The vector space of absolutely continuous functions $V$ on $[0,1]$ endowed with a special norm $||\cdot ||$ as $||f||=\int_0^1 |f(t)|dt +\int_0^1 |f'(t)dt|=||f||_1+||f'||_1, \quad f\in V$. Then the space is complete.
Here is what I tried: we want to prove $V$ is complete, so it suffices to prove that $V$ is summable, i.e for any $\sum_{n=1}^\infty ||f_n||<\infty$, we can get $\sum_{n=1}^\infty f_n$ converges in $V$.
Define $F:=\sum_{n=1}^\infty f_n, f:=\sum_{n=1}^\infty f'_n$, then $||F||_1\leq \sum_{n=1}^\infty ||f_n||_1<\infty$ so that $f\in L^1[0,1]$ and $||f||_1\leq \sum_{n=1}^\infty ||f'_n||_1<\infty$ so that $f\in L^1[0,1]$.
$\int_0^x f'=\int_0^x\sum_{n=1}^\infty f'_n=\sum_{n=1}^\infty \int_0^x f'_n=\sum_{n=1}^\infty (f_n(x)-f_n(0))=F(x)-F(0)$. So $F$ is absolutely continuous, i.e $F\in V$, we get $\sum_{n=1}^\infty f_n$ converges to $F$ in $V$.
Is this process right? or there are other correct proof?
Best Answer
Yes, your proof is correct.
Another way to use the completeness of $L^1$ is to set up an isomorphism between your space (usually denoted $AC$ or $W^{1,1}$) and $L^1\oplus \mathbb R$. Such an isomorphism is given by $$f\mapsto \left(f',\int_0^1 f\right)$$ This map is bounded and surjective. It also has trivial kernel. Being lazy, I'd just invoke open mapping theorem to get this over with.