1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ What is the question? $\endgroup$ – lcv Apr 27 '17 at 14:06
  • 1
    $\begingroup$ Did you try Faa di Bruno's explicit formula for derivatives of compositions, followed by an elementary combinatorial bound? $\endgroup$ – Abdelmalek Abdesselam Apr 27 '17 at 14:14
  • $\begingroup$ @Icv I did formulate a question in a new version. $\endgroup$ – Bazin Apr 27 '17 at 16:21
  • $\begingroup$ @Abdelmalek Abdesselam I tried Faa di Bruno's formula but it got messy. $\endgroup$ – Bazin Apr 27 '17 at 16:21
5
$\begingroup$

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 $\rho=3e$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Bazin May 2 '17 at 16:51
  • $\begingroup$ The $n_i$ are constrained to be $\ge 1$ instead of $\ge 0$ so I think my formula is correct. $\endgroup$ – Abdelmalek Abdesselam May 2 '17 at 18:12
  • $\begingroup$ Yes, you are right, I have to replace my $n$ by $n-k$ then it gives your formula. Nice job. $\endgroup$ – Bazin May 2 '17 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.