Question if an operator is compact

Question

So here is my problem,

Let $$J_p:\ell^p\rightarrow c_o$$ be the canonical embedding where $c_0:=\{x_n\subseteq\mathbb C:x_n\rightarrow 0\quad n \rightarrow\infty\}$. I have to decide whether the upper operator is compact. I thought that is indeed compact since one can define, $$J_p^{(k)}(x_n)=\begin{cases}x_n\quad \text{if}\;n\leq k\\ 0\quad\text{else} \end{cases}$$ and the following holds, $$||J_p(x_n)-J_p^{(k)}||_p=\sup_{k>n}|x_n|\rightarrow 0\quad k\rightarrow \infty$$ we can conlude that a sequence of finite ranked operators which in particular are comapact, converges to $J_p$ in the operator norm.

Is my argumentation correct?

Thanks!

Answer 1

More generally, none of the natural embeddings $\ell^p\to \ell^q$ for $1\le p<q\le \infty$ are compact. The reason is the same as in the comment by John: the infinite sequence of standard basis vectors $e_n$ is bounded in the $\ell^p$ norm but is uniformly separated with respect to $\ell^q$ norm, hence has no convergent subsequence.