Given a sequence of bounded operators $\{T(t)\}_{t\ge0}$ defined on the Banach space $C_0(\mathbb R^d)$ equipped with the supremum norm $|f|_0:=\sup_{x\in\mathbb R^d}|f(x)|$. Suppose $$\int_0^\infty \|T(t)\|dt<\infty,\tag{1}$$ where $\|\cdot\|$ is the operator norm.
Now I have three operators
\begin{align} A &:=\int_0^\infty T(t) dt, \\ Bf &:=\int_0^\infty T(t)f dt, \quad \forall f\in C_0(\mathbb R^d), \\ Cf(x) &:=\int_0^\infty T(t)f(x) dt,\quad \forall f\in C_0(\mathbb R^d), x\in\mathbb R^d, \end{align}
where the integral in $A$ is the Bochner integral defined on the Banach space $(\mathcal L(C_0(\mathbb R^d)),\|\cdot\|)$ of all bounded operators on $C_0(\mathbb R^d)$, the integral in $B$ is the Bochner integral defined on the Banach space $(C_0(\mathbb R^d),|\cdot|_0)$.
I know that these three operator are all well-defined bounded operators on $C_0(\mathbb R^d)$ by the condition $(1)$, but are they equal: $A=B=C$?
Any Bochner integral is also a Pettis integral. Since $T \to Tf$ is a continuous linear map on $L(C_0(\mathbb R^{d})$ it follows that $A=B$. Similarly, continuity of $T \to T(f(x))$ proves that $A=C$.