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

**Definition** A function , if , and the nth order distributional derivative is a finite Radon measure.

**Theorem** If , 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** .

**Remark** If 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.

### Like this:

Like Loading...