# $(\int_0^\infty T(t) dt) f(x) = (\int_0^\infty T(t)f dt)(x) = \int_0^\infty T(t)f(x) dt$?

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: $$A=B=C$$?

• $A = B = C$ when $T$ is simple. The general case follows by the usual approximation argument. – Sangchul Lee Nov 9 '18 at 4:16

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 $$A=C$$.