The following materials are exercise 8.10 and 8.11 from the functional analysis book by Prof. Brezis. The questions arise from whether we can well define a composition of two functions or not. From Corollary 8.11 we know if , is well defined. However bad things might happen if . For instance, , then might be bad on the set . I thought we could still define things in the sense of distribution, but Prof. Brezis said in this special case we could have something more than that, because of the following propositions. The proof is not hard by following the hints, but the idea is really clever (IMO)!

**Proposition 1** Let and with . Then a.e. on the set .

*Proof:* Fix a function such that , for some constant , and

Set

Since , converges to 0 uniformly.By Corollary 8.11(differentiation of a composition) , we then have . Hence . Define

We can see that as a.e. on by the construction of . And , hence in . By Lebesgue dominated convergence theorem we will have in . According to the definition of weak derivative, we have

For LHS we have in , for RHS uniformly. Passing the limit we will have

Which implies a.e. And hence a.e. on .

**Proposition 2** Let and assume that with , for some constant . Let . Then for every , we will have and

Furthermore, if a sequence in such that in , then in .

*Proof:* Let be a sequence of mollifiers, define , then , and . Hence . Denote , by Corollary 8.11 again, we have and . Since , , uniformly, and pointwise on the set where . Define

Then

The first term goes to 0 by Lebesgue dominated convergence theorem since we have a uniform bound for and , while the second term is identically zero by the previous proposition. Therefore in . By the definition of weak derivative, passing the limit again we will have

This implies and .

Now assume we have a sequence in and converges to in , we want to show in .

Assume does not converges to in , which means there exist an and a subsequence such that . Since in , there exists a sub-subsequence such that a.e. with , a.e. with . It is easy to verify that in , which leads to the contradiction.

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