Problem 16 states to discuss the sense in which $u(\cdot,t) \rightarrow g $ as $t\rightarrow 0^+$ defined by $$ u(x,t) = \frac{1}{(4\pi i t)^{n/2}} \int_{\mathbb{R}^n} e^{\frac{i |x-y|^2}{4t}}g(y)dy \ \ \ (x\in \mathbb{R}^n, \ t >0) $$ using some Lemma. Schrödinger equation given by: $$ \begin{cases} iu_{t} + \Delta u = 0 & \quad \text{in } \ \mathbb{R}^{n} \times (0, \infty), \label{schrgl}\\ u = g & \quad \text{on } \ \mathbb{R}^{n} \times \{t = 0\}. \end{cases} $$
I have tried to show the pointwise convergence $u(\cdot,t) \rightarrow g$ for $t \rightarrow 0$, given $g$ is smooth, compactly supported and am almost there, but a factor of $(-1)^n$ is bothering me.
There is a lemma from the book L.Evans PDE page 221.
Lemma: Let $a \in C^\infty_c(\mathbb{R}^n)$, then $$ \frac{1}{(4 t i \pi)^{n/2}} \int_{\mathbb{R}^n} e^{\frac{i}{4t} |y|^2} a(y) dy = a(0)+ O(t) \quad \quad \text{ for } t \rightarrow 0. $$
If $g \in C^\infty_c(\mathbb{R}^n)$, then the shifted $y \mapsto g(x-y)=:a(y)$ is also in $C^\infty_c(\mathbb{R}^n)$. It holds $a(0) = g(x)$ and $a(x-y)=g(y)$. It makes sense to make the substitution $z=x-y$ in the integral of the lemma so that it is equal to $u(x,t)$, but the Jacobian determinant of this transformation is $(-1)^n$.
Overall, I have: $$ \begin{aligned} u(x,t) &= \frac{1}{(4 t i \pi)^{n/2}} \int_{\mathbb{R}^n} e^{\frac{i}{4t} |x-y|^2} g(y) dy\\ &= \frac{1}{(4 t i \pi)^{n/2}} \int_{\mathbb{R}^n} e^{\frac{i}{4t} |x-y|^2} a(x-y) dy\\ &= \frac{(-1)^n}{(4 t i \pi)^{n/2}} \int_{\mathbb{R}^n} e^{\frac{i}{4t} |y|^2} a(y) dy \\ &= (-1)^na(0)+O(t)\\ &= (-1)^ng(x)+O(t) \end{aligned} $$ which does not necessarily provide the pointwise convergence for odd $n$. Where is my mistake? Thank you.
