Let $E=l^p=\left\{x=\left(x_n\right): x_n\in\mathbb{R}, \sum_{1}^{\infty}\left|x_n\right|^p<\infty\right\}$ with $1\le p<\infty$, $\left\|x\right\|_E=\left(\sum_{1}^{\infty}\left|x_n\right|^p\right)^{1/p}.$ Let $\left(\lambda_n\right)$ is a bounded sequence in $\mathbb{R}$ and consider the operator $T$ defined by $$T(x)=\left(\lambda_1x_x,\ldots,\lambda_nx_n,\ldots\right)$$ where $x\in l^p$.
Prove that $T$ is a compact operator from $E$ to $E$ iff $\lambda_n\rightarrow 0$.

Can anyone give me some hints to solve this problem? Thank you!

  • $\begingroup$ Maybe show it is an SOT limit of finite rank operators. $\endgroup$ – Cameron Williams Jun 14 '15 at 0:51
  • $\begingroup$ What finite rank operators should I use ? $\endgroup$ – Omega Jun 14 '15 at 1:00
  • $\begingroup$ Your finite rank operators should be like finite versions of your operator. What would this correspond to in your case? $\endgroup$ – Cameron Williams Jun 14 '15 at 1:02
  • 2
    $\begingroup$ I think $T_k(x)=(\lambda_1x_1,\ldots,\lambda_kx_k,0\ldots,0,\ldots)$. Is it right ? $\endgroup$ – Omega Jun 14 '15 at 1:10
  • $\begingroup$ That's exactly what I was going for. $\endgroup$ – Cameron Williams Jun 14 '15 at 3:02

Hint for $\implies$: If $\lambda_n \not \to 0,$ then for some $\epsilon>0,|\lambda_{n_k}| > \epsilon$ along a subsequence $n_k.$ Letting $e_n$ denote the usual "basis" vector, consider the sequence $e_{n_k}$ in the unit ball of $l^p$ and its images under $T.$

  • $\begingroup$ If there is a sub-sequence of $T(e_{n_k})$ that tend to $u\in l^p$, then $u$ must be equal to zero. However, $||T(e_{n_k})||=\lambda_{n_k}>\epsilon$, so there is no sub-sequence of $T(e_{n_k})$ that tends to zero. Is it what you mean ? $\endgroup$ – Omega Jun 14 '15 at 2:25
  • $\begingroup$ I was thinking $T(e_{n_k})$ has no Cauchy subsequence. $\endgroup$ – zhw. Jun 15 '15 at 1:27
  • $\begingroup$ Ok! I understood your goal. $\endgroup$ – Omega Jun 16 '15 at 1:48

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