Series estimate

Question

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.

Answer 1

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 $\theta\to 0^+$.