[논문 리뷰] On Copson's inequalities for $0< p<1$

Peng Gao, Huayu Zhao|arXiv (Cornell University)|2020. 03. 16.

Point processes and geometric inequalities참고 문헌 9인용 수 1

한 줄 요약

이 논문은 $0 < p < 1$ 인 경우 Copson 유형 부등식의 날카운 상수를 확립하며, 시퀀스 $\lambda_n$ 에 대한 충분 조건을 도출하여 다음 부등식이 성립하도록 한다: $$ \sum_{n=1}^\infty \left( \frac{1}{\Lambda_n} \sum_{k=n}^\infty \lambda_k x_k \right)^p \geq \left( \frac{p}{L - p}\right)^p \sum_{n=1}^\infty x_n^p $$ 여기서 $\Lambda_n = \sum_{i=1}^n \lambda_i$ 이고, $L > p$ 는 $\Lambda_n / \lambda_n$ 의 증가율에서 유도된 매개변수이다. 주요 기여는 최적 상수를 보장하는 $\lambda_n$ 에 대한 일반 조건을 제시함으로써, 기존의 $p \leq 1/3$ 에 대한 결과를 확장하고, $\lambda_n = n^{\alpha} - (n-1)^\alpha$ 또는 $\lambda_n = n^{\alpha-1}$ 인 경우에 대해 새로운 명시적 범위의 $p$ 를 제공한다.

ABSTRACT

Abstract Let $(\lambda_{n})_{n \geq1}$ (λn)n≥1 be a positive sequence and let $\varLambda_{n}=\sum^{n}_{i=1}\lambda_{i}$ Λn=∑i=1nλi . We study the following Copson inequality for $0< p<1$ 0p$ L>p : $$\begin{aligned} \sum^{\infty}_{n=1} \Biggl(\frac{1}{\varLambda_{n}} \sum^{\infty }_{k=n}\lambda_{k} x_{k} \Biggr)^{p} \geq \biggl( \frac{p}{L-p} \biggr)^{p} \sum^{\infty}_{n=1}x^{p}_{n}. \end{aligned}$$ ∑n=1∞(1Λn∑k=n∞λkxk)p≥(pL−p)p∑n=1∞xnp. We find conditions on $\lambda_{n}$ λn such that the above inequality is valid with the constant being the best possible.

연구 동기 및 목표

• The authors aim to determine conditions on the sequence $\lambda_n$ such that the Copson-type inequality $$ \sum_{n=1}^\infty \left( \frac{1}{\Lambda_n} \sum_{k=n}^\infty \lambda_k x_k \right)^p \geq \left( \frac{p}{L - p}\right)^p \sum_{n=1}^\infty x_n^p $$ holds with the best possible constant for $0 < p < 1$, where $\Lambda_n = \sum_{i=1}^n \lambda_i$.
• To resolve the open problem of identifying the best constant in Copson's inequality for $0 < p < 1$, particularly when the classical constant $p^p$ is not optimal for $p > 1/3$, and to extend known results for $p \leq 1/3$.
• To generalize the classical Hardy and Copson inequalities to the case $0 < p < 1$ by introducing a parameter $L > p$ derived from the sequence $\Lambda_n / \lambda_n$, and to establish sharpness of the resulting constant.
• To provide explicit ranges of $p$ for which the inequality holds with the constant $\left(\frac{p}{1-p}\right)^p$ when $\lambda_n = n^\alpha - (n-1)^\alpha$ or $\lambda_n = n^{\alpha-1}$, particularly for $\alpha > 1$.
• To develop a new method based on duality and convexity analysis to derive sufficient conditions on $\lambda_n$ ensuring the inequality holds, using auxiliary functions and inequalities involving $L$ and $p$.

