# Absolutely continuous and integration by parts

Let $f,g\in L^1([0,1])$. Suppose for all $\phi\in C^\infty[0,1]$ such that $\phi(0)=\phi(1)$ we have $$\int_0^1f(t)\phi^{\prime}(t)dt=-\int_0^1g(t)\phi(t)dt$$ Show that $f$ is absolutely continuous.

I tried to prove as this: absolutely continuous. However, the problem is not exactly same as we now require $\phi$ is smooth. So I want to know how to modify the construction of such $\phi_n$ to get the conclusion. Thanks for any hint.

• You can use almost the same construction as in the answer, but you have to use a Lipschitz function. This can be computed by using a smooth cut-off function. – gerw Jan 19 '16 at 20:26
• Yes. I tried leave a space such as $(1/n,2/n]$ and $(x-1/n,x]$ and etc to make function $\phi_n$ smooth. However, we also need to let $\int_{(1/n,2/n]}f\phi_n^{\prime}=0$ and etc. I am not sure how to write down them explicitly or show that such $\phi_n$ exist for all $f$. – Syoung Jan 19 '16 at 20:35