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Show that there exists a solution for the heat operator (in one spatial variabe) that doesn't belong to the Gevrey class of order $s$ for all $s<2$

I already defined the Gevrey class here: $\sup_K |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s$ then $u$ is analytic for $s\le 1$ but here is the definition again:

A function $u\in C^{\infty}$ belongs to the Gevrey Class of order $s$ if for every compact $K$ of $\Omega$ there is a constant $C$ such that

$$\sup_K |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s, \ \alpha\in\mathbb{Z}_+^N$$

This exercise comes after it asks me to prove that a function on the Gevrey class is analytic, so maybe it helps in the solution of this exercise.

I've studied the general form of the solution for the wave equation but I couldn't find it to be something such that its derivatives are bound by that constant in the Gevrey Class definition.

PPP
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  • Which variable are you referring to? Gevrey class in x, or in t, or in both (x,t)? – timur Dec 04 '18 at 01:06
  • @timur I don't understand. I thought the gevrey class was in $s$ – PPP Dec 04 '18 at 01:27
  • Gevrey class is a set of functions $u$. What is $u$ a function of? – timur Dec 04 '18 at 01:30
  • @timur just noticed that my notes don't speciy this. On wikipedia, https://en.wikipedia.org/wiki/Gevrey_class, it says $u$ is a function on $\mathbb{R}^n$ . I'd say it's on $x$ only, but I'm not sure – PPP Dec 04 '18 at 01:38

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