Linear operator on sequences $M_y$ given by $M_y(x) = (y_jx_j)_{j \in \mathbb{N}}$ preserves $l^p$ iff $y \in l^{\infty}$

Question

Let $y = (y_j)_{j \in \mathbb{N}} \subset \mathbb{K}$ where $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$ be a fixed sequence. For each sequence $x = (x_j)_{j \in \mathbb{N}}$ define a new sequence $M_yx := (y_jx_j)_{j \in \mathbb{N}}$. Then $M_y$ is a linear operator on the space of all sequences in $\mathbb{K}$.
Prove that TFAE:

i) $M_y(l^p) \subset l^p$ and $M_y$ is a bounded operator
ii) $y \in l^{\infty}$

I showed the direction ii) $\implies$ i) using Hölder. But have no idea for the opposite direction.

Answer 1

Assume $y \notin l^\infty$, then there is sequence $k_i$ such that $|y_{k_i}| > 2^{i}$.

Let $x_j = 2^{-i}$ if $j = k_i$ and $x_j = 0$ otherwise.

Then $x \in l^p$: $\sum\limits_{j=0}^\infty |x_j|^p = \sum\limits_{i=0}^\infty 2^{-ip} < \infty$, but $M_y(x) \notin l^p$ as it has infinitely many coordinates with absolute value of $1$.