Generator of $C_0$ Group of Shift Operators Let $X$ be the space of all continuous functions, $f$ with supremum norm such that 
$f(t) \rightarrow 0$ as $t \rightarrow \pm\infty$
Let $(T(t)f)(s) = f(s+t)$, the forms a $C_0$ group.
As far as I can tell, the generator for this group is simply the differentiation operator with domain of all functions which are differentiable everywhere. This seems to be a 2 line proof. But the question asks for detailed justification which makes me think I'm missing something.
 A: Maybe your short proof proves that if $f$ belongs to the generator's domain then $f$ is everywhere differentiable (which is a true claim whose converse is false) with $Af=f'$.
Actually, the generator of this group is the differentiation operator with domain
$$D:=\left\{f\in X\mid f\text{ is differentiable and } f'\in X\right\}.$$
Proof: Let $A$ be the infinitesimal generator of the group. Take $f\in D(A)$. Then
$$\sup_{x\in\mathbb{R}} \left|\frac{f(x+t)-f(x)}{t}-Af(x)\right|\overset{t\to 0}{\longrightarrow}0$$
and thus $f$ is differentiable with $\frac{df}{dx}(x)=Af(x)$ for all $x\in \mathbb R$. 
This shows that $D(A)\subset D$ with $Af=\frac{d}{dx}f$ for all $f\in D(A)$. In other words, this shows that $A$ is a restriction of the differentiation operator $\frac{d}{dx}:D\to X$. To complete the proof, it remains to show that $D(A)=D$.
By the Hille-Yosida Theorem we have $1\in\rho(A)$. We also can prove that $1\in\rho(\frac{d}{dx})$. So
$$(I-\tfrac{d}{dx})(D(A))=(I-A)(D(A))=X, \qquad D=(I-\tfrac{d}{dx})^{-1}(X)$$
and thus $D=(I-\tfrac{d}{dx})^{-1}(I-\tfrac{d}{dx})(D(A))=D(A)$. $\square$
