Bounding norm in $\ell_p$ by the norm in $\ell_{\infty}$ using multiplication by a vector

Question

Let $p \in [1, \infty)$. Is there a vector $y \in \mathbb{R}^{\mathbb{N}}$ such that for every $x \in \ell_p$ we have $\|x\|_p \leq \|xy\|_{\infty}$?

The multiplication is pointwise, and the norm on the right might be infinite.

Thank you!

Answer 1

Some observations:

• if $y$ works, then considering $x$ as the vector whose $n$-th coordinate is $1$ and all the others $0$, we get that $1\leqslant \left\lvert y_n\right\rvert$.
• Consider $x_n= \left\lvert y_n\right\rvert^{-1}$ for $0\leqslant n\leqslant N$, and zero for the others $n$. Then the $\ell^p$ norm of $x$ is $\left(\sum_{n=0}^N \left\lvert y_n\right\rvert^{-p}\right)^{1/p}$ while $\left\lVert xy\right\rVert_\infty =1$. Consequently, we should have $\sum_{n=0}^N \left\lvert y_n\right\rvert^{-p}\leqslant 1 $ and since $N$ is arbitrary, we get $$\tag{*} \sum_{n=0}^{+\infty} \left\lvert y_n\right\rvert^{-p}\leqslant 1.$$

Actually, any sequence $\left(y_n\right)_{n\geqslant 1}$ satisfying (*) does the job, since $$\sum_{n=0}^{ +\infty}\left\lvert x_n\right\rvert^p=\sum_{n=0}^{ +\infty}\left\lvert x_n\right\rvert^p \left\lvert y_n\right\rvert^p \frac 1{\left\lvert y_n\right\rvert^p}\leqslant\left\lVert xy\right\rVert_\infty^p\sum_{n=0}^{ +\infty} \frac 1{\left\lvert y_n\right\rvert^p}\leqslant \left\lVert xy\right\rVert_\infty^p . $$