My question is from Cazenave's book "Semilinear Schrödinger Equation", page 35. I am stuck with Step 2 of the Strichartz's estimates.

The book says that $||\Phi_f(t)||_{L^2}^2=\left(\int_0^t \mathcal{T}(t-s)f(s)ds,\int_0^t \mathcal{T}(t-\sigma)f(\sigma)d\sigma\right)_{L^2}=$

$\underbrace{\int_0^t\int_0^t\left( \mathcal{T}(t-s)f(s), \mathcal{T}(t-\sigma)f(\sigma)\right)_{L^2}d\sigma ds=\int_0^t \int_0^t \left( f(s),\mathcal{T}(s-\sigma)f(\sigma)\right)_{L^2}d\sigma ds=}_?$

$=\int_0^t\left(f(s),\Theta_{t,f}(s)\right)_{L^2}ds$,

where $\Theta_{t,f}(s)$ was defined earlier in the chapter, but it is not important for the purpose of this question.

$\mathcal{T}(t)$ is defined as the group of isometries on $L^2(\Omega)$ generated by the skew-adjoint operator $iA$, where $A$ is the Laplacian with the Dirichlet boundary conditions on $\partial\Omega$. On the other hand there is a theorem which says that $\mathcal{T}(t)\phi(x)=(4\pi it)^{-\frac{N}{2}}\int_{\mathbb{R}^N}e^\frac{i|x-y|^2}{4t}\phi(y)dy$ for suitable functions $\phi$.

I would really appreciate if someone could give me a hand. I actually struggle with the third equality, namely how the identity below is determined:

$\int_{0}^{t}\int_{0}^{t} (\mathcal{T}(t-s)f(s),\mathcal{T}(t-\sigma)f(\sigma))_{L^2} d\sigma ds=\int_{0}^{t}\int_{0}^{t} (f(s),\mathcal{T}(s-\sigma)f(\sigma))_{L^2} d\sigma ds$.

If I am not wrong, I need to prove that

$\int\mathcal{T}(t-s)f(s)\overline{\mathcal{T}(t-\sigma)f(\sigma)}=\int f(s)\overline{\mathcal{T}(s-\sigma)f(\sigma)}$,

where the $L^2$ inner product is defined by $(u,v)_{L^2}$=Re $\int_{\Omega}u(x)\overline{v(x)}dx$.

Thanks in advance!