A Tricky Weak Derivatives question

Asked Aug 25 '14 at 11:37

Active Aug 25 '14 at 11:52

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I recently came across the following statement and am having trouble proving it correct. I wonder if you could help.

Given a weak derivative, $x'$, there exists an absolutely continuous representative of $x$ (which we shall still call $x$, for convenience), such that $x(b)-x(a)= \int_{a}^{b}x'(v)\mathrm{d}v$, $\forall a,b \in\Bbb R^{+}$. Please, this has me somewhat stumped.

edited Aug 25 '14 at 11:52

asked Aug 25 '14 at 11:37

Maybe somewhat related to this question, which has, sadly, not received much attention. – Jean-Claude Arbaut Aug 25 '14 at 11:43
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possible duplicate of Characterisation of one-dimensional Sobolev space – Tomás Aug 25 '14 at 11:56
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Please be precise about your particular definition of 'weak derivative.' – Disintegrating By Parts Aug 25 '14 at 19:56
Indeed, so it seems, that it is a repeat as Tomas cited. Forgive my ignorance, but is there any snags as we are dealing with the representative (?); possibly not. Thank you Tomas. – Aug 26 '14 at 12:51
Given, I am new to this subject (but wish to learn), perhaps Tomas night be kind enough to comment which Lemma/Thm he used (from Brezis presumably) to generate his solution in the link. Thank you – Aug 26 '14 at 15:56
Dear MrNemo, when you want talk to someone in particular, you have to write @nickname, so the guy will be notified. Now, concerning your request, could you be more specific please? Is this related with your last question that I alread answered? – Tomás Aug 28 '14 at 15:19

A Tricky Weak Derivatives question

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