An estimation of a sum by $\ell_p$-norm

Question

A few days ago my professor gave me a homework. After wasting about twenty sheets of paper I essentially reduced the task to the following

Question. Given a real $p\ge 1$, whether there exists $C=C(p)>0$ such that for each natural $n$, each non-negative numbers $x_1,\dots, x_n$, and $y=2^{-1/p}$, we have $$\sum_{i=1}^n \left(\sum_{j=i}^n x_jy^{j-i}\right)^p\le C \sum_{i=1}^n x_i^p.$$

An equivalent formulation is

Question’. Given a real $p\ge 1$, whether a linear operator $$(x_i)_{i=1}^\infty\to \left(\sum_{j=i}^\infty x_iy^{j-i}\right)_{i=1}^\infty,$$

is a continuous map from $\ell_p$ to $\ell_p$ (where $(\ell_p,\|\cdot\|_p)$ it the normed space of all real-valued sequences $x=(x_i)_{i=1}^\infty$ such that $\|x\|^p=\sum_{i=1}^\infty |x_i|^p<\infty$)?

Thanks.

My try. It seems that when $p$ equals $1$, $2$, or $3$ then we can obtain an affirmative answer, expanding the left-hand side (when $p=1$ this is especially easy) and then estimating products of $p$ distinct $x_i$’s by sum of their $p$th powers. But there is an abyss between these particular cases and the general answer.

I hope I can answer the question after some work. But this can be a bicycle invention, because this result can be known (but hard to find). Also it looks nice, so I decide to share the question with the community. Maybe it has a nice solution based on a known (but not by me) inequality. Remark that the straightforward application of Hölder’s inequality to the left-hand side provides too weak bound.

Motivation. If the inequality does not hold for some $p\ge 2$ then we expect to apply a construction of an unconditionally convergent series from [VK] to show that for each infinite set $X$, a Banach ring $\ell_p(X)$ does not have a property required in this dch’ question.

References

[VK] N. Vakhania, V. Kvaratskhelia, On unconditional convergence of series in Banach spaces with unconditional basis, Bull. Georgian Acad. Sci. 3:1 (2009) 20–23.

Answer 1

Let $i\in\{1,\dots,n\}$ be fixed and $c_i:= \sum_{j=i}^n2^{-j/p}$ and $\alpha_j=2^{-j/p}/c_i$. Then $$ \left(\sum_{j=i}^n x_j2^{(i-j)/p} \right)^p=2^i\left(\sum_{j=i}^n x_j\alpha_j c_i \right)^p=2^ic_i^p\left(\sum_{j=i}^n x_j\alpha_j \right)^p $$ and since $\sum_{j=i}^n\alpha_j=1$, Jensen's inequality gives $$ \left(\sum_{j=i}^n x_j2^{(i-j)/p} \right)^p\leqslant 2^ic_i^p\sum_{j=i}^n x_j^p\alpha_j $$ and it follows that $$ \sum_{i=1}^n\left(\sum_{j=i}^n x_j2^{(i-j)/p} \right)^p\leqslant \sum_{i=1}^n2^ic_i^{p-1}\sum_{j=i}^n x_j^p2^{-j/p}. $$ Bounding $c_i$ by $\kappa_p 2^{-i/p}$ and switching the sums in $i$ and $j$ gives the wanted result.