Let $L^2$ denote $L^2([0,T];\mathbb{R}^N)$ and let $A:[0,T]\times L^2 \to (L^2)^\prime =L^2$ be a Carathéodory function satisfying the following estimate: $$\exists \alpha \in L^2([0,T];\mathbb{R}_+):\quad \forall x\in L^2,\ |A(t,x(t))|_N\leq \alpha (t) + | x(t) |_N \quad \text{a.e. in } [0,T]$$ (where $|\cdot |_N$ denotes the Euclidean norm in $\mathbb{R}^N$).
Under such assumptions it is easy to prove that the application $t\mapsto \langle A(t,x^*(t)), x(t) - x^*(t)\rangle$ (here $x,x^* \in L^2$ and $\langle \cdot ,\cdot \rangle$ is the scalar product in $L^2$) is an $L^1([0,T];\mathbb{R})$ function, because of the Cauchy-Schwarz inequality (for scalar products in $\mathbb{R}^N$ and in $L^2$): in fact: $$\begin{split} \int_0^T \left| \langle A(t,x^*(t)), x(t) - x^*(t)\rangle \right|&\stackrel{\text{C-S}}{\leq} \int_0^T \left( \alpha(t) + |x^*(t)|_N\right)\cdot |x(t) - x^*(t)|_N\\ &\stackrel{\text{C-S}}{\leq} \| \alpha + |x^*|_N\|_{L^2}\cdot \| x-x^*\|_{L^2} \\ &< \infty \; .\end{split}$$
I read that it is also possible to prove that $t\mapsto \langle A(t,x^*(t)), x(t) - x^*(t)\rangle$ is an $L^2([0,T];\mathbb{R})$ function... But at the moment I don't find any reasonable way to do this.
Any hint?