제안 방법

• The authors define a dual inequality involving $q = p/(p-1) < 0$, transforming the original Copson-type inequality into a dual form that is easier to analyze via convexity and monotonicity arguments.
• They introduce a parameter $L > p$ defined as $L = \sup_n \left( \frac{\Lambda_{n+1}}{\lambda_{n+1}} - \frac{\Lambda_n}{\lambda_n} \right)$, which governs the growth of the sequence $\Lambda_n / \lambda_n$, and use it to control the sharpness of the constant.
• The main result is derived by analyzing the sign of a function $h_{L,p}(x)$ and its derivative, showing that $h_{L,p}(x) \geq 1$ under certain conditions on $L$ and $p$, which implies the inequality holds.
• For the case $L < 1$, they introduce a refined condition involving a second parameter $M$ such that $\frac{\Lambda_{n+1}}{\lambda_{n+1}} - \frac{\Lambda_n}{\lambda_n} \leq L + M \frac{\lambda_n}{\Lambda_n}$, and derive a new range for $p$ in terms of $L$ and $M$.
• They define two auxiliary functions $a_1(L,p)$ and $a_2(L,p)$ to characterize the parameter space where the inequality holds, and use asymptotic analysis to show that these functions impose non-trivial restrictions on $L$ and $p$.
• The proofs rely on convexity and concavity estimates of rational functions in $x$, and on bounding the derivative of a key auxiliary function $h_{L,M,p}(x)$ from below by a concave function $v_{L,M,p}(x)$, whose minimum is analyzed at endpoints.

실험 결과

연구 질문

• RQ1 What conditions on the sequence $\lambda_n$ ensure that the Copson inequality $$ \sum_{n=1}^\infty \left( \frac{1}{\Lambda_n} \sum_{k=n}^\infty \lambda_k x_k \right)^p \geq \left( \frac{p}{L - p}\right)^p \sum_{n=1}^\infty x_n^p $$ holds with the best possible constant for $0 < p < 1$, where $L > p$ is a parameter derived from $\lambda_n$?
• RQ2 For $\lambda_n = n^\alpha - (n-1)^\alpha$ or $\lambda_n = n^{\alpha-1}$, what is the largest range of $p \in (0,1)$ for which the inequality holds with the constant $\left(\frac{p}{1-p}\right)^p$?
• RQ3 Can the best constant $\left(\frac{p}{1-p}\right)^p$ be achieved for $p > 1/3$ when $\lambda_n = 1$, and if so, under what conditions on the sequence $\lambda_n$?
• RQ4 How does the parameter $L = \sup_n \left( \frac{\Lambda_{n+1}}{\lambda_{n+1}} - \frac{\Lambda_n}{\lambda_n} \right)$ control the sharpness of the constant in the inequality for $0 < p < 1$?
• RQ5 Can a refined condition involving a second parameter $M$ improve the range of $p$ for which the inequality holds when $L < 1$?

주요 결과

• The inequality $$ \sum_{n=1}^\infty \left( \frac{1}{\Lambda_n} \sum_{k=n}^\infty \lambda_k x_k \right)^p \geq \left( \frac{p}{L - p}\right)^p \sum_{n=1}^\infty x_n^p $$ holds for all non-negative sequences $x_n$ if $L = \sup_n \left( \frac{\Lambda_{n+1}}{\lambda_{n+1}} - \frac{\Lambda_n}{\lambda_n} \right) > p$, $L \geq 1$, $0 < p \leq 1/3$, and $a_1(L,p) \geq 0$, where $a_1(L,p)$ is a defined function of $L$ and $p$.
• When $0 < L < 1$, the inequality holds for all non-negative $x_n$ if $p \leq L^2/4$ and $a_2(L,p) \geq 0$, where $a_2(L,p)$ is a second auxiliary function of $L$ and $p$, and this range is sharp in the sense that $a_2(L,p) < 0$ for small $p$.
• For the case $\lambda_n = 1$, the condition $L = 1$ and $p \leq 1/3$ recovers the known sharp constant $\left(\frac{p}{1-p}\right)^p$ from Levin and Stečkin, confirming the result is best possible in this case.
• For $\lambda_n = n^\alpha - (n-1)^\alpha$ with $\alpha > 1$, the inequality holds with constant $\left(\frac{\alpha p}{1 - \alpha p}\right)^p$ for $0 < p \leq p_{1/\alpha} = \frac{1}{4\alpha}$, extending known results to $\alpha > 1$.
• For $\lambda_n = n^{\alpha-1}$ with $\alpha \geq 2$, the inequality holds with the same constant for $0 < p \leq p_{1/\alpha} = \frac{1}{4\alpha}$, and the constant is best possible under the derived conditions.
• A refined condition involving a second parameter $M$ allows for a larger range of $p$ when $0 < L < 1$, with the inequality holding for $p \leq \min\left\{ \frac{L(2L - 1)}{4(2L + M)}, \frac{L(1 - L - 2M)}{2(1 - L - M)} \right\}$, improving upon the $p \leq L^2/4$ bound.