0
$\begingroup$

From definition, a locally integrable $ u \in BV(\Omega) $ if its distribution derivative is given by a signed Radon measure. That is there exists $ \mu $ such that for any $ \phi \in C^\infty_c(\Omega) $ we have $$ -\int_\Omega u\phi' dx = \int_\Omega \phi d\mu $$ Now for any $ u \in L^1_{loc}(\Omega) $ it is clear that $ \Lambda(\phi) = \int_\Omega u\phi' dx $ is in dual of $ C^\infty_c(\Omega) $ hence from Riesz Representation Theorem such a signed measure $\mu$ should always exist, this implies $ L^1_{loc}(\Omega ) = BV(\Omega) $. Where am I mistaken ?

$\endgroup$
2
  • $\begingroup$ The Riesz Representation theorem is only about the dual of $C_c(\Omega)$. $\endgroup$
    – gerw
    Commented May 9, 2013 at 12:52
  • $\begingroup$ Oh yes...Thank you $\endgroup$
    – smiley06
    Commented May 9, 2013 at 15:01

1 Answer 1

1
$\begingroup$

It is true that for any $u\in L^1_{\rm loc}$ the formula $\Lambda(\phi)=\int_{\Omega} u\phi'\,dx$ defines a linear functional on the space of test functions. The functional is controlled by the norm of $\phi'$ and therefore is a distribution of order 1. However, this makes it more singular (in general) than measures, which are distributions of order $0$.

(CW answer to fill this box).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .