I have to prove that $(AC([a,b]), ||\cdot||_{AC}) \hookrightarrow (C([a,b]),|| \cdot ||_\infty)$.
I'm suggested to use fundamental theorem of calculus. I have to prove that there exists a positive constant M such that $$|| f ||_\infty \le M|| f ||_{AC} \ \ \forall f \in AC([a,b]) $$
I have: \begin{align} || f ||_\infty &= \max_{t \in [a,b]} |f| \\&= \max_{t \in [a,b]}|f(a)+\int_a^tf'(\tau)d\tau| \\ &\le |f(a)| +\int_a^b |f'(t)|dt= |f(a)| +||f'||_1 \\ ||f||_{AC}&= ||f||_1+||f'||_1 \end{align} And I' stuck at this point. According to my notes $ |f(a)|+||f'||_1$ is equivalent to $ ||f||_1+||f'||_1$ in $AC([a,b])$ which would imply my result, but I cannot prove that.
Clearly, if $f$ is absolutely continuous, then $f'\in L^1$ and $$ f(x)-f(a)=\int_a^x f(t)\,dt, $$ for all $x\in [a,b]$. Hence $$ |f(a)|\le|f(x)|+\int_a^x |f'(t)|\,dt\le |f(x)|+\int_a^b |f'(t)|\,dt=|f(x)|+\|f'\|_1 $$ and integrating from $a$ to $b$ we obtain $$ (b-a)|f(a)|\le \int_a^b |f(x)|\,dx+(b-a)\|f'\|_1=\|f\|_1+(b-a)\|f'\|_1 $$ and thus $$ |f(a)|\le \frac{1}{b-a} \|f\|+\|f'\|_1. $$ Finally, for all $x\in [a,b]$, $$ |f(x)|\le |f(a)|+\int_a^x |f(t)|\,dt\le |f(a)|+\|f\|_1\le \left(\frac{1}{b-a} \|f\|+\|f'\|_1\right)+\|f\|_1 \\ \le\left(1+\frac{1}{b-a}\right)\big(\|f\|_1+\|f'\|_1\big) $$ and consequently $$ \|f\|_\infty \le\left(1+\frac{1}{b-a}\right)\big(\|f\|_1+\|f'\|_1\big)= \left(1+\frac{1}{b-a}\right)\|f\|_{AC} $$
Note. $AC[a,b]$ is more than continuously imbedded in $C[a,b]$. It is compactly imbedded due to Arzelà-Ascoli.