
the minimum value of h*M_2 + (M_0)/h over (0,+inf),
(M_1)^2 <= 4*(M_0)*(M_2).
From: David C. Ullrich on 11 May 2010 08:17
On Tue, 11 May 2010 05:56:57 EDT, Maury Barbato
<mauriziobarbato(a)aruba.it> wrote:

>Hello,
>let R be the field of real numbers and f:(c,+inf) -> R^n
>a twice-differentiable function, where c is a real number
>and n an integer greater than 1.
>Let M_0, M_1, M_2 be the least upper bounds of |f(x)|,
>|f'(x)| and |f''(x)|, respectively, on (c, +inf).
>Can we have (M_1)^2 > 4*(M_0)*(M_2) for some f?
>
>Thank you very much for your attention.
>My Best Regards,
>Maury Barbato
>
>PS The question is from Rudin, Principles of Mathematical
>Analysis. If n = 1, the answer is negative, that is
>we have (M_1)^2 <= 4*(M_0)*(M_2) for every f.

There's a little trick that allows one to deduce lots of
things about vector-valued functions from the
corresponding result for scalar-valued functions.

Fix a unit vector v (ie |v| = 1) and define

g(t) = <f(t), v>,

the standard scalar product or "dot product".

Note that |g(t)| <= M_0 and |g''(t)| <= M__2,
hence |g'(t)|^2 <= 4 M_0 M_2. But

|f'(t)| = sup_v |g'(t)|.


>This is
>the proof. If h>0, then Tayolr's theorem shows
>
>f'(x) = [f(x+2h) - f(x)]/(2h) - h*f''(y),
>
>for some y in (x, x+2h). Hence
>
>|f'(x)| <= h*M_2 + (M_0)/h,
>
>and since h is arbitrary, you obtain, choosing
>the minimum value of h*M_2 + (M_0)/h over (0,+inf),
>(M_1)^2 <= 4*(M_0)*(M_2).

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