Someone has a proof that the anti-derivative of a piecewise function does not exist
Subject: Someone has a proof that the anti-derivative of a piecewise function does not exist
From: Barton Willis
Date: Fri, 24 Dec 2010 07:53:53 -0600
(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.