what is the smallest p such the following is an element of $l_p$ Suppose that $\{x_k\}_{k = 1}^{\infty}$ and $\{y_k\}_{k = 1}^{\infty}$ are elements of $l_s$. Then what is the smallest p such that $h_k$ defined by $x_k y_k = h_k$ is an element of $l_s$. I solved this problem for $l_4$ and I have found it is $l_2$, so I was wondering how can I do the general version of this question.
 A: Let's write $\odot$ for the componentwise multiplication of two sequences, so $(x\odot y)_k = x_k y_k$.
I interpret the question as

For a given $s > 0$, what is the smallest $p > 0$ such that $x\odot y \in l_p$ for all $x,y \in l_s$?

If we do not universally quantify, the answer depends on $x$ and $y$ of course.
With the universal quantification, the key is the inequality
$$ab \leqslant \frac{1}{2}(a^2 + b^2)\tag{$\ast$}$$
for $a, b \in [0,+\infty)$. Thus, we have
$$\lvert (x\odot y)_k\rvert^p = \lvert x_ky_k\rvert^p = \lvert x_k\rvert^p \lvert y_k\rvert^p \leqslant \frac{1}{2}\bigl(\lvert x_k\rvert^{2p} + \lvert y_k\rvert^{2p}\bigr)$$
and it follows that $x\odot y \in l_p$ for all $p \geqslant s/2$. Specialising to $y = x$, we note that
$$x\odot x \in l_p \iff x \in l_{2p},$$
and since
$$\bigcup_{r < s} l_r \subsetneqq l_s$$
it follows that for every $p < s/2$ we can find $x,y \in l_s$ such that $x\odot y \notin l_p$.
Hence the answer is $p = s/2$. (The argument above is for $s < +\infty$, but the conclusion also holds for $s = +\infty$.)
