Let $u \in C^{1}(0,1)$. Prove that the following are equivalent:
(a) $u \in W^{1,1}(0,1)$
(b) $u' \in L^{1}(0,1)$ (where $u'$ means the derivative in the usual sense)
(c) $u \in BV(0,1)$
My try: For $(a) \implies (b)$: Since $u \in W^{1,1}(0,1)$, there exists $g \in L^{1}(0,1)$ such that $\int_{I}u\phi'=-\int_{I} g\phi=-\int_{I}u'\phi$ (since $u \in C^{1}(0,1)$), for all $\phi \in C_c^1(0,1)$. Thus $u'=g a.e$. Hence the result.
$(b) \implies (c)$: Here I am using the equivalent formulation of $BV(0,1)$ which says that there exists a $c \gt 0$ such that $|\int_{I}u\phi'| \le c||\phi'||_{L^1(I)}$. Now since $u \in C^{1}(0,1)$, $\int_{I}u\phi'=-\int_{I}u'\phi$. Taking modulus on both sides we have the result.
$(c) \implies (a)$: I am trying to use the equivalent formulation that $u$ has as distributional derivative as a bounded measure. Does it mean that it has derivative in terms of distribution?? If it says so, then I am done.
Thanks for the help!!