# Technical question about Strichartz estimate's proof.

I was studying the proof of Strichartz estimates from the book "Semilinear Schrödinger equation" of T. Cazenave. The proof is divided in several steps. Here we can assume

$$\Theta_{t,f}(t)=\Phi(t)=\Phi_f(t)=\int_0^t e^{i(t-s)\Delta}f(s)ds.$$

I have to prove the following Strichartz estimate:

$$\left\|\Phi_f\right\|_{C\left(0,T;L^2(\mathbb{R}^n)\right)}\leq\left\|f\right\|_{L^{q'}\left(0,T;L^{r'}(\mathbb{R}^n)\right)},$$

where $(q,r)$ is any admissible pair, i.e. $-\frac{2}{q}=n\left(\frac1r-\frac12\right)$, and $f\in L^{q'}\left(0,T;L^{r'}(\mathbb{R^n})\right)$.

In the following picture I see that in order to use Fubini theorem to exchange the integrals I need to prove that $\Phi_f\in L^2$. I don't understand how to prove it.

I don't understand which part of the argument uses Fubini's theorem. Given $v\in H$ and $g\in C([0,T],H)$, where H is a Hilbert space, then $$\left(\int_0^t g(s)ds, v\right)_H = \int_0^t (g(s),v)_H ds.$$
The key observation seems to be that $C_c([0,T), L^{r'}\cap L^2)$ is actually dense in $L^{q'}((0,T),L^{r'})$. But if you assume that $f\in C_c([0,T), L^2)$, then also $$g: t\mapsto \int_0^t T(t-s)f(s)ds \in C([0,T],L^2)$$ and you can perform the calculation as Cazenave writes.