This is an exercise in Evans's PDE book:
$f\in W^{1,p}((0,1))$ is absolute continuous where $1\leq p < \infty $.
Try : By definition of sobolev space, $f$ has weak derivative $f'$
So $$\ f(x)= f(0) +\int_0^x f'(t)dt \tag{*}$$
By fundamental theorem of calculus, $f$ is absolutely continuous.
But I cannot show ($\ast $). How can we complete the proof ?
Best Answer
The context of the exercise is the very very beginning, i.e., there is only a definition of Sobolev space $W^{1,p}(0,1)$ in terms of weak derivatives and Lebesgue space $L^p(0,1)$, while there is nothing else, e.g., there are no Sobolev embedding theorems, there are no approximations by smooth functions. Of course, one could try to establish first the seems-to-be-required approximation by smooth functions in conjunction with the embedding $W^{1,p}\hookrightarrow C$, jumping somehow over the representation $(\ast)$. This would be an extremely complicated approach to a rather trivial but still most fundamental exercise. Under the circumstances, the only known correct simple approach to proving $(\ast)$ reduces to just two elementary exercises, being by itself most instructive for beginners. First, taking $$ g(x)\overset{\rm def}{=}\int\limits_0^x f'(t)\,dt, $$ find its weak derivative $g'$ using the standard definition of a weak derivative, namely, $$ \begin{align} \int\limits_0^1 g(t)\varphi'(t)\,dt= \int\limits_0^1 \varphi'(t)\,dt\int\limits_0^t f'(s)\,ds= \int\limits_0^1 f'(s)\,ds\int\limits_t^1 \varphi'(t)\,dt\\ =\int\limits_0^1 f'(s)\bigl(\varphi(1)-\varphi(s)\bigr)\,ds =-\int\limits_0^1 f'(s)\varphi(s)\,ds\quad\forall\,\varphi\in C_0^{\infty}(0,1), \end{align} $$ whence follows that $\,g'=f'\,$ or $\,(f-g)'=0\,$ a.e. on $\,(0,1)$. Second, taking a function $h\in L^1_{loc}(0,1)$ that does possess a weak derivative $\,h'=0\,$ a.e. on $\,(0,1)$, prove that there is some constant $C$ such that $\,h=C\,$ a.e. on $\,(0,1)$. To achieve this, start with the definition of $\,h'=0\,$, i.e., $$ \int\limits_0^1 h(x)\varphi'(x)\,dx=0\quad\forall\,\varphi\in C_0^{\infty}(0,1).\tag{1} $$ Choosing a function $\eta\in C_0^{\infty}(0,1)$ such that $\int_0^1 \eta(x)\,dx=1$, which is easy to construct explicily, consider a function $$ \Phi(x)=\psi(x)-\eta(x)\int\limits_0^1\psi(s)\,ds $$ with an arbitrary given $\psi\in C_0^{\infty}(0,1)$. It is clear that $\Phi\in C_0^{\infty}(0,1)$ and $\int_0^1 \Phi(x)\,dx=0$, hence a test function $$ \varphi(x)\overset{\rm def}{=}\int\limits_0^x\Phi(s)\,ds\in C_0^{\infty}(0,1).\tag{2} $$ Now, substituting $(2)$ in $(1)$ results in the identity $$ \int\limits_0^1 h(x)\psi(x)\,dx=\int\limits_0^1 C\psi(x)\,dx\quad \forall\,\psi\in C_0^{\infty}(0,1) $$ with the constant $$ C\overset{\rm def}{=}\int\limits_0^1 h(x)\eta(x)\,dx. $$ Therefore, the locally integrable function $\,h=C\,$, and hence $\,f=C+g\,$ a.e. on $\,(0,1)$. Since $g$ is continuous on $[0,1]$, the function $f$ can be redefined on a subset of Lebesgue measure zero on $[0,1]$ to become everywhewe continuous, in wich case $C=f(0)$ with $f$ satifying $(\ast)$. Interestingly enough, Paul Du Bois-Reymond was the first to have done the second exercise yet in 1879 for a function $h\in C(0,1)$, i.e., was the first to find a weak solution $h\in C(0,1)$ of the simplest differential equation $h'=0$. This remarkable but little known fact was dicovered by Jean Dieudonné (see pages 221-222 in his History of Functional Analysis).