Proof that a function is quasi-analytic. Error in paper?

Question

Let $$b(x)= \sum_{n=1}^\infty \exp(-2^{n}/n) \cos(2^nx).$$

I am reading a proof where it is claimed that there is a constant $k >0$ such that $$|b^{(r)}(x)| \leq k^r (r \log r)$$ for every integer $r \geq 0,$ where $b^{(r)}$ is the $r$th derivative.

Here is the proof:

Iunderstand what is going on until’ the sentence, on pg.32 that there exists a $r_0 >2/log 2$ such that...

Why does an $r_0$ as they claim exist? Why is $$A(r) < \sum_{n \leq N_r} 2^{rn} \exp{(-2^n/n)}$$

I do not see how this inequality could possibly hold.

Could someone help me in making sense of this proof? Im at loss and have thought about it some time and would be grateful for your help.

Answer 1

I haven't read this in detail, but if I'm reading it correctly then the paper is totally messed up. I suspect that the idea of the argument is to cunningly choose some $N$ such that $$\begin{align}A(r) &= \sum_{n=1}^{N} u_n + \sum_{n=N+1}^\infty u_n \\ &= \sum_{n=1}^{N} u_n + u_{N+1}\sum_{n=N+1}^\infty e^{\theta_{n-1} + \theta_{n-2} + \cdots + \theta_{N+1}} \\ &< N \sup u_n + u_{N+1}\sum_{n=0}^\infty e^{n\theta} \end{align}$$ by proving that $\theta_n < \theta$ for all $n \ge N_r$ for some $N_r$, chosen such that $\theta$ is sufficiently negative that this gives a useful bound on the answer.

I think that

• their argument about $\theta$ is wrong as written, as it only proves the useless result that one can take $\theta = {\bf\color{red}+} \frac{1}{2} r \log{2}$ for their choice $N=N_r$, although I suspect the stated result with minus sign is probably true anyway; and
• they miss out the geometric sum term in the last line of $A(r)$ above.

I suggest trying to

• numerically check then prove the correct, negative bound for an appropriate $\theta < 0$; and
• check that the geometric term fits in nicely to the bound which is being aimed for.