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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!

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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.

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  • $\begingroup$ @ Willie Wong: Thanks a lot for the answer! It is really helpful. I have one more question. Sorry if it sounds stupid! Let us use the equality from your post. Since $\mathcal{T}(\cdot)$ is a group of isometries, we have that $\mathcal{T}(s-t)\mathcal{T}(t-\sigma)$ will be another element of the same group. How do we know that this element will be exactly $\mathcal{T}(s-\sigma)$ or it is just a matter of notation? $\endgroup$ – Candidate Jun 13 '16 at 23:20
  • $\begingroup$ That's just the standard notation of a one parameter group. See also en.wikipedia.org/wiki/C0-semigroup ; If you think in terms of the solution operator: solving forward first by time $t-\sigma$ then by time $s - t$ must equal to solving forward by time $t - \sigma + s - t = s-\sigma$... $\endgroup$ – Willie Wong Jun 14 '16 at 1:53
  • $\begingroup$ @ Willie Wong: Thanks for helping me again! Actually my first attempt was to plug everything into the formula for $\mathcal{T}(t)\phi(x)$ and then derive that $\mathcal{T}(s-t)\mathcal{T}(t-\sigma)=\mathcal{T}(s-\sigma)$. This turned out to be not an easy task. $\endgroup$ – Candidate Jun 14 '16 at 9:05

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