0
$\begingroup$

The Riesz representation is very useful in study BV space. There is a lot of version of it and one of the good one can be found in this book, page 49.

Here I come up with a question which has similar assumption as Riesz representation but different in some way, and I am wondering that could I make some conclusion similar to Riesz representation.

Suppose $\mu\in\mathcal M(I)$ where $I\subset \mathbb R^1$ is open interval and $\mathcal M(I)$ denotes all fnite Rodan measures. Assume $$ \sup\left\{\int_I \phi'\,d\mu, \,\phi\in C_c^\infty(I),\,\|\phi\|_{L^{\infty}(I)}\leq1,\,\|\phi'\|_{L^\infty}\leq 1\right\}<\infty \tag 2 $$ Then I am wondering that can I have some similar result compare to Riesz representation? Since in the integration is $\phi'$ but not $\phi$ and the constraint is on $W^{1,\infty}$ norm and it is much stronger then the $L^\infty$ norm in usual Riesz representation, I am not expecting I will have same result as in Riesz. But would it be possible to conclude something similar to integration by parts formula? i.e., to have another Radon measure $\nu$ such that $$ \int_I\phi'\,d\mu = -\int_I\phi\,d\nu\tag 1 $$ Any help is really welcome!

Improve this question
$\endgroup$
12
  • $\begingroup$ What form of Sobolev norm are you using, i. e. how are the sup-norm of the function and it's derivative combined? $\endgroup$
    – Dirk
    Commented Jul 17, 2015 at 13:38
  • $\begingroup$ @Dirk Let me just put $\|\phi\|_{L^\infty}\leq 1$ and $\|\phi'\|_{L^\infty}\leq 1$. This question actually come from the study of $TGV^2$ space. $\endgroup$
    – JumpJump
    Commented Jul 17, 2015 at 13:41
  • $\begingroup$ One more thing: Your functional is not linear but only positively homogeneous, so what kind of result do you expect? $\endgroup$
    – Dirk
    Commented Jul 17, 2015 at 13:46
  • 1
    $\begingroup$ This defines a distribution ("generalized function") that is the derivative of the measure $\mu$ (which is itself a distribution. $\endgroup$
    – Deane Yang
    Commented Jul 17, 2015 at 13:53
  • 1
    $\begingroup$ $\nu$ is not a measure. It is a distribution (as defined by Laurent Schwartz). See, for example, en.wikipedia.org/wiki/Distribution_(mathematics) $\endgroup$
    – Deane Yang
    Commented Jul 17, 2015 at 13:58

0

Reset to default

You must log in to answer this question.