Finding $\sup\limits_{\lambda\ge 0}\{\lambda^ke^{-a\lambda^2/2}\}$ 
c) Let $a>0$. Find, for each $k=0,1,\cdots$,
  $$\sup_{\lambda\ge 0}\{\lambda^ke^{-a\lambda^2/2}\}.$$
d) Define, for $x\in\mathbb{R}$,
$$v(x) = \int_{\mathbb{R}}e^{ix\lambda-a\lambda^2} dλ.$$
Show that $v$ belongs to the Gevrey class of order $1/2$ on $\mathbb{R}$.

This question arises in context with $\sup_K |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s$ then $u$ is analytic for $s\le 1$ where the Gevrey class is defined.
I believe they are closely reated. I know that for d) I have to show that $\sup\limits_K |\partial^{\alpha}v|\le C^{|\alpha|+1}\alpha!^s$ and I believe that the $1/2$ from c) will appear in the $s$ of the Gevrey class.
UPDATE:
$$\frac{d}{dx}\lambda^ke^{-a\lambda^2/2} =e^{-(a \lambda^2)/2} \lambda^{-1 + k} (k - a \lambda^2)=0\implies k = a\lambda^2\implies \lambda = \pm\frac{\sqrt{k}}{\sqrt{a}},$$
so the sup is the value of $\lambda^ke^{-a\lambda^2/2}$ at $-\dfrac{\sqrt{k}}{\sqrt{a}}$? I think I need a stronger argument. How to show it is actually the sup?
 A: $\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\R{\mathbb{R}}\def\peq{\mathrel{\phantom{=}}{}}$For (c), if denoting $f(λ) = k\ln λ - \dfrac{1}{2}aλ^2$ ($λ > 0$), then $f'(λ) = \dfrac{k}{λ} - aλ$ implies that $f$ is increasing on $\left( 0, \sqrt{\dfrac{k}{a}} \right]$ and decreasing on $\left[ \sqrt{\dfrac{k}{a}}, +∞ \right)$. Thus,\begin{align*}
&\peq \sup_{λ \geqslant 0} λ^k \exp\left( -\frac{1}{2} aλ^2 \right) = \sup_{λ > 0} λ^k \exp\left( -\frac{1}{2} aλ^2 \right)\\
&= \sup_{λ > 0} \exp(f(λ)) = \exp\left( f\left( \sqrt{\frac{k}{a}} \right) \right) = \left( \frac{k}{\e a} \right)^{\frac{k}{2}}.
\end{align*}
For (d), the dominated convergence theorem and induction imply that for $k \geqslant 0$,$$
v^{(k)}(x) = \i^k \int_{\R} λ^k \exp\left( \i xλ - aλ^2 \right) \,\d λ, \quad \forall x \in \R
$$
thus for $x \in \R$,\begin{align*}
|v^{(k)}(x)| &= \left| \int_{\R} λ^k \exp\left( \i xλ - aλ^2 \right) \,\d λ \right| \leqslant \int_{\R} |λ^k \exp\left( \i xλ - aλ^2 \right)| \,\d λ\\
&= \int_{\R} |λ|^k \e^{-aλ^2} \,\d λ = 2 \int_0^{+∞} λ^k \exp\left( -\frac{1}{2} aλ^2 \right) · \exp\left( -\frac{1}{2} aλ^2 \right) \,\d λ\\
&\leqslant 2 \left( \frac{k}{\e a} \right)^{\frac{k}{2}} \int_0^{+∞} \exp\left( -\frac{1}{2} aλ^2 \right) \,\d λ = 2 \left( \frac{k}{\e a} \right)^{\frac{k}{2}} \sqrt{\frac{a}{2π}}. \tag{1}
\end{align*}
It is easy to prove by induction that $\left( \dfrac{k}{\e} \right)^k \leqslant k!$ for $k \geqslant 0$, thus (1) implies that$$
|v^{(k)}(x)| \leqslant \sqrt{\frac{2a}{π}} · (a^{-\frac{1}{2}})^k · \sqrt{k!}. \quad \forall x \in \R
$$
Therefore, $v$ belongs to the Gevrey class of order $\dfrac{1}{2}$ on $\R$.
A: I have nothing to add for (c) but for (d), given that the other problem ( Solution for Cauchy Problem $u_t-u_{xx} = 0$ belongs to the Gevrey class of order $1/2$ ) appears first, its not a bad idea to use that problem to solve this. The function in (d) is the inverse fourier transform of
$$ e^{-a \lambda^2} $$
It is well known that this is of the form
$$ v(x) = B e^{-A x^2}$$
for some other constants $A,B$. There exists a fixed $t_0>0$ depending on $A$ such that we can recover this as a constant in $t_0$ times the solution to the heat equation with dirac mass at 0 initial condition(cf fundamental solution), evaluated at $t=t_0$:
\begin{align}
u_{t} = u_{xx} , t>0 \quad u|_{t=0} = \delta_0\\u(t_0,x) = v(x)
\end{align} The dirac mass isn't Schwartz, but since the heat equation doesnt depend explicitly on $t$, we can instead take $U_0(x) := u(x,t_0/2)$ as the initial condition which is Schwartz. $U_0$ generates a solution $U$ to the heat equation, and then by uniqueness of solutions $$U(x,t_0/2) = u(x,t_0) = c(t_0) v(x)$$
so by the previous problem, $v$ is Gevrey of order $1/2$.
