## One dimensional trace of a Sobolev function

Check that the mapping from to is a continuous linear functional on . Deduce that there exists a unique such that

Show that is the solution of some differential equation with appropriate boundary conditions and compute explicitly.

*H. Brezis, Functional Analysis, Ex 8.18*

**Solution:** Theorem 8.2 from the same book states that for with ( is an interval) then there exists a unique function such that a.e. on and

The proof is presented in the book, but the main idea is that every function has a continuous representative and we have a nice integral representation for . (Note that this does not happen in higher dimensions) In the following I will use always the continuous representative of a function in . Suppose we have with . Using the above formula we get

and therefore we have

Using Holder’s inequality we get

Integrating the last inequality on and using again Holder’s inequality we obtain

This last inequality proves that the considered application is continuous. The linearity is trivial to prove, therefore is a continuous functional. Because is a Hilbert space we can apply Riesz representation theorem to conclude that there exists a unique such that

It’s not hard to see that is the solution of

which is not hard to solve.