Inequality between permutations of an infinite weighted sum

Question

Let $p\geq 1$ and let $(w_n)_{n\in\mathbb N}$ a non-increasing sequence of positive numbers such that: $$ \lim_{m\in\mathbb N} w_n = 0 \qquad \text{and}\qquad \sum_{n=1}^\infty w_n = \infty $$

We denote $S_\infty = \{ f:\mathbb N\to \mathbb N|\, f \text{ is a bijection}\}$. If $(x_n)_{n\in \mathbb N}$ is a sequence such that:

$$ \sup_{\pi\in S_\infty} \left( \sum_{n=1}^\infty |x_{\pi(n)}|^p w_n \right) <\infty, $$ we say that $(x_{\pi(n)}^*)_{n\geq 1}$ is a non-increasing rearrangement of $(x_n)_{n\geq 1}$ when there exists a permutation of integers, $\sigma$ such that $(x^*_n)_{n\geq 1}=(|x_{\sigma(n)}|)_{n\geq 1}$ is non-increasing.

I want to prove the following inequality:

Suppose that $(x_n)_{n\geq 1}$ is a sequence such that $\sup\limits_{\pi\in S_\infty} \left(\sum\limits_{n=1}^\infty |x_{\pi(n)}|^p\, w_n \right)^{\frac{1}{p}}\!\!<\infty$, then the following inequality holds true: $$ \sup\limits_{\pi\in S_\infty} \left(\sum\limits_{n=1}^\infty |x_{\pi(n)}|^p\, w_n \right)^{\frac{1}{p}}\!\!\leq \left(\sum\limits_{n=1}^\infty |x_n^*|^p\, w_n \right)^{\frac{1}{p}} $$

What I have done so far:

If $\sigma$ is a permutation such that $(x^*_n)_{n\geq 1}=(|x_{\sigma(n)}|)_{n\geq 1}$ and $\pi\in S_\infty$, if suffices to show that: $$ \sup\limits_{\pi\in S_\infty} \sum\limits_{n=1}^\infty |x_{\pi(n)}|^p\, w_n \leq \sum\limits_{n=1}^\infty |x_n^*|^p\, w_n $$

Each one of the series is the limit of an (increasing) sequence of positive terms that is bounded from above, therefore both limits are the suprema of the finite sums. If, for $k\in \mathbb N$ I can find $k'\in\mathbb N$ such that: $$ \sup\limits_{\pi\in S_\infty} \sum\limits_{n=1}^k |x_{\pi(n)}|^p\, w_n \leq \sum\limits_{n=1}^{k'} |x_n^*|^p\, w_n $$

then the desired inequality follows. The problem is that I can't find a good way to choose such $k'$. I tried to look for $\sigma \circ \pi^{-1}(j)$ for $1\leq j\leq k$ but I couldn't figure what to do with $(w_n)_{n\geq 1}$, since they are not affected by $\pi$ or $\sigma$.

I would appreciate if anyone can suggest an elegant argument to finish my proof or provide a different argument.

Answer 1

For $0\le a\le b$ and $0\le v\le w$ we have $$aw+bv\le av+bw\quad (*)$$ Denote $a_n=|x_n|^p.$ Let $a_{n_1}=\max a_n$ (if $n_1$ is not unique we choose the smallest index). Apply the tranposition $\pi_1=(1,n_1)$ and $(*)$ to get $$\sum_{n=1}^\infty a_nw_n\le \sum_{n=1}^\infty a_{\pi_1(n)}w_n$$ Let $a_{n_2}=\displaystyle\max_{n\neq n_1}a_n$ and $\pi_2=(2,n_2).$ Then $$\sum_{n=1}^\infty a_nw_n\le \sum_{n=1}^\infty a_{\pi_1(n)}\le \sum_{n=1}^\infty a_{\pi_2\pi_1(n)}$$ By continuing this process we will get $$\sum_{n=1}^\infty a_nw_n\le \sum_{n=1}^\infty a_{\pi(n)}w_n $$ where $\{a_{\pi(n)}\}$ is the nonincreasing rearrangment of $\{a_n\}.$

Answer 2

Given $\pi\in S_\infty$ and $k\geq 1$, in what follows I will prove that $\sum\limits_{n=1}^k |x_{\pi(n)}|^p w_n \leq \sum\limits_{n=1}^k |x_n^*|^pw_n$.

Lemma 1: Given $n\geq 1$, for all $\sigma \in S_n$ and decreasing sequences $(x_j)_{j=1}^n$ and $(w_j)_{j=1}^n$: $$\sum_{j=1}^n x_{\sigma(j)}w_j \leq \sum_{j=1}^n x_jw_j $$

Proof of Lemma 1: We proceed by induction in $n$.

The case $n=1$ is trivial and for $n=2$ the argument follows from $(x_1 - x_2)(w_1-w_2)\geq 0$.

Suppose the Lemma is true for $n\geq 1$. If $\sigma \in S_{n+1}$, we consider the following two cases:

• If $\sigma(n+1) = (n+1)$: Then it is trivial.
• If $\sigma(n+1) = k <n+1$: There exists $i<n+1$ such that $\sigma(i)=n+1$.

Observe that $x_k\geq x_{n+1}$ and $w_i\geq w_{n+1}$. Applying the case $n=2$, in conjunction with $\tau\in S_{n+1}$ defined by: $$ \tau(j) =\begin{cases} k,& \text{if } j=i\\ n+1,& \text{if }j=n+1\\ \sigma(j),& \text{otherwise} \end{cases} $$ it follows that $\sum\limits_{j=1}^{n+1} x_{\sigma(j)}w_j \leq \sum\limits_{j=1}^{n+1} x_jw_j \hspace{10cm} \blacksquare$

Let $\pi': \{1,\ldots,k\}\rightarrow \{\pi(1),\ldots,\pi(k)\}$ be a bijection such that $(|x_n|^p)_{j=1}^k$ is decreasing. From Lemma 1 it follows: $$\sum_{j=1}^k |x_{\pi(x)}|^p w_n \leq \sum_{j=1}^k |x_{\pi'(x)}|^p w_n \tag{1}$$

Furthermore, it is easy to prove by induction that: $$1\leq j\leq k \implies |x_{\pi'(j)}|^p \leq |x_j^*|^p \tag{2}$$

Finally, from (1) and (2) we have that $\sum\limits_{n=1}^k |x_{\pi(n)}|^p w_n \leq \sum\limits_{n=1}^k |x_n^*|^pw_n$. Therefore, taking the supremum on both sides (and the supremum over $\pi\in S_\infty$) we have: $$ \sup_{\pi\in S_\infty} \left(\sum\limits_{n=1}^\infty |x_{\pi(n)}|^p w_n\right)^{\frac{1}{p}} \leq \left(\sum\limits_{n=1}^\infty |x_n^*|^pw_n\right)^{\frac{1}{p}}. $$