$T_{a}(\text{ball}(\ell^p))$ is compact, where $T_a\colon\ell^p\to\ell^p$, $(T_a x)_{n}:=a_{n}x_{n}$ and $\mathbb{C}\ni a_n\to0$ as $n\to\infty$ Let $a\in\ell^{\infty}=\ell_{\mathbb{C}}^{\infty}$ and define $T_a\colon\ell^p\to\ell^p$, $(T_a x)_{n}:=a_{n}x_{n}$ for $1\leq p<\infty$. Suppose that $a_n\to0$. How do I prove that $T_{a}(\text{ball}(\ell^p))$ is compact? I have shown that $T_{a}$ is a compact operator. So $\overline{T_{a}(\text{ball}(\ell^p))}$ is compact. If $p\neq1$, then I think I know how to prove that $T_{a}(\text{ball}(\ell^p))$ is compact (by using reflexivity of $\ell^{p}$ and weak topology arguments). But I'm stuck on the case $p=1$. Any suggestions are greatly appreciated! Maybe there is a method that proves it for all $1\leq p<\infty$ at once?

We can use the following criterion of compacity in a Banach space.
A subset $A$ is precompact (namely $\bar A$ is compact) if (and only if) it is bounded, and for every $\epsilon$ there is a finite dimensional vector subspace  $V$ such that for every $y\in A$ $d(y,V)\leq \epsilon$
Let $ V_n$ the subspace of sequences such that $\forall k\geq n, x_k=0$ . 
Then, for every sequence $x$ in the unit ball, $d(T_ax, V_n)\leq (\Sigma_{k>n} \vert a_k u_k \vert ^p )^{1/p}\leq Sup_{k>n}  \vert a_k\vert $. This proves the result.
To prove the criterion, note that if a set $A$ satisfy it it is at a the distance $\epsilon $ of a closed ball of this finite dim. vector space, a set wich is covered by a finite number of ball of radius $\epsilon$. So $A$ is contained in a finite union of balls of radius 2$\epsilon$. 
