# A clarification about BV functions.

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 ?

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

## 1 Answer

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).