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Let $p\in[1,\infty)$ and consider $\ell_p$. Let $A=\{x=(x_n)\in\ell_p: x_n\ge0\text{ for all }n\in \mathbb{N}\}$.

Is there a sequence $u=(u_n)\in\ell_p$ such that $\inf\{x, n\cdot u\}\uparrow x$ as $n\to\infty$ for all $x=(x_n)\in A?$ Here $n\cdot u$ denotes the pointwise product that is, $n\cdot u=(n\cdot u_1, n\cdot u_2,\ldots)$

I feel like such a $u=(u_n)$ cannot exists, because given an $\ell_p$ sequence, I can always find a sequence that converges at a faster rate. But I don't know how to process this into a correct mathematical argument.

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What's $n . u$? Is it each element in the sequence $u$ scaled by $n$? – Host-website-on-iPage Jul 10 '12 at 8:18
How to understand $\inf\{x,n\cdot u\}\uparrow x$ as $n\to\infty$ ? – Norbert Jul 10 '12 at 9:34
Doesn't every $u$ for which $u_n \gt 0$ for all $n$ have the property you want? (I suppose $\inf\{u,k\cdot u\} = u \wedge k\cdot u$ must refer to the lattice operation $\wedge$ = pointwise minimum of $x$ and $k\cdot u$. Then $x \wedge ku \to x$ pointwise, monotonically and hence also in the $\ell^p$-norm as $k \to \infty$ whenever $x \in A = \{x \in \ell^p\,:\, x \geq 0\}$.) – t.b. Jul 10 '12 at 15:38

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