Bochner integrability I have a question, which is related to the notion of Bochner integrability. In the course of solving some particular problem, I need to show that
\begin{align}
 \sqrt{\sum_{k=1}^\infty (\lambda_k + \gamma)\left(\int_0^t e^{-\lambda_k (t-s)}\left\langle g(v(s)), {\varphi}_k\right\rangle_{L^2}ds\right)^2} \leq \nonumber\\
 \leq \int_0^t \sqrt{\sum_{k=1}^\infty (\lambda_k + \gamma) e^{-2\lambda_k (t-s)}\left\langle g(v(s)), {\varphi}_k\right\rangle_{L^2}^2}ds
\end{align} 
holds. Here $\gamma$ is some given positive constant, $g:L^2(0,1)\to L^2(0,1)$ is a given continuous function, $\lambda_k$ are the eigenvalues of the linear operator $A$, which is symmetric with respect to $L^2$-norm and $(\varphi_k)_{k\in \mathbb{N}}$  represent the orthonormal basis of $L^2(0,1)$ such that
\begin{equation*}
A\varphi_k = \lambda_k \varphi_k, \quad k = 1,2,\dots
\end{equation*}
In addition, $\lambda_1>0$.
My question is the following: if I define a sequence $f(s) = (f_k(s))_{k\in\mathbb{N}}$ such that
\begin{equation}
f_k(s):= \sqrt{\lambda_k +\gamma}\; e^{-\lambda_k (t-s)}\left\langle g(v(s)), {\varphi}_k\right\rangle_{L^2},
\end{equation}
would it be correct to assume, that the required inequality takes the form
\begin{equation}
\left\|\int_0^t f(s)\; ds \right\|_{l^2}\leq \int_0^t \left\|f(s)\right\|_{l^2} \;ds.
\end{equation}
If the answer is yes, then I need to show, that sequence $f(s)$ is Bochner integrable. I think, that I need to show, that $f(s)$ is strongly measurable and the mapping $s\mapsto \left\|f(s)\right\|_{l^2}$ is summable, i.e. $\int_0^t\left\|f(s)\right\|_{l^2} <\infty$ holds. My thoughts on the strong measurability of $f(s)$: it probably follows from the fact, that $f_k(s)$ is a continuous function for each $k$.  
I would appreciate any hints! Thanks!
 A: Recall the Pettis measurability theorem:
Let $(X,\Sigma,\mu)$ be a measure space, $B$ be a Banach space, then $f:X\to B$ is strongly measurable iff


*

*There exist $A\in\Sigma$ such that $\mu(X\setminus A)=0$ and $f(A)$ is separable.

*For all $\varphi\in B^*$ the function $\varphi\circ f:X\to\mathbb{R}$ is measurable.
In your case $B=\ell_2$ which is separable, so the first requirement of your theorem is satisfied. For the second requirement recall that every $\varphi\in \ell_2^*$ is of the form
$$
\varphi: B^*\to\mathbb{R} : (x_k)_{k\in\mathbb{N}}\mapsto\sum\limits_{k=1}^\infty x_k z_k
$$
for some $z=(z_k)_{k\in\mathbb{N}}\in \ell_2$. Then for all $s\in[0,t]$ we have 
$$
(\varphi\circ f)(s)=\sum\limits_{k=1}^\infty f_k(s)z_k
$$
Since $(f_k)_{k\in\mathbb{N}}\subset C([0,t])$, then $\varphi\circ f$ is measurable. Hence the second requirement of Pettis theorem is satisfied. Thus $f$ is strongly measurable.
Moreover $f$ is strongly integrable. Here is the proof. Since $A$ is a Bounded linear operator, then there exist $C_1>0$ such that $\lambda_k<C_1$ for all $k\in\mathbb{N}$. This implies that there exist $C_2>0$ such that $e^{-2\lambda_k(t-s)}<C_2$ for all $s\in[0,t]$ and $k\in\mathbb{N}$. Given this inequalities we get
$$
\Vert f(s)\Vert_{\ell_2}
=\left(\sum\limits_{k=1}^\infty(\lambda_k+\gamma)e^{-2\lambda_k(t-s)}\langle g(v(s)),\varphi_k\rangle\right)^{1/2}
\leq\left(\sum\limits_{k=1}^\infty(C_1+\gamma)C_2\langle g(v(s)),\varphi_k\rangle^2\right)^{1/2}
$$
$$
\leq\sqrt{C_2(C_1+\gamma)}\left(\sum\limits_{k=1}^\infty\langle g(v(s)),\varphi_k\rangle^2\right)^{1/2}
$$
Since $(\varphi_k)_{k\in\mathbb{N}}$ is an orthonormal system from Bessel's inequality, we get
$$
\Vert f(s)\Vert_{\ell_2}
\leq\sqrt{C_2(C_1+\gamma)}\left(\sum\limits_{k=1}^\infty\langle g(v(s)),\varphi_k\rangle^2\right)^{1/2}
\leq\sqrt{C_2(C_1+\gamma)}\Vert g(v(s))\Vert_{\ell_2}
$$
Since $g\circ v\in C([0,t])$, then its integral over $[0,t]$ is finite and we get
$$
\int\limits_{0}^{t}\Vert f(s)\Vert_{\ell_2}ds
\leq\sqrt{C_2(C_1+\gamma)}\int\limits_{0}^{t}\Vert g(v(s))\Vert_{\ell_2}ds<+\infty
$$
Thus $f$ is strongly integrable.
