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Let $e^itDelta$ be the fundamental Schrodinger solution. If $u_0$ is the corresponding initial data to the problem associated to Schrodinger free equation $u_t = iDelta u$ and $S(mathbbR^N)$ 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...

$$a) u_0in S(mathbbR^N)Rightarrow e^itDeltau_0in S(mathbbR^N) $$
$$ b) e^i(t+s)Deltau_0 = e^it Delta (e^isDeltau_0) quad forall uin L^2(mathbbR^N) $$
$$ c) e^i 0 Deltau_0 = u_0 quad forall uin L^2(mathbbR^N)$$

Any help or hint would be appreciated.

Thanks in advance!

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 $hatu_0 in mathcalS$ and, letting $D_j = frac1ipartial_j$, we have
$$D_x^alpha(e^itDeltau_0) = (2pi)^-dint e^i(-tlvert xi rvert^2 + xcdotxi) xi^alpha hatu_0(xi) dxi.$$
Noticing that $mathcalF$ is in fact an automorphism on $mathcalS$ will let us see that $e^itDeltau_0 in mathcalS$.
For part c),
$$e^i0Deltau_0 = (2pi)^-dint e^-i(0lvert xi rvert^2 + xcdotxi) hatu_0(xi) dxi = mathcalF^-1hatu_0.$$
If, as mentioned in the comments, we want to consider the limit $tto 0$, the underlying logic should be similar (once one justifies passing the limit through the integral).
Lastly, b):
$$e^i(t+s)Deltau_0 = (2pi)^-dint e^i(-(t+s)lvert xi rvert^2 + xcdot xi)hatu_0(xi) dxi = (2pi)^-dint e^-itlvert xi rvert^2[e^i(-slvert xi rvert^2 + xcdot xi)hatu_0(xi)] dxi = e^itDelta(e^isDeltau_0).$$

For these problems, it will be easier to work in Fourier space. Notice that $mathcalF : L^2 to L^2$ and $mathcalF: mathcalS to mathcalS$ are isometries, so it is sufficient to prove the results for $mathcalFleft(e^itDelta u_0right)$.

For part (a), we have that
$$mathcalFleft(e^itDelta u_0right)(xi) = e^^2 hatu_0(xi)$$
so it suffices to prove that the term ont he right is in the Schwartz space. Consider the norm $[f]_M,K := sup_ leq K |xi|^alpha |partial_xi^beta f(xi)|$, where $alpha, beta$ are multi-indices. Notice that if we differentiate $e^it hatu_0(xi)$ with respect to $xi_j$, the derivative can either 'hit' the oscillating phase $e^it$ to give $-2itxi_j e^it$, or it can hit the function $hatu_0(xi)$. Iterating this, we can see that
$$[e^it hatu_0]_M,Kleq C_Kt^K[hatu_0]_M+K, K$$
since taking the derivative will give us a sum of terms involving polynomials in $xi|$ of order at most $M+K$ and derivatives of $hatu_0$ 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 $e^-i(t+s) = e^^2e^xi$.

For part (c) (assuming you want the limit version), follows from the dominated convergence theorem. Observe that we would like to show that
$$ 0 = lim_t to 0 lVert e^^2 hat u_0 - hat u_0rVert_L^2^2 = lim_t to 0int left|e^^2 - 1right|^2 |hatu_0(xi)|^2;dxi$$
Notice that the integrand is bounded by $4|hatu_0(xi)|^2$, 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 $D_x = frac1i partial_x$ and $P(D_x)$ is obtained by formally substituting the differential operator for $x$ in $P(x)$. Such equations are known as dispersive equations.