So firstly we defined Gevrey Class as following:
"A function $f$ is in the Gevrey class of order $θ$ if for every $r > 0,$ there are some constants $M,\; a$ such that $$|f(m)^{(t)}| ≤ Ma^m(m!)^θ,\;\, |t| < a"$$
Problem:
Show that
$$ϕ(t) := \left\{\begin{matrix} e^{−1/t^2}, & t>0\\ 0, & t\leq 0 \end{matrix}\right.,$$ is in the Gevrey class of order $θ =\frac{3}{2}.$
My Approach:
So I tried using Cauchy Integral $$ \phi^{(m)}(t)=\frac{m!}{2\pi i}\oint_{\gamma}^{}\frac{e^{-z^{-2}}}{(z-t)^{m+1}}dz $$ with $$ \gamma=\left\{ t+\frac{t}{2}e^{i\omega} \Bigg|0\leq\omega \leq2\pi \right\}. $$
Firstly I tried to calculate the minimum of $\Re(z^{-2}):$
I substituted $\omega=z-2$, which means I needed to minimize $\Re(\omega + 2)^{-2}$ on the unit circle, $x$ is a minimum for it, one must have $xf′(x)<0$, where $f(\omega)=(\omega+2)^{−2}$ so $\frac{x}{(x+2)^3}>0$ or reversing $x^2+6x+12+8\bar{x}=c>0$ since $|x|=1$ and taking imaginary parts we get $\Im x^2−2\Im x=0$ which easily gives $sinθ=0$ or $cosθ=1$, where $x=e^{iθ}$ so $θ=0,π$ and only $θ=0$ gives the expected $\frac{1}{9}$ minimum.
My Problem: So this is the part, where I got stuck and couldn't go further because I am not sure wheater my calculation of the minimum was the good one. I would appreciate any kind of help!
Noting as above that the minimum depends on $t$, so $1/9$ is for $t=2$ but for general $t$ it is $4/(9t^2)$, the problem reduces to maximize $|t^{-m}|e^{-4/(9t^2)}$ or equivalently $x^me^{-4x^2/9}$ for $x>0$
By calculus it is clear that the maximum happens at $x^2=9m/8$ so we need to estimate $m^{m/2}e^{-m/2}$ as the $(9/8)^{m/2}$ gets comingled into the constant $a^m$ with the extra $2^m$ from the integral ${2\pi i}\oint_{\gamma}^{}\frac{1}{(z-t)^{m+1}}dz=(2/t)^m$ once we estimate the numerator of the original integral away
But by Stirling $m!\ge \sqrt {2\pi m}(m/e)^m$ so $m^{m/2}e^{-m/2} \le (m!)^{1/2}$ and we finally get our required approximatiom:
$|\phi^{(m)}(t)| \le (2\sqrt {9/8})^m (m!)^{3/2}$