Re: Differentiability of f everywhere but |f| nowhere, f: R - > R?

On 23 Nov 2003 21:40:33 -0800, [EMAIL PROTECTED] (painch1207)
wrote:

>I ran across an interesting problem in Spivak's Calculus on Manifolds
>- I am unaware of whether or not it is solvable.  Maybe someone
>perusing this board will know the answer.
>
>The problem:
>
>Exhibit a differentiable function f:R->R such that the function |f|
>defined by
>
>|f|(t) = |f(t)|
>
>is not differentiable.  (Spivak, Calculus on Manifolds 2-13d)
>
>It is a little bit mystifying to me exactly what he is looking for
>here.  Unless the problem is requiring that the derivative exists
>nowhere on R it seems uniquely unchallenging (i.e. very unlike a
>Spivak problem, for the most part).
>
>So I was wondering if anyone knew of such a function that is as
>described above, with the assumption that |f|'(t) exists nowhere on R.

It's clear that there is no such function: If f(t) <> 0 then |f| is
either f or -f in a neighborhood of t, so f is differentiable at t.
So the only possible example is identically zero, which is not
an example.

(For second I thought that the existence of such a function f:R->R^n
might be more interesting, but it's just as easy: If f(t) <> 0 then
there is a smooth function g such that |f| = g o f near t...)

>Full context of the problem follows, and it seems to me that they
>should be related to this.  Just assume each of parts (a),(b), and (c)
>to be true (or solve them if you wish but I am only concerned with the
>above - part (d)).
>
>2-13 Define IP:R^n X R^n -> R by IP(x,y) = <x,y>
>
>a)  Find D(IP)(a,b) and (IP)'(a,b) [Note: here Spivak is making a
>distinction between the linear transformation D(IP)(a,b) and it's
>matrix with respect to the usual basis of R^n, which he generally just
>calls f'(a), in this case(IP)'(a,b)]
>
>b)  If f,g:R -> R^n are differentiable and h:R->R is defined by 
>h(t) = <f(t),g(t)>, show that h'(a) = <f'(a)^T,g(a)> + <f(a),g'(a)^T>
>(here the T indicates the transpose).
>
>c)  If f:R -> R^n is differentiable and |f(t)| = 1 for all t, show
>that
><f'(t)^T,f(t)> = 0


David C. Ullrich