Someone has a proof that the anti-derivative of a piecewise function does not exist

-----Original Message----- 
From: Barton Willis
Sent: Friday, December 24, 2010 8:53 AM
To: Richard Hennessy
Cc: Maxima List
Subject: Re: [Maxima] Someone has a proof that the anti-derivative of a 
piecewise function does not exist

(1) "piecewise" describes a way of defining a function, not a property of a 
function.
    A function might be piecewise continuous or piecewise defined, but never 
just
    piecewise.

(2) Some definitions require that an antiderivative be differentiable 
everywhere,
    but some do not:

      Let f : [a,b] --> R be Riemann integrable and let F  : [a,b] --> R. We 
say F is an
      antiderivative of f provided (i) F is continuous on [a,b] and (ii) 
F'(x) = f(x) for all x
      such that f is continuous at x.

   Using this definition, an antiderivative of the unit step function is x 
in R |--> max(0,x).
   The function x in R |--> max(0,x) is continuous on R, but not 
differentiable at 0.

   Some qualification about the function f is needed--otherwise every 
continuous function
   would be an antiderivative of a nowhere differentiable function, for 
example.

--Barton

-----maxima-bounces at math.utexas.edu wrote: -----


>I found this on the web.
>
>http://www.mathhelpforum.com/math-help/f6/showing-there-no-anti-derivative
>-piecewise-function-154607.html
>
>Curiously the proof is wrong.

Oh.  I only make the requirement that F is differentiable almost everywhere 
in [a,b], meaning only a finite number of exceptions are allowed.

Rich