Bochner integral of a uniformly continuous semigroup

Question

Let's consider an uniformly continuous semigroup $\left\{T(t)\right\}_{t\,\geqslant\,0} \subset \mathscr{L}(X)$ (being $\left\{T(t)\right\}_{t\,\geqslant\,0}$ an UC-semigroup, one has $\displaystyle\lim_{t\,\longrightarrow\,0^+} \left\|T(t) - I\right\| = 0$, where $I$ is the identity operator), in which $X$ is a Banach space and $\mathscr{L}(X)$ is the Banach algebra of bounded linear operators in $X$.
Now, I wanna prove that the Bochner integral $\displaystyle\int_0^t T(s) \, \mathrm{d} s$ defines a bounded linear operator in $X$ such that $$\left(\displaystyle\int_0^t T(s) \, \mathrm{d}s\right)(x) = \displaystyle\int_0^t T(s)(x) \, \mathrm{d}s \quad (*),$$ for all $x \in X$, for every compact interval $[0,\,t]$, $t \geqslant 0$.
Ok, it's pretty straightforward to prove the Bochner-integrability of the semigroup (i.e., the family $\left\{T(t)\right\}_{t\,\geqslant\,0}$); since there exists constants $\omega \geqslant 0$ and $M \geqslant 1$ such that $\left\| T(t)\right\| \leqslant Me^{\omega t}$, for every $t \geqslant 0$, the right side of the inequality is Lebesgue-integrable and, hence, the right side is Lebesgue-integrable as well.
Thus, the map $t \in [0,\,+\,\infty) \longmapsto T(t) \in \mathscr{L}(X)$ is Bochner-integrable, by Bochner's theorem (a Bochner-measurable map $f \colon I \longrightarrow X$, where $I$ is a non-empty set and $X$ is an arbitrary Banach space, is Bochner-integrable iff its norm$\left\|f\right\| \colon X \longrightarrow \mathbb{R}$ is Lebesgue-integrable). A similar argument can be applied to prove that, for all $x \in X$, the map $t \in [0,\,+\,\infty) \longrightarrow T(t)(x) \in X$ is Bochner-integrable as well.
That said, I don't know how to prove the $(*)$ equality above, i.e., that the integral defines a bounded linear operator in $X$.
I tried to use the following result (Hille's Theorem, page 47 of "Vector Measures" by Diestel and Ühl), unsuccessfully:
Let $T$ be a closed linear operator between $X$ and the Banach space $Y$. If $f$ and $T \circ f$ are Bochner-integrable with respect to the measure $\mu$, then $$T\left(\int_E f \, \mathrm{d} \mu \right) = \int_E (T \circ f) \, \mathrm{d} \mu,$$ for all measurable $E$.
I'm a bit stuck, and any help will be appreciated!

Answer 1

Assume $f:[a,b]\to Y$ is a continuous function with values in a Banach space $Y.$ Then there exists a unique element $I_f\in Y$ such that for any $\varepsilon>0,$ there exists $\delta>0$ such that for any partition $\mathcal{P}=\{t_0,t_1,\ldots, t_n\}$ of the interval $[a,b],$ with mesh less than $\delta,$ and any intermediate points $t_{i-1}\le s_i\le t_i,$ we have $$\|S(\mathcal {P},f)-I_f\|<\varepsilon, \qquad S(\mathcal {P},f)=\sum_{k=1}^nf(s_k)(t_k-t_{k-1})$$ The proof is pretty straightforward, nearly identical as for $Y=\mathbb{R}$ (see for example notes). We call $I_f$ the Riemann integral and use the notation $$I_f=\int\limits_a^b f(t)\,dt$$

The integral shares all basic properties of the classical Riemann integral, in particular $$\left \|\int\limits_a^b f(t)\,dt\right \|\le \int\limits_a^b \|f(t)\|\,dt$$ Partition into equal parts gives $$\lim_n{b-a\over n}\sum_{k=1}^n f\left (a+k{b-a\over n}\right )=\int\limits_a^b f(t)\,dt\qquad (**)$$ The formula $(**)$ applied to $Y=\mathcal{L(X)}$ and $Y=X$ and a continuous function $f:[a,b]\to \mathcal{L}(X),$ implies the formula $(*)$ in question, as for any $x\in X$ we have $$\displaylines{\left(\int\limits_a^b f(t) \,dt\right )(x)=\left (\lim_n{b-a\over n}\sum_{k=1}^n f\left (a+k{b-a\over n}\right )\right)(x)\\ = \lim_n{b-a\over n}\sum_{k=1}^n f\left (a+k{b-a\over n}\right )(x)= \int\limits_a^b f(t) (x)\,dt}$$ Chainging the order of the limits and pointwise evaluation was possible as operator norm convergence implies the pointwise convergence.

Remark The Riemann integral is convenient for continuous functions. If one prefers applying the Bochner integral, then $(**)$ is also valid, provided the function $f$ is continuous. Indeed, for $t_k=a+k{b-a\over n}$ we have $$\int\limits_a^bf(t)\,dt -{b-a\over n}\,\sum_{k=1}^n f(t_k)={b-a\over n}\,\sum_{k=1}^n\,\int\limits_{t_{k-1}}^{t_k} [f(t)-f (t_k)]\,dt$$ Hence $$ \left\| \int\limits_a^bf(t)\,dt -{b-a\over n}\sum_{k=1}^n \,f(t_k)\right \|\le {b-a\over n}\sum_{k=1}^n\,\int\limits_{t_{k-1}}^{t_k}\|f(t)-f(t_k)\|\,dt\le \sup_{|t-s|\le {b-a\over n}}\|f(t)-f(s)\|$$