The space of absolutely continuous functions on $[0,1]$ is equipped with norm $\|f\|:=|f(0)|+V_0^1(f)$.
($V$ is the variation, here is the definition if needed: https://en.wikipedia.org/wiki/Bounded_variation)
Check that this is indeed a norm and check if the space is complete.
I did the first part, I don't know how to prove it is complete or not…
Best Answer
Note that if $f$ is ac. then $f$ is differentiable ae., $f' \in L^1[0,1]$ and $f(t) = f(0)+ \int_0^t f'(\tau) d \tau$. We have $\|f\|_{AC} = |f(0)|+\int_0^1 |f'(\tau)| d \tau$.
Suppose $f_n$ is Cauchy, then you need to produce a candidate $f$, show that it is ac. and that $f_n \to f$ in the given norm.
Since $f_n$ is Cauchy, we see that $f_n(0)$ is Cauchy in $\mathbb{R}$ and $f_n'$ is Cauchy in $L_1[0,1]$. Hence there is some $f_0 \in \mathbb{R}$ and $g \in L_1[0,1]$ such that $f_n(0) \to f_0$ and $f_n' \to g$ (in $L_1[0,1]$).
Define $f(t) = f_0+\int_0^t g(\tau) d \tau $. Since $g $ is integrable, we see that $f$ is ac, and we have $\|f-f_n\|_{AC} \le |f_0-f_n(0)| + \int_0^1 |g(\tau)-f_n'(\tau) | d \tau $, hence $f_n \to f$ in the given norm.