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!