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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!

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    Yes, $v$ is continuous. Should have mentioned that.2012-10-03

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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

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

  2. 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 for all $k\in\mathbb{N}$. This implies that there exist $C_2>0$ such that $e^{-2\lambda_k(t-s)} 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.

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    @Dina I agree with everything you said in your second comment. Thanks for accepting the answer.2012-10-04