Proposition:
Fix $1\leq p\leq \infty$ and a bounded sequence of real number $\lambda=(\lambda_i)_{i\in\mathbb{N}}$ and $e_i=\delta_{ij}\in l^p$. Defined the bounded linear operator \begin{equation} K_{\lambda}:l^p\to l^p \quad K_{\lambda}x:=(\lambda_i x_i)\quad \forall x \in l^p\end{equation} Then $K$ is compact if and only if $\lim_{i\to \infty}\lambda_i=0$.
Proof: The condition $\lim_{i\to \infty} \lambda_i = 0$ is necessary for compactness because, if there exists a constant $\delta>0$ and a sequence $1\leq n1 <n2 <n3 <···$ such that $\vert \lambda_{ nk}\vert \geq \delta$ $\forall k\in \mathbb{N}$ ,then the sequence $Ke_{nk} =\lambda_{nk}e_{nk}$, $k\in \mathbb{N} \in l^p$ has no convergent subsequence. The condition $\lim_{i\to \infty} \lambda_i = 0$ implies compactness because then $K$ can be approximated by a sequence of finite rank operators $K_i$ in the norm topology. And since every finite rank operators is compact, we can conclude.
Question: How can we approximate $K$ with those $K_i$? how are those $K_i$ defined?