For each $r\geq0$, let $\Omega(r):C((-\infty,0])\rightarrow C((-\infty,0])$ be a map defined by $\Omega(r)(x(t))=x(t-r)$ for $t\leq0$. How can I prove that $\Omega (0)=I\ $?
My attempt: If $r=0 \Rightarrow\Omega(0)(x(t))=x(t)$. I could note that $\Omega^n(0)(x(t))=\Omega(0)(\Omega(0)\circ...\circ\Omega(0)(x(t)))=...=\Omega(0)\left ( \Omega(0)(x(t)) \right )=\Omega(0)(x(t))=x(t)$, ie $\Omega^n(0)(x(t))=x(t),\ n\in \mathbb{N}$. Is that correct? Can someone help me to continue please?
No, I'm sorry but your proof is nonsense.
The chief problem here is that you have misplaced parentheses. $\Omega(r)$ has the following action on each $x$:
$$(\Omega(r)x)(t)=x(t-r)$$
It is an $r$-shift operator. If $r=0$, the shift is null. In symbols:
$$(\Omega(0)x)(t)=x(t-0)=x(t)$$
In other words, $\Omega(0)x$ is just $x$ itself.
There is not much here to prove.