Can you help me showing that $(AC([a,b]),||.||_{ac})$ is a continuous injection in $(C([a,b]),||.||_{\infty})$?
With $||f||_{ac}=||f||_{1}||+||f^{'}||_{1}$
I tried to show that $ ||.||_{ac} > M||.||_{\infty} $ using the fundamental theorem of calculus but I didn't get the hoped result.
The optimal inequality is $\|f\|_\infty\leq \min(b-a,1)^{-1}\|f\|_{ac}.$ I assume you can translate this into the required statement about the inclusion operator.
Consider an $f\in AC([a,b])$ and a value $c$ such that assume $|f(c)|=M=\|f\|_\infty.$ Let $m=\min_{x\in[a,c]}|f(x)|.$ Then $\int_{[a,c]}|f|\geq (c-a)m$ and $\int_{[a,c]}|f'|\geq M-m.$ The sum of these terms satisfies $(c-a)m+M-m\geq \min(1,c-a)M.$ Similarly $\int_{[c,b]}|f|+|f'|\geq \min(1,b-c)M.$ So $$\|f\|_{ac}\geq (\min(1,c-a)+\min(1,b-c))M\geq\min(1,b-a)\|f\|_\infty.$$
To see this is optimal, for $b-a\leq 1$ take $f$ to be a constant function, and for $b-a\geq 1$ consider a function that rapidly decreases from its maximum at $x=a$ to zero, so $\int|f|\approx 0$ but $\int|f'|= \|f\|_\infty.$