# 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?

• c) looks fairly easy by differentiating this function of $\lambda$. Nov 27, 2018 at 19:15
• @OlivierMoschetta please take a look at my update
– PPP
Nov 27, 2018 at 19:55
• Well $\lambda\geq 0$ by assumption so only the root $+\sqrt{k/a}$ is of interest. You can show by examining the derivative that $f$ is increasing on $[0,\sqrt{k/a}]$, then decreases so that the maximum is taken there. Nov 27, 2018 at 21:46

$$\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\R{\mathbb{R}}\def\peq{\mathrel{\phantom{=}}{}}$$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 $$\R$$.

• thank you. I just opened two more bounties on thigns related to this question, would be nice if you could take a look math.stackexchange.com/questions/3018750/… math.stackexchange.com/questions/3018740/…
– PPP
Dec 1, 2018 at 15:25
• why $\sup_{λ \geqslant 0} λ^k \exp\left( -\frac{1}{2} aλ^2 \right) = \sup_{λ > 0} λ^k \exp\left( -\frac{1}{2} aλ^2 \right)$?
– PPP
Dec 4, 2018 at 19:26
• I cannot see $|v^{(k)}(x)|$ as less than $C^{k+1}k!^{1/2}$
– PPP
Dec 4, 2018 at 20:09
• I think the sup on $\lambda\ge 0$ is the same as sup on $\lambda>0$ is because of the open interval on the derivative, right?
– PPP
Dec 4, 2018 at 20:11
• @LucasZanella For your last question, the two sups are equal because the function equals $0$ for $λ=0$ and $\sup A∪B=\max(\sup A,\sup B)$.
– Ѕааԁ
Dec 4, 2018 at 23:47

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 $$1/2$$.