sequence space question

Question

Consider a sequence $x := (x_n)_{n=1}^{\infty} \subseteq \mathbb{R}$, with $\sup_{n \ge 1}|x_n| < \infty$.

Can we always find $y := (y_n)_{n=1}^{\infty} \subseteq \mathbb{R}$ and such that:

1. $\sum_{n=1}^{\infty}|y_n| = 1$
2. $\sum_{n=1}^{\infty}|x_n y_n| = \sup_{n \ge 1}|x_n|$

I really have no clue how to solve this question, I will appreciate any help!

Answer 1

Notice that you always have $$ \sum_n|x_ny_n|\leq\sup_n|x_n|\,\sum_n|y_n|=\sup_n|x_n|. $$ So the question is whether the inequality is an equality.

If $\{x_n\}$ has a maximum, say $x_k$, then you can take $y_k=1$, $y_j=0$ when $j\ne k$. The problem arises when the supremum is not a maximum. If we denote, as usual, $\|x\|_\infty=\sup_n|x_n|$, and $|x_n|<\|x\|_\infty$ for all $n$, then for any $y_n\ne0$, $$ |x_ny_n|<\|x\|_\infty\,|y_n|. $$ It follows that $$\sum_n|x_ny_n|<\|x\|_\infty\sum_n|y_n|=\|x\|_\infty. $$ You cannot have equality because you would get $\sum_n|(\|x\|_\infty-|x_n|)\,|y_n|=0$, forcing $|x_n|=\|x\|_\infty$ whenever $y_n\ne0$.

Answer 2

No. You can come within $\epsilon$, but you can't do it exactly.

Say $x_n=1-1/n$. Then $\sup |x_n|=1$, but since $x_n<1$ for all $n$ if $\sum|y_n|=1$ then $\sum|x_ny_n|<\sum|y_n|=1$.