Linearity property of integral involving a bounded linear operator Suppose $X$ is a Hilbert space and $T\in\mathcal{B}(X)$. For $f\in\mathcal{C}([a,b],X)$, where $a\leq b$, we have
\begin{equation}
  Tf\in\mathcal{C}([a,b],X)\quad\text{and}\quad T\int_{a}^{b}f(x)\,\mathrm{d}x=\int_{a}^{b}Tf(x)\,\mathrm{d}x.
\end{equation}
Is the latter merely a consequence of linearity? If so, how do I apply this here to show the identity?
 A: Writing the integral as a Riemann sum:
$$ \int_a^b f(x)dx = \lim_{n\to\infty} \sum_{k=0}^n \frac{b-a}{n}f\Big(a + \frac{k}{n}(b-a)\Big) $$
which converges for $f\in\mathcal{C}([a,b])$, you can apply the fact that $T$ is both linear (exchange sum and $T$) and continuous (exchange $T$ and limit), and that $Tf$ is again continuous for the convergence of the Riemann sum for $\int Tf(x) dx$.
A: I am ssuming you mean that $f:[a,b]\to X;$ that is, that $f$ takes values in $X$. If not, please advise and I will delete this answer. 
In case $f:[a,b]\to X$, then the usual way to define the integral is by means of regulated functions. That is, if $a\le\cdots\le x_i\le x_{i+1}\le \cdots \le b,\ c_i\in X,$ one considers simple functions of the form $s(t)=\sum^n_{i=1}c_i\chi_{[x_i,x_{i+1}]}(t)$  and defines $\int s$ to be $\sum^n_{i=1}|x_{i+1}-x_i|c_i$.
With this definiton, $\int$ becomes a bounded linear function on the space $S$ of simple functions, with values in $X$ and so extends to the closure $\overline S$. One proves that $\overline S$ contains the continuous functions $\mathcal{C}([a,b],X)$ that is, the set of continuous  $f:[a,b]\to X$. 
Then, if $T$ is a bounded linear operator from  $X$ to $X$, the claim is true:
if $s(t)=\sum^n_{i=1}c_i\chi_{[x_i,x_{i+1}]}(t)$ is a simple function, then 
$\tag 1 T\int s=T\left(\sum^n_{i=1}|x_{i+1}-x_i|c_i\right)=\sum^n_{i=1}|x_{i+1}-x_i|T(c_i)$.
On the other hand,  for each $t\in [a,b],\  s(t)=\sum^n_{i=1}c_i\chi_{[x_i,x_{i+1}]}(t)\in X$ so $Ts$ makes sense and is equal to $\sum^n_{i=1}T(c_i)\chi_{[x_i,x_{i+1}]}(t)$, and from this we have, by definition,  
$\tag 2\int Ts=\sum^n_{i=1}|x_{i+1}-x_i|T(c_i)$.
Comparing $1).$ and $2).$, we get the result on $S$. Extending this to $\overline S$ is routine. 
