The aim of this paper is to establish weighted Hardy type inequality in a broad family of means. In other words, for a fixed vector of weights $(\lambda_n)_{n=1}^\infty$ and a weighted mean $\mathscr{M}$, we search for the smallest number $C$ such that $$\sum_{n=1}^{\infty} \lambda_n \mathscr{M} \big((x_1,\dots,x_n),(\lambda_1,\dots,\lambda_n)\big) \le C \sum_{n=1}^{\infty} \lambda_nx_n \text{ for all admissible }x.$$ The main results provide a definite answer in the case when $\mathcal{M}$ is monotone and satisfies the weighted counterpart of the Kedlaya inequality. In particular, if $\mathcal{M}$ is symmetric, Jensen-concave, and the sequence $\big(\tfrac{\lambda_n}{\lambda_1+\cdots+\lambda_n}\big)$ is nonincreasing. In addition, it is proved that if $\mathcal{M}$ is a symmetric and monotone mean, then the biggest possible weighted Hardy constant is achieved if $\lambda$ is the constant vector.
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