Upper bound of the k-th derivative of $g(t) = e^{-t^{-a}}$

Asked Feb 04 '22 at 18:54

Active Mar 29 '24 at 01:45

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This is part of Stein and Shakarchi's Fourier Analysis, Chapter 5, Problem 4.

For $a > 0$, consider the function deﬁned by $$g(t) = \begin{cases} e^{-t^{-a}} & \text{if } t > 0\\ 0 & \text{if } t \leq 0. \end{cases}$$

One can show that there exists $0 < \theta < 1$ depending on $a$ so that

$$| g^{(k)} (t) | \leq \frac{k!}{(\theta t)^k} e^{-\frac{1}{2}t^{−a}}$$

for $t > 0$.

I cannot figure out to do it.

asked Feb 04 '22 at 18:54

Petra Axolotl

Have you tried induction. – hamam_Abdallah Feb 04 '22 at 19:04
Fourier analysis, so the Cauchy integral formula isn't allowed? – reuns Feb 04 '22 at 20:21
@hamam_Abdallah Yes. That was the first thing that I tried. I wanted to get some kind of recursive formula for $g^{(k)}$ It got too complicated so I gave up. – Petra Axolotl Feb 05 '22 at 02:58
@reuns Correct. – Petra Axolotl Feb 05 '22 at 02:59

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