# $A$ is compact operator

Let $A : l_p \to l_p, 1 \le p \le \infty$ be an operator defined by $A\left ( x_n \right ) = \left ( a_n x_n \right )$ where $\left \{ a_n \right \}$ is a bounded real sequence. Prove that $A$ is compact if and only if $\lim_{n \to \infty} a_n =0$.

$A$-compact if and only if $A(B)$ is relatively compact, where $B$ is the unit sphere in $l_p$. I don't know how to use this hypothesis. Is there another standard to verify the compactness of an operator?

• Note that finite rank operators (meaning that the dimension of the image is finite) are compact and that the compact operators are norm-closed. So if you can show that an operator is a (norm-) limit of finite rank operators, you know it's compact... – This Is Me Dec 1 '15 at 23:06

Hint for $\implies$: $(e_n)_{n=1}^\infty$ (the standard basis) is a bounded sequence in $\ell^2$, as is any subsequence of $(e_n)$. The image of any such sequence must have a convergent subsequence.
Hint for $\Longleftarrow$: If $a_n \to 0$, show that $A(B)$ is compact.