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!



Step 2 of The Strichartz's Estimates in Cazenave's Book
On the operator level: if $U,V: L^2 \to L^2$, denote by $U^\*$ the adjoint operator of $U$, you have that


$$ \langle U f, V g\rangle = \langle f, U^\* V g\rangle $$


by definition. (Note, this is *outside* any time integration.)


Now: $\mathcal{T}(s)^\* = \mathcal{T}(-s)$ (as evident from the explicit formula you gave, or from abstract considerations as a group of isometries). 


So before even integrating you have


$$ \langle \mathcal{T}(t-s) f(s), \mathcal{T}(t-\sigma) f(\sigma)\rangle = \langle f(s), \mathcal{T}(s-t) \mathcal{T}(t-\sigma) f(\sigma) \rangle $$


Now use the group property and you are done. 


