# Gevrey estimate of derivatives

Let $\phi$ be the $C^\infty$ function defined on $\mathbb R_+^*$ by $e^{-t^{-2}}$ and by $0$ on $\mathbb R_-$.

Question: I think that there exists $\rho>0$ such that $$\forall t\in \mathbb R,\forall n\in \mathbb N,\quad\vert{\phi^{(n)}(t)}\vert\le (n!)^{3/2} \rho^{1+n}, \tag{\ast}$$ but I do not have a simple proof. I would like to know if an elementary argument could provide the above estimates.

As an interesting byproduct of this global Gevrey estimate of order $3/2$, there is this nice counter-example by A.N. Tychonov violating Cauchy uniqueness for the heat equation with $$u(x,t)=\sum_{n\ge 0}\phi^{(n)}(t)\frac{x^{2n}}{(2n)!},\quad \partial_t u-\partial_x^2 u=0, \quad u_{\vert t\le 0}=0.$$ Of course to prove convergence in $C^\infty$ of the series defining $u$, some estimates are needed and $(\ast)$ is sufficient.

• What is the question? – lcv Apr 27 '17 at 14:06
• Did you try Faa di Bruno's explicit formula for derivatives of compositions, followed by an elementary combinatorial bound? – Abdelmalek Abdesselam Apr 27 '17 at 14:14
• @Icv I did formulate a question in a new version. – Bazin Apr 27 '17 at 16:21
• @Abdelmalek Abdesselam I tried Faa di Bruno's formula but it got messy. – Bazin Apr 27 '17 at 16:21

Faa di Bruno's formula for derivatives of compositions of functions says $$(f\circ g)^{(n)}(t)=n!\sum_{k\ge 0}\ \sum_{n_1,\ldots,n_k\ge 1} \mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\} \ \frac{f^{(k)}(g(t))\ g^{(n_1)}(t)\cdots g^{(n_k)}(t)}{k!\ n_1!\cdots n_k!}$$ where $\mathbf{1}\{\cdots\}$ stands for the indicator function of the condition within braces. For $f(x)=e^{-x}$ and $g(t)=\frac{1}{t^2}$, we of course have $f^{(n)}(x)=(-1)^n e^{-x}$ and $g^{(n)}(t)=(-1)^n (n+1)!\ t^{-(n+2)}$ for the derivatives. When $n\ge 1$, we thus get $$\phi^{(n)}(t)=e^{-t^{-2}} n! \sum_{k\ge 1} \sum_{n_1,\ldots,n_k\ge 1} \mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\} \frac{(-1)^{n+k}t^{-(n+2k)}}{k!} (n_1+1)\cdots (n_k+1)$$ Now use the trivial bound $\frac{x^m}{m!}\le e^x$ for $x=t^{-2}$ (followed by taking the square root) and for $x=\frac{1}{2}t^{-2}$ in order to get $$t^{-n}\le \sqrt{n!}\ e^{\frac{1}{2}t^{-2}}$$ and $$t^{-2k}\le 2^k\ k!\ e^{\frac{1}{2}t^{-2}}\ .$$ This results in $$|\phi^{(n)}(t)|\le n!^{\frac{3}{2}}\times \sum_{k\ge 1}\ \sum_{n_1,\ldots,n_k\ge 1} \mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\} \ 2^k\ (n_1+1)\cdots (n_k+1)\ .$$ By the arithmetico-geometric inequality and the elementary bound $1+x\le e^x$, we have $$(n_1+1)\cdots (n_k+1)\le \left(\frac{n+k}{k}\right)^k\le e^n\ .$$ Therefore, $$|\phi^{(n)}(t)|\le n!^{\frac{3}{2}}\ e^n\times \sum_{k\ge 1}\ 2^k\ \sum_{n_1,\ldots,n_k\ge 1}\mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\}$$ $$=n!^{\frac{3}{2}}\ e^n\times \sum_{k=1}^{n}\ 2^k\ \left( \begin{array}{c} n-1 \\ k-1 \end{array} \right) =2 n!^{\frac{3}{2}} e^n 3^{n-1}\ .$$ So your bound holds with $\rho=3e$.
• About the penultimate equality: I believe that the sum should be replaced by $\sum_{1\le k\le n} 2^k\binom{n+k-1}{k-1}$, which increases geometrically wrt $n$ anyway. – Bazin May 2 '17 at 16:51
• The $n_i$ are constrained to be $\ge 1$ instead of $\ge 0$ so I think my formula is correct. – Abdelmalek Abdesselam May 2 '17 at 18:12
• Yes, you are right, I have to replace my $n$ by $n-k$ then it gives your formula. Nice job. – Bazin May 2 '17 at 20:41