Cfrgtkky

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

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

$$alphainmathbb{Z}_+^N$$

I've found this thing $u$ harmonic then $|D^{alpha} u(x_0)|le frac{C_k}{r^{n+k}}||u||$ or $sup_{xin B}|partial^{alpha} f(x)|le C^{|alpha|+1}alpha!$ 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:

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

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 $sup_{xin B}|partial^{alpha} f(x)|le C^{|alpha|+1}alpha!$ that for any ball that:

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

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?

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
$$ frac1R = limsup_{ntoinfty} sqrt[n]{frac {u^{(n)}(x_0)}{n!}}le lim_{ntoinfty} Csqrt[n]{frac {1}{(n!)^{1-s}}} = 0,$$
since $n!sim sqrt{2pi n} (n/e)^n$. 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
$$ sup_{xin B}|partial^{alpha} f(x)|le
C^{|alpha|+1}alpha!le C^{|alpha|+1}alpha!^s$$
starting from
$$ sup_{xin B}|partial^{alpha} f(x)| le C^{|alpha|+1}alpha!^s.$$

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 $frac{d^n}{dx^n} exp(-1/x)$.