Sufficient conditions that guarantee the integrate by part formula?

Question

Suppose, for simplicity, that $f,g$ are functions defined on $[0,1]$. Under suitable hypotheses on $f$ and $g$, the integrate by parts formula yields $$ \int_0^1 f'(t)g(t)dt = f(t)g(t)|_{t=0}^1 - \int_0^1 f(t)g'(t)dt\ . $$ If $f$ and $g$ are assumed to be continuously differentiable then the formula holds, but this condition proves to be too strict in most cases. I know that absolutely continuous is a sufficient condition but is there any strictly weaker conditions that still give the same result?

More generally, what I really want to know is for $f,g$ that belong to a Sobolev space, says $H^k([0,1])$, is the formula $$ \int_0^1 f(t)g^{(k)}(t)= \big[ \sum_{j=0}^{k-1}(-1)^jf^{(j)}(t)g^{(k-j-1)}(t)|_{t=0}^1\big] + (-1)^k\int_0^1f^{(k)}(t)g(t)dt $$ always justified and why?

I don't have any good book about Sobolev space at hand. I would really appreciate if someone could suggest me a good book on the subject.

Answer 1

The following result tells that $u \in H^1$ is not less restrictive than absolutely continuous.

Theorem: Let $\Omega \subset \mathbb{R}$ be an open set, let $1\le p < \infty$. Then $u \in W^{1,p}(\Omega)$ if and only if it admits an absolutely continuous representative whose classical derivative belongs to $L^p(\Omega)$.

A proof of this theorem can be found, for example, in Sobolev Spaces by G. Leoni.