5

A Gevrey function of order $\rho \geq 1$ is a smooth function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ that satisfies the estimate $$|\partial^\alpha f|\leq AL^{|\alpha|}\alpha!^\rho $$ for some positive constants $A,L$, where $\alpha$ is an arbitrary multi-index.

For $\rho=1$, such functions are easily seen to be analytic and so $\mathcal{G}^1$ does not admit bump functions. On the other hand it is well known that for $\rho>1$ that $\mathcal{G}^\rho$ does admit bump functions.

For a project of mine, I need to construct Gevrey bump functions in arbitrary such classes, and to this end I would like to know:

Q/ How can one find the optimal $\rho$ for the function $\exp(-x^{-\alpha})$?

For $\alpha=1$, we can compute that it suffices to estimate $\sup_{t>0}p_k(t)e^{-t}$, where $p_{k+1}(t)=t^2(p_k(t)-p_k'(t))$ and $p_0=1$. The crude estimate of replacing $p_k$ by its leading term $t^{2k}$ does seem to yield the optimal $\rho=2$ but I have been unable to make this argument rigorous.

Alternatively, I would also welcome either a reference to such a computation or a method of bump function construction where the Gevrey regularity is more transparent.

goonfiend
  • 2,759

1 Answers1

5

Use the Cauchy integral formula with $\gamma$ a circle around $x$ and radius $x/2$ to estimate the derivatives: $$ f^{(m)}(x) = \frac{m!}{2\pi i} \oint_\gamma \frac{e^{-z^{-2}}}{(x-z)^{m+1}} dz. $$ On $\gamma$ the function $Re(z^{-2})$ is given by $$ Re(z^{-2}) = Re\left( \frac{1}{(t+t/2 e^{i\vartheta})^2} \right) = 4\frac{4+4\cos(\vartheta)+\cos(2\vartheta)}{t^2(5+4\cos(\vartheta))^2} $$ and its minimum is $4/(9x^2)$ and hence $$ |f^{(m)}(x)| \le m! \left(\frac{2}{x}\right)^m e^{-4/(9x^2)}\le m! \left(\frac{9 m}{2e}\right)^{m/2}. $$ So you get $\rho=3/2$. For a general power it seems harder to get an exact value for the minimum, but it clearly is of the from $c_\alpha/x^\alpha$ and hence you get $\rho=1+\frac{1}{\alpha}$. You also need to take a radius $x/\alpha$ such that $c_\alpha>0$.

gerald
  • 568