Re: how to prove that f^2+f ' ^2 <=1 if ...

From: Peter L. Montgomery (Peter-Lawrence.Montgomery_at_cwi.nl)
Date: 08/26/04 

Date: Thu, 26 Aug 2004 04:37:46 GMT

In article <200408191114.i7JBEjf18741@proapp.mathforum.org>
eric.detourre@laposte.net (eric detourre) writes:
>Here is the following problem :
>
>Prove that : f^2 + f ' ^2 <= 1 when f is twice differentiable over R
>and : f^2 <= 1 and : f ' ^2 + f " ^2 <= 1 (1 is greater than the sum
>of the squre of the first and seconf derivative of f).
>
>I don't have the slightest idea on how to assert this, and this result
>(which happened to be an exercise in a french engineering school)
>may be false (but i found no counter-example) as it is false
>when replacing R with any subset of R.

    Denote g(x) = f(x)^2 + f'(x)^2. The function g
is bounded by f(x)^2 + f'(x)^2 + f''(x)^2 <= 2.

    If we can argue that there exists x0 with g(x0) = sup g(x)
(as would exist if g were defined on a compact set)
then we can argue that g'(x0) = 0. However

       g'(x0) = 2 f'(x0) f(x0) + 2 f''(x0) f'(x0)
              = 2 f'(x0) * (f(x0) + f''(x0))

Whether f'(x0) = 0 or f(x0) + f''(x0) = 0, we find g(x0) <= 1.
By selection of x0, we find g(x) <= 1 everywhere.

     Looking at this argument more carefully,
let x0 be any real value for which g(x0) > 1. Then f'(x0) <> 0.
Assume f'(x0) > 0 (other case is symmetric). Then either

       1) g achieves a local maximum at some x1 > x0.
          We can apply the earlier argument to x1.
or
       2) g is monotonic non-decreasing for x >= x0.
          Let C = sqrt(g(x0) - 1) > 0. Then, for x > x0,

               1 + f'(x)^2 >= g(x) >= g(x0) = 1 + C^2.

          Since f' is continuous, f'(x) >= C for x > x0.
          This implies f(x) >= f(x0) + C*(x - x0)
          for x >= x0, which is inconsistent with f(x)^2 <= 1. 

-- 
During a wedding ceremony, members of the audience cried.  
Scientists examined the tears.  They determined the liquid was eye dew.
pmontgom@cwi.nl    Microsoft Research and CWI      Home: Bellevue, WA