On the Schrodinger fundamental solution Let $e^{it\Delta}$ be the fundamental Schrodinger solution. If $u_0$ is the corresponding initial data to the problem associated to Schrodinger free equation $u_t = i\Delta u$ and $S(\mathbb{R}^N)$ denotes the Schwarz class of functions. I wonder how to prove the following properties. I have proved another ones but these three got me stucked...
$$a) \ \ u_0\in S(\mathbb{R}^N)\Rightarrow e^{it\Delta}u_0\in S(\mathbb{R}^N) $$
$$ b) \ \ e^{i(t+s)\Delta}u_0 = e^{it \Delta}  (e^{is\Delta}u_0) \quad \forall u\in L^2(\mathbb{R}^N)  $$
$$ c) \ \ e^{i 0 \Delta}u_0 = u_0 \quad \forall u\in L^2(\mathbb{R}^N)$$
Any help or hint would be appreciated.
Thanks in advance!
 A: I'm actually trying to learn some of this type of analysis myself, so I'm going to take a stab at this and hopefully it will be helpful. If not, then my sincere apologies. Let's start with part a). Observe that $\hat{u}_0 \in \mathcal{S}$ and, letting $D_j = \frac{1}{i}\partial_j$, we have
$$D_x^\alpha(e^{it\Delta}u_0) = (2\pi)^{-d}\int e^{i(-t\lvert \xi \rvert^2 + x\cdot\xi)} \xi^\alpha \hat{u}_0(\xi) \ d\xi.$$
Noticing that $\mathcal{F}$ is in fact an automorphism on $\mathcal{S}$ will let us see that $e^{it\Delta}u_0 \in \mathcal{S}$.
For part c),
$$e^{i0\Delta}u_0 = (2\pi)^{-d}\int e^{-i(0\lvert \xi \rvert^2 + x\cdot\xi)} \hat{u}_0(\xi) \ d\xi = \mathcal{F}^{-1}\hat{u}_0.$$
If, as mentioned in the comments, we want to consider the limit $t\to 0$, the underlying logic should be similar (once one justifies passing the limit through the integral).
Lastly, b):
$$e^{i(t+s)\Delta}u_0 = (2\pi)^{-d}\int e^{i(-(t+s)\lvert \xi \rvert^2 + x\cdot \xi)}\hat{u}_0(\xi) \ d\xi = (2\pi)^{-d}\int e^{-it\lvert \xi \rvert^2}[e^{i(-s\lvert \xi \rvert^2 + x\cdot \xi)}\hat{u}_0(\xi)] \ d\xi = e^{it\Delta}(e^{is\Delta}u_0).$$
A: For these problems, it will be easier to work in Fourier space.  Notice that $\mathcal{F} : L^2 \to L^2$ and $\mathcal{F}: \mathcal{S} \to \mathcal{S}$ are isometries, so it is sufficient to prove the results for $\mathcal{F}\left(e^{it\Delta} u_0\right)$.  
For part (a), we have that
$$\mathcal{F}\left(e^{it\Delta} u_0\right)(\xi) = e^{-it|\xi|^2} \hat{u_0}(\xi)$$
so it suffices to prove that the term ont he right is in the Schwartz space.  Consider the norm $[f]_{M,K} := \sup_{\xi\in \mathbb{R}^N, |\alpha| \leq M, |\beta| \leq K} |\xi|^\alpha |\partial_\xi^\beta f(\xi)|$, where $\alpha, \beta$ are multi-indices.  Notice that if we differentiate $e^{it|\xi|^2} \hat{u_0}(\xi)$ with respect to $\xi_j$, the derivative can either 'hit' the oscillating phase $e^{it|\xi|^2}$ to give $-2it\xi_j e^{it|\xi|^2}$, or it can hit the function $\hat{u_0}(\xi)$.  Iterating this, we can see that
$$[e^{it|\xi|^2} \hat{u_0}]_{M,K}\leq C_{K}t^K[\hat{u_0}]_{M+K, K}$$
since taking the derivative will give us a sum of terms involving polynomials in $\xi|$ of order at most $M+K$ and derivatives of $\hat{u_0}$ of order at most $K$.  (This is just a sketch, I'll let you work out the details).
Part (b) is quite simple: it follows from the fact that $e^{-i(t+s)|\xi|^2} = e^{-it|\xi|^2}e^{-is|\xi|^2}$.
For part (c) (assuming you want the limit version), follows from the dominated convergence theorem.  Observe that we would like to show that 
$$ 0 = \lim_{t \to 0} \lVert e^{-it|\xi|^2} \hat u_0 - \hat u_0\rVert_{L^2}^2 = \lim_{t \to 0}\int \left|e^{-it|\xi|^2} - 1\right|^2 |\hat{u_0}(\xi)|^2\;d\xi$$
Notice that the integrand is bounded by $4|\hat{u_0}(\xi)|^2$, which is integrable, that that it converges pointwise to $0$.  By the dominated convergence theorem, we conclude that the integral is indeed $0$.

As an addendum, notice that the same reasoning holds (possibly with minor adjustments) if we replace $|xi|^2$ by any polynomial $P(\xi)$.  The propagator $e^{itP(\xi)}$ is the fundamental solution to
$$u_t = iP(D_x) u$$
where $D_x = \frac{1}{i} \partial_x$ and $P(D_x)$ is obtained by formally substituting the differential operator for $x$ in $P(x)$.  Such equations are known as dispersive equations.
