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c) Let $a>0$. Find, for each $k=0,1,cdots$,
$$sup_{lambdage 0}{lambda^ke^{-alambda^2/2}}.$$

d) Define, for $xinmathbb{R}$,

$$v(x) = int_{mathbb{R}}e^{ixlambda-alambda^2} dλ.$$

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 $sle 1$ where the Gevrey class is defined.

I believe they are closely reated. I know that for d) I have to show that $suplimits_K |partial^{alpha}v|le C^{|alpha|+1}alpha!^s$ and I believe that the $1/2$ from c) will appear in the $s$ of the Gevrey class.

UPDATE:

$$frac{d}{dx}lambda^ke^{-alambda^2/2} =e^{-(a lambda^2)/2} lambda^{-1 + k} (k - a lambda^2)=0implies k = alambda^2implies lambda = pmfrac{sqrt{k}}{sqrt{a}},$$
so the sup is the value of $lambda^ke^{-alambda^2/2}$ at $-dfrac{sqrt{k}}{sqrt{a}}$? I think I need a stronger argument. How to show it is actually the sup?

$defd{mathrm{d}}defe{mathrm{e}}defi{mathrm{i}}defR{mathbb{R}}defpeq{mathrel{phantom{=}}{}}$For (c), if denoting $f(λ) = kln λ - dfrac{1}{2}aλ^2$ ($λ > 0$), 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 expleft( -frac{1}{2} aλ^2 right) = sup_{λ > 0} λ^k expleft( -frac{1}{2} aλ^2 right)\
&= sup_{λ > 0} exp(f(λ)) = expleft( fleft( 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 $k geqslant 0$,$$
v^{(k)}(x) = i^k int_{R} λ^k expleft( i xλ - aλ^2 right) ,d λ, quad forall x in R
$$
thus for $x in R$,begin{align*}
|v^{(k)}(x)| &= left| int_{R} λ^k expleft( i xλ - aλ^2 right) ,d λ right| leqslant int_{R} |λ^k expleft( i xλ - aλ^2 right)| ,d λ\
&= int_{R} |λ|^k e^{-aλ^2} ,d λ = 2 int_0^{+∞} λ^k expleft( -frac{1}{2} aλ^2 right) · expleft( -frac{1}{2} aλ^2 right) ,d λ\
&leqslant 2 left( frac{k}{e a} right)^{frac{k}{2}} int_0^{+∞} expleft( -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 $R$.