I'm afraid I'm very confused by how to correctly apply the Fubini theorem to simplify integrals?
I have some integral
$$ \sum_{k = 0}^{2}\int_{0}^{T} dt_2 \int_{0}^{t_1} dt_1 \bigg[ \big(\psi_{0}(t_2) , V(t_2)\psi_{k}(t_2)\big) \hspace{3 mm} \big(\psi_k(t_1),V(t_1)\psi_0 (t_1)\big)\bigg] \hspace{15 mm} (i) $$
Where $ (f,\hat{O}g) = \overline{f} \cdot \hat{O}g $ where $f,g$ are some vectors with coefficients in $\mathbb{C}$ and $\hat{O}$ and operator.
So similar to a Euclidean dot product but with the first vector conjugated.
Now the problem I'm having is with applying Fubini's theorem. Below is my derivation
$$ \sum_{k = 0}^{ 2}\int_{0}^{T} dt_2 \int_{0}^{t_1} dt_1 \bigg[ \big(\psi_{0}(t_2) , V(t_2)\psi_{k}(t_2)\big) \hspace{3 mm} \big(\psi_k(t_1),V(t_1)\psi_0 (t_1)\big)\bigg] \\= \sum_{k = 0}^{ 2}\int_{0}^{T} dt_2 \big(\psi_{0}(t_2) , V(t_2)\psi_{k}(t_2)\big) \hspace{3 mm}\int_{0}^{T}dt_1 \big(\psi_k(t_1),V(t_1)\psi_0 (t_1)\big) $$ $$ = \sum_{k=0}^{2} \bigg| \int_{0}^{T} ( \psi_0(t) , V(t) \psi_k(t)) \ dt \bigg|^2 \hspace{15 mm} (\star) $$
Sorry if that's a mess, I suppose I'm really just curious as to how to get from $(i)$ to $(\star)$ as I'm quite sure the last line is correct.
Thank You.
Fubini just tells you that you can interchange things: $$f\in\mathcal{L}:\quad\int_{\Omega\times\Omega'} f(x,x')\mathrm{d}x\otimes x'=\int_\Omega\left(\int_{\Omega'}f(x,x')\mathrm{d}x\right)\mathrm{d}x'$$ So what you have to check in order to freely interchange the order of integration is integrability: $$f\in\mathcal{L}:\iff\int_{\Omega\times\Omega'}|f(x,x')|\mathrm{d}x\otimes x'$$ In your case take the absolute value of the expression in the integral apply Cauchy-Schwarz to split these nasty inner products and see wether the remaining thing is integrable.
(Let me know when you reach that step.)