Show hat $A:c_{00}\to c_{00}$ is compact where $A((a_n) _{n=1}^\infty)=(\frac{a_n}{n}) _{n=1}^\infty$

Question

We define an operator $A:c_{00}\to c_{00}$ by: $$A((a_n) _{n=1}^\infty)=(\frac{a_n}{n}) _{n=1}^\infty$$

Show that $A$ is a compact operator under the max norm.

My try:

I have tried to take $(a^{(m)}) _{m=1}^\infty\subseteq B_{c_{00}}$ where $B_{c_{00}}$ is the unit ball.

Thus, $\forall m \in \mathbb{N}: |(a^{(m)})|\leq1$.

I'm struggling to construct a cauch subsequence of $A(a^{(m)}) _{m=1}^\infty$. Any help would be appreciated.

Answer 1

I suppose you are using the sup norm on $c_{00}$.

Define $A_N((a_n))=(a_1/1,a_2/2,\dots,a_N/N,0,0...)$. It is fairly easy to check that $\|A-A_N\|\leq \max \{\frac 1 {N+1}, \frac 1{N+2},\dots\}$. Each $A_N$ is compact because it has finite rank. Limit of compact operators in operator norm is compact, so $A$ is compact.