$\sup_K |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s$ then $u$ is analytic for $s\le 1$

Question

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

$$\sup_K |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s$$

$$\alpha\in\mathbb{Z}_+^N$$

I've found this thing $u$ harmonic then $|D^{\alpha} u(x_0)|\le \frac{C_k}{r^{n+k}}||u||$ or $\sup_{x\in B}|\partial^{\alpha} f(x)|\le C^{|\alpha|+1}\alpha!$ which is close but has no $s$. It's known that for $s=1$ (Evans) such functions are analytic, but I don't know how to prove it for $s <1$.

It looks like that for $s>1$ I can simply insert $C^{|\alpha|+1}\alpha!^s$ in the end of the inequality like this:

$$\sup_{x\in B}|\partial^{\alpha} f(x)|\le C^{|\alpha|+1}\alpha!\le C^{|\alpha|+1}\alpha!^s$$

so for $s>1$ every funtion that satisfies this is analytic?

UPDATE:

I have from $u$ harmonic then $|D^{\alpha} u(x_0)|\le \frac{C_k}{r^{n+k}}||u||$ or $\sup_{x\in B}|\partial^{\alpha} f(x)|\le C^{|\alpha|+1}\alpha!$ that for any ball that:

Given any closed ball $B\subset\Omega$, there exists $C>0$ such that $$\sup_{x\in B}|\partial^{\alpha} f(x)|\le C^{|\alpha|+1}\alpha!$$

and

$f$ is real analytic in $\Omega$

are equivalent statements. Therefore, if we pick a function $u$ from the Gevrey class, it is true that

$$\sup_{\overline{B}} |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s $$

if we pick ${\overline{B}}$ as our compact $K$

therefore

$$\sup_{\overline{B}} |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s \le C^{|\alpha|+1}\alpha! $$

for $0<s<1$. Therefore, by our theorem, it is true that $u$ is real analytic in $\Omega$

is it true?

Answer 1

You are right for $s<1$, these functions are analytic. In fact, such a $u$ is entire. Expanding $u$ at any point $x_0$ where it is originally defined, we have the Taylor series $u(x) = \sum_{n=0}^\infty \frac {u^{(n)}(x_0)}{n!} (x-x_0)^n$. The radius of convergence $R$ of this series satisfies $$ \frac1R = \limsup_{n\to\infty} \sqrt[n]{\frac {u^{(n)}(x_0)}{n!}}\le \lim_{n\to\infty} C\sqrt[n]{\frac {1}{(n!)^{1-s}}} = 0,$$ since $n!\sim \sqrt{2\pi n} (n/e)^n$. So the series converges on $\mathbb R,$ and defines an analytic extension past $K$.

Bur for $s>1$, you cannot get the chain of inequalities $$ \sup_{x\in B}|\partial^{\alpha} f(x)|\le C^{|\alpha|+1}\alpha!\le C^{|\alpha|+1}\alpha!^s$$ starting from $$ \sup_{x\in B}|\partial^{\alpha} f(x)| \le C^{|\alpha|+1}\alpha!^s.$$

It is the other way, an analytic function(i.e. $s=1$) is also Gevrey class of order $s>1$. The Gevrey class for $s>1$ contains non-analytic functions: for instance $\exp(-1/x)\mathbb1_{x>0}$ is Gevrey but not analytic, see my computation in Bounding the extrema of polynomials from $\frac{d^n}{dx^n} \exp(-1/x)$.