If $x_n\in l^1(\mathbb{K})$ and $x_n\rightarrow 0$ in $X_p$, then $x_n\rightarrow 0$ in $X_1$?

Question

Is it true that if $x_n\in l^1(\mathbb{K})$ and $x_n\rightarrow 0$ in $X_p$, then $x_n\rightarrow 0$ in $X_1$?

Where $X_1= (l^1(\mathbb{K}),\|.\|_1)$ and $X_p=(l^1(\mathbb{K}),\|.\|_p)$

I know that $l^1(\mathbb{K})$ is subspace of $l^p(\mathbb{K})$.

Answer 1

(I assume $p\in \mathbb{N}$ and $p \geq 2$)

It is not true. Let $x_n \in \ell^1$ be the sequence $$ x_{n,k} := \begin{cases} n^{-\frac{2}{p}} & , ~k \leq n \\ 0 &, ~k > n \end{cases}. $$ Then $$ \lVert x_n \rVert_p^p = \sum_{k = 1}^{\infty} \lvert x_{n, k} \rvert^p = \frac{1}{n} \overset{n \rightarrow \infty}{\longrightarrow}0. $$ But: $$ \lVert x_n \rVert_1 = n^{1 - \frac{2}{p}} \geq 1 $$