Show that the sum operator is not compact

Question

Define $S:\mathcal{l}_1 \rightarrow \mathcal{l}_{\infty}$ as $$ S((a_n)_{n \in \mathbb{N}}=\{a_1 +a_2 + \ldots + a_n\}_{n \in \mathbb{N}}=\{ \sum_{i=1}^{n} a_i \}_{n \in \mathbb{N}} $$ Show that $S$ is continous but not compact Proof of continuity Let $\{a_n\} \in \mathcal{l}_{1}$, then $$||\{a_n\}||_{\infty}=\sup_{n} | \sum_{i=1}^{n} a_i | \leq \sup_{n} \sum_{i=1}^{n} |a_i | \leq \sum_{i=1}^{\infty} |a_i|=1 ||\{a_n \}||_{1}$$ and therefore $S$ was bounded and continous. I try to found a sequence in $\mathcal{l}_1$ such that the sequence of images no have a convergent subsequence but i not sure to these is the right way. Any suggestion or hel i will very grateful.

Answer 1

As @Feng mentioned in the comments, the sequence $e_k=(0,\dots,0,1,0,\dots)$ is such that $S(e_k)$ does not hae convergent subsequences and this proves the result.

Another way to prove it is to notice that $S(\ell_1)=\ell_\infty$ (to see this, notice that $S((0,\dots,0, 1,-1,0,\dots))=e_k$ and so $S(\ell_1)=\overline{\text{span}}(\{e_k\})=\ell_\infty$): since compact operators have separable image and $\ell_\infty$ is not separable, $S$ cannot be compact.

Yet another way to prove the result is to notice that $S$ commutes with the shift operator $s$ and so it cannot be compact (it's a nice exercise to try and see by yourself why this is the case. I'll add a solution if needed)