Two basic questions on $p-$summable sequences

Question

Let $X$ be a Banach space and $(x_{n})_{n=1}^{\infty}$ be a $p-$summable sequence in $X$. My basic questions are the following:

1. For any $\epsilon>0$, is there a sequence $(\xi_{n})_{n=1}^{\infty}$ such that $1\leq \xi_{n}\rightarrow \infty(n\rightarrow \infty)$ and $\|(\xi_{n}x_{n})_{n=1}^{\infty}\|_{p}\leq \|(x_{n})_{n=1}^{\infty}\|_{p}+\epsilon$;

2. Is there a sequence $(\xi_{n})_{n=1}^{\infty}$ such that $1\leq \xi_{n}\rightarrow \infty(n\rightarrow \infty)$, $\sum_{n=1}^{\infty}\frac{1}{\xi_{n}^{p}}<\infty$ and $(\xi_{n}x_{n})_{n=1}^{\infty}$ is a $p-$summable sequence in $X$?

Answer 1

I assume "$p$-summable" means "absolutely $p$-summable", i.e., $\sum \|x_n\|^p < \infty$. Since all that matters about $x_n$ is its norm, the question reduces to the scalar case.

1. Yes. Find a strictly increasing sequence of indices $(n_k)$, $k \geq 1$, such that $\sum_{i \geq n_k} \|x_i\|^p < \frac{\epsilon}{2^k}$. Then define $\xi_n = 1$ for $n < n_1$ and $\xi_n = (k+1)^{1/p}$ for $n_k \leq n < n_{k+1}$. Now we have $$\sum_{n=1}^\infty \xi_n^p\|x_n\|^p = \sum_{n=1}^\infty \|x_n\|^p + \sum_{k=1}^\infty\sum_{i=n_k}^\infty \|x_i\|^p \leq \sum_{n=1}^\infty \|x_n\|^p + \epsilon,$$ which is good enough.

2. Not for general $(x_n)$. For example, let $x_n = n^{-2/p} \in \mathbb{R}$ and let $(\xi_n)$ be any sequence of positive numbers. For each $n$ we either have $\xi_n^p \geq n$ or $\xi_n^p < n$. In the first case, $\frac{\xi_n^p}{n^2} \geq \frac{1}{n}$, and in the second case, $\frac{1}{\xi_n^p} > \frac{1}{n}$. So $\sum\big(\frac{1}{\xi_n^p} + \frac{\xi_n^p}{n^2}\big) = \infty$ and therefore either $\sum\frac{1}{\xi_n^p} = \infty$ or $\sum \xi_n^p x_n^p = \sum\frac{\xi_n^p}{n^2} = \infty$.