How to prove that, $T\left(\int_a^b f(t)dt\right) =\int_a^b T\circ f(t)dt $

Question

Let $T:E\to F$ be a linear and continuous function between the Banach spaces $E$ and $F$.

We consider $f:\to [a,b]\to E$ (with $a<b $) be an integrable function in the sense of Bochner

How do I prove that $$T\left(\int_a^b f(t)dt\right) =\int_a^b T\circ f(t)dt $$ I don't know from which argument $T$ cross the integral. Please Help.

Answer 1

You can prove that in exactly the same way you would do it in conventional measure theory. First you can prove it with

$$\phi=\displaystyle\sum_{i=1}^n b_i 1_{E_i}$$

A simple function, then

$$\int T \phi=\int \displaystyle\sum_{i=1}^n Tb_i1_{E_i}=\int \displaystyle\sum_{i=1}^n b_i 1_{TE_i}=\displaystyle\sum_{i=1}^n b_i \mu(TE_i)=T\displaystyle\sum_{i=1}^n b_i \mu(E_i)=T\int \phi$$

And so $f_n\uparrow f$ then

$$\int Tf=\lim \int Tf_n=T \lim \int f_n=T\int f$$