1
$\begingroup$

Let $\theta\in(0,1)$ be given. I define for $a>0$ and $\lambda \ge 1$, $ S(\lambda,a )=\sum_{k\ge 1} k^{\frac12-\theta}e^{-a\vert k-\lambda\vert}. $ I want to prove that $$ \sup_{\substack{a\in(0,1)\\\lambda \ge 1}}S(\lambda,a)\lambda^{-\frac{1}{2}+\theta} a^{1-\theta}=C(\theta)<+\infty. $$ It looks like a rather trivial question, but I was only able to prove a somewhat weaker estimate.

$\endgroup$

1 Answer 1

2
$\begingroup$

I think the desired inequality is false. Suppose $\theta<\frac 12$; fix $a$ and set $\lambda=2N$ for a large integer $N$. Then \begin{align*} \lambda^{-1/2+\theta}S(\lambda,a)&\ge \lambda^{-1/2+\theta}\sum_{k\le\lambda} k^{1/2-\theta}e^{-a|k-\lambda|}\\ &\ge (2N)^{-1/2+\theta}\sum_{N\le k\le 2N}k^{1/2-\theta}e^{-a(2N-k)}\\ &\ge 2^{-1/2}\sum_{N\le k\le 2N}e^{-a(2N-k)}\\ &\ge 2^{-1/2}\sum_{k\le N}e^{-a(N-k)}\\ &=2^{-1/2}\sum_{k\le N}e^{-ak}\\ &\sim \frac{1}{a\sqrt 2}, \end{align*}

Multiplying by $a^{1-\theta}$, you see the lower bound explodes as $\theta\to 0^+$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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