Theorem Let be a closed ideal of a C* algebra , then for any self-adjoint element , there exists an such that
Proof: Since is a closed ideal, is also a C* algebra and one can define a natural projection homomorphism , with . Fix an element , define a continuous function as follow:
Let , then is identical on . By continuous functional calculus, . And since can be approximated by polynomials, we know . The spectral radius , so we know is identical on the spectrum of , which implies . Therefore is in the kernel of , and hence in the ideal . Since and by the Gelfand Naimark isomorphism theorem, we know and hence . So is the we are looking for.
Note: this is a homework exercise given by Dr. Carlen.
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 ,
This gives us the desire inequality. Then by density result we know that admits continuous representative for every element.
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.