Compact operator on function spaces Given an operator $T : l^p \rightarrow l^p$ ($1 \leq p < \infty$) s.t. $Ty= (\beta_1 y_1, \beta_2 y_2, \dots)$ for $y=(y_1, y_2, \dots) \in l^p$, where $\beta= (\beta_1, \beta_2, \dots)$ is bounded, show that if $T$ is compact, then $\lim \beta_n \rightarrow 0$.
I have shown this in the opposite direction, but this way is giving me some trouble. Could someone point me in the right direction?
 A: Let $e_n$ be the standard basis of $\ell^p(\Bbb N)$. You find $T(e_n)=\beta_n\,e_n$. Suppose that $\beta_n\not\to0$, then there is a subsequence and an $\epsilon>0$ $\beta_{n_k}$ so that $\beta_{n_k}>\epsilon>0$ for all $k$. It follows for any $k,l$ so that $k\neq l$:
$$\|T(x_{n_k})-T(x_{n_l})\|=\sqrt[p]{|\beta_{n_k}|^p+|\beta_{n_l}|^{p}}≥\sqrt[p]2 \,\epsilon$$
Thus $T(x_{n_k})$ admits no subsequence that is a Cauchy sequence, and so also no convergent subsequence. But if $T$ is compact $T(\overline{B_1(0)})$ must be pre-compact, thus every sequence in it must have a convergent subsequence. Since $\|x_n\|=1$ it follows that $T(x_{n_k})$ is a sequence in $T(\overline{B_1(0)})$ that has no convergent sub-sequence, contradicting compactness of $T$.
A: $T$ is compact, so the closure of $\left \{ T\vec e_i=\beta_i\vec e_i :i\in \mathbb N\right \}$ is compact, and so contains a Cauchy subsequence, 
$\left \| \beta_{m_k}\vec e_{m_k}- \beta_{n_k}\vec e_{n_k}\right \|=\sqrt[p]{|\beta_{n_k}|^p+|\beta_{m_k}|^{p}}$ 
and since the the LHS is as small as we want for $n,m$ large enough, it is clear that RHS$\to 0$ as $k\to \infty,\ $ which implies that $\beta_{n_k}\to 0$ as $k\to \infty.$ This means that $\left \{ \beta_i \right \}$ is not finite. 
To finish, observe that $\beta_i\in P\sigma (T)$ for each $i\in \mathbb N,\ $ and so that the only possible limit point for $\left \{ \beta_i \right \}$ is $0$.
