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Re: Zero Function
Posted:
May 25, 2011 2:33 PM
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Stretto wrote:
>
> "William Elliot" wrote in message
> news:20110517020633.U8095@agora.rdrop.com...
>
> If f:R -> R is differentiable, f(0) = 0 and for all x, |f'(x)| <= |f(x)|,
> then f = 0.
>
> How is this proposition proved? Here's what's been suggested.
>
> If there's an a with f(a) /= 0, then
> there's some r between 0 and a (including end points)
> for which |f(r)| = max{ |f(x)| : x between 0 and a } and
>
> |f(r)| = |integral(0,r) f'(x) dx| <= integral(0,r) |f'(x)| dx
> <= integral(0,r) |f(x)| dx <= integral(0,a) |f(x)| dx
> <= integral(0,a) |f(r)| dx = |af(r)|.
> Since 0 < |f(r)|, 1 <= |a|.
>
> What now?
>
> ====
>
> Try induction,
>
> Assume the kth step is f^(k)(0) = 0 and |f^(k+1)(x)| <= |f^(k)(x)| and use
> that to prove that f^(k+1)(0) = 0 and |f^(k+2)(x)| <= |f^(k+1)(x)|. From
> this is follows from the base step that |f^(k)(0)| <= |f(0)|, from which it
> follows that f(x) = 0(the Taylor series of f at 0 is identically 0).
f(x) = e^{-1/x^2} when x =/= 0
= 0 else
has f^(k)(0) = 0 for all k, but it isn't the zero function.
--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting
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