# Operator is continuous by boundedness?

I have a fairly simple question, I just want to make sure I don't oversee something:

Let $l^\infty$ denote the vector space of all bounded sequences equipped with the $\sup$-norm and $c_0$ the subspace of all sequences converging to $0$. Define $$A:l^\infty \to c_0; \quad Ax:= (k^{-1}x_k)_{k \in \mathbb N}$$ Clearly, $A$ is a well-defined linear operator. Show that $A$ is continuous.

Now my solution is simply the observation that $$\lVert Ax \rVert = \sup_k \lvert k^{-1}x_k \rvert \leq \lVert x \rVert_\infty$$ which means $A$ is bounded and, therefore, continuous. In the solution they make use of sequential continuity but essentially use the same inequality. Did I miss something?

• No, I think you are fine. $||Ax|| \leq ||x||$ for all $x$ is clear enough. A good question, though, would be to find the norm itself. Or no, maybe that is easy : take $x_k = 1,0,0,0,0, \ldots$,then $||Ax|| = ||x|| = 1$. Sep 15, 2017 at 23:38
• Agreed. You are right on. Sep 16, 2017 at 0:13