1
$\begingroup$

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: $A=B=C$?

$\endgroup$
  • $\begingroup$ $A = B = C$ when $T$ is simple. The general case follows by the usual approximation argument. $\endgroup$ – Sangchul Lee Nov 9 '18 at 4:16
0
$\begingroup$

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 $A=C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.