As known, bounded variation (BV) functions are not continuous, but evidently higher order BV functions on corresponding dimension have continuous representative. As a consequence, .

DefinitionA function , if , and the nth order distributional derivative is a finite Radon measure.

TheoremIf , then has a continuous representative.

*Proof:* Note that is dense in , i.e. for any , there exists a sequence such that in , and . (See for example, book by Evans and Gariepy). So it suffice to show for all ,

Indeed,

This gives us the desire inequality. Then by density result we know that admits continuous representative for every element.

Corollary.

RemarkIf the dimension is greater than the order of BV function, then the above theorem will fail. For example, let be defined in a neighborhood around 0, with smooth boundary in . and . So . Then we can extend on , since is smooth (See for example, PDE book by Evans). But then as , which doesn’t admit a continuous representative.