Let $v,g \in L^{p}(a,b)$ such that $v(x)-v(y)=\int^x_yg(t)dt$

Question

This exercise is about the continuity of functions in Sobolev spaces in dimension one.

Let $v \in L^{p}(a,b)$ and $g \in L^{p}(a,b)$ such that $v(x)-v(y)=\int^x_yg(t)dt$ and this $\forall x,y \in (a,b)$

Show that $v \in W^{1,p}$ and that $v'=g$ in the sense of distributions.

So far I can only prove that $v$ is continuous using Holder's identity, but I don't know how to proceed.

Any hints? Thanks.

Answer 1

Let $c \in (a, b)$. We can take $$v(x) = \int_c^x g(t)dt.$$ We have, for $\varphi \in C^\infty_0$ \begin{align} \int_a^b v \varphi' &= \int_a^b \left(\int_c^x g(t)dt\right) \varphi'(x)dx \\ &= -\int_a^c dx\int_x^cg(t)\varphi'(x)dt + \int_c^b dx\int_c^x g(t)\varphi'(x)dt\\ &= -\int_a^c g(t)dt\int_a^t\varphi'(x)dx + \int_c^b g(t)dt\int_t^b \varphi'(x)dx\\ &= -\int_a^b g(t)\varphi(t)dt \end{align} and you deduce the desired result. The third equality comes from Fubini's theorem.