4
$\begingroup$

Let us say I have an expression $$\frac{d}{dt}f = g$$ where the derivative is taken in a weak sense (of distributions). Can I integrate this from $0$ to $t_0$ and get $$f(t_0) - f(0) = \int_0^{t_0}g?$$

Does the derivative need to be a classical one to do that?

$\endgroup$
4
$\begingroup$

On the face of it, $g$ would have to be a function, and preferably in $L^1$ (or at least, in $L^1_{\mathrm{loc}}$). The condition for this turns out to be that $f$ be absolutely continuous. (Edited to add: Absolute continuity here means that for any $\varepsilon>0$ there is some $\delta>0$ so that, for any finite collection of non-overlapping intervals $[a_k,b_k]$ with $\sum_k|b_k-a_k|<\delta$, we have $\sum_k|f(b_k)-f(a_k)|<\varepsilon$.)

With a suitable interpretation, $g$ could be a signed measure, and $f$ a function of bounded variation. You need to be careful about the ends of the interval at points where $f$ has a jump discontinuity (or $g$ has a point mass), but these points are only countably many.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. So you say if $g \in L^1$ this should work. In the wiki, absolute continuity condition is written in an integral form. Do you have a citation for why the integral cancels out the derivative on the LHS though? $\endgroup$ – matt.w May 9 '13 at 19:48
  • $\begingroup$ I added the definition of absolute continuity. As for citations, I'm at home without access to my bookshelf, but any decent text on integration theory should have it. Royden, for example, has it for sure. Absolute continuity is essential: The Cantor function has derivative zero almost everywhere, yet it is not constant. $\endgroup$ – Harald Hanche-Olsen May 9 '13 at 20:01
  • $\begingroup$ Just to clarify: if $g \in L^1(0,t_0)$ then the weak derivative $f'$ is absolutely continuous, so the integration carries through. Thanks, I'll have a look at Royden $\endgroup$ – matt.w May 9 '13 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.