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Infinite series involving eigenvalues of the Beltrami-Laplace operator on Riemannian manifolds as well as $L^p$-estimates of eigenfunctions arise in the study of the nonlinear Schrödinger equation (NLS) on compact manifolds. Denote by $\mu_{k} := k(k + 1)$ the eigenvalue of the operator $- \Delta_{\mathbb{S}^2}$ associated to the eigenfunction $e_k \in C^{\infty}(M)$. Now, consider the summation $$ \sum_{k = 0}^{\infty} \frac{1}{ \langle \mu_k - \alpha \rangle \langle \mu_k \rangle^{\varepsilon}}$$ where $\alpha > 0$ (is a positive arbitrary constant), $\varepsilon > 0$ and $\langle x \rangle : = 1 + |x|$. My question is the following:

$$\langle \mu_k - \alpha \rangle^{-1} \langle \mu_k \rangle^{- \varepsilon} \in \ell^{1}_{k}(\mathbb{N}) \mbox{ }, $$

independently of the choice of $ \alpha$?

My failed attempt was to consider two cases: (Case 1) $\mu_k \geq 4 \alpha$. In this case we have $|\mu_k - \alpha| \geq \frac{3}{4} \mu_k$ and one can obtain the desired conclusion. (Case 2) $\mu_k \leq 4 \alpha$. In this case, we have a finite sum, but I would like to prove that the summation is bounded by a constant which does not depends on $\alpha$. Thanks in advance !!!

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  • $\begingroup$ What do you mean by "independence in relation to $\alpha$"? Looks to me like you've shown that for any $\alpha \neq \mu_k$, the sequence is summable. Are you trying to draw a stronger conclusion? $\endgroup$ – Neal Aug 17 '19 at 18:43
  • $\begingroup$ It means that the summation is bounded by a constant which does not depend on \alpha $\endgroup$ – Marcelo Ng Aug 17 '19 at 18:46
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    $\begingroup$ In that case don't you need to restrict $\alpha$ more? Otherwise seems like you could choose $\alpha$ arbitrarily close to some $\mu_k$ to make the sum arbitrarily large. $\endgroup$ – Neal Aug 17 '19 at 18:48
  • $\begingroup$ I would like to prove that if $\alpha >0$ is arbitrary, then there exists $C >0$ such that $\sum < C $ where $C$ is not dependent of $\alpha$. $\endgroup$ – Marcelo Ng Aug 17 '19 at 18:54
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Neal pay attention on that $\langle \mu_k-\alpha\rangle \ge 1$.

Let $l\in \mathbb{N}$ be such that $\mu_l $ is closest to $\alpha$. Consider $k=l+i$, $i\ge 0$ and $k=l-i$, $0\le i\le l$. The rest seems to be easy.

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  • $\begingroup$ Yes, you're correct: I mistook the brackets as parentheses. $\endgroup$ – Neal Aug 20 '19 at 1:40

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