For every infinite sequence $x = (x_1, x_2, x_3, ...)$ of complex numbers define $S(x)$ by $S(x_1, x_2, x_3, ...) = (x_1, 2x_2, 3x_3, ...)$. Is $S$ in $\mathcal{L}(\mathcal l^1, \mathcal l^\infty)$?

I argue that $S$ is unbounded and hence not in $\mathcal{L}(\mathcal l^1, \mathcal l^\infty)$.

Proof: Firstly, using $|| x||_\infty \geq || x||_1$, we have that $$||S(x)||_\infty = \sup_n|S(x_n)| = \sup_n|n\cdot x_n|= n\cdot|| x||_\infty \geq n\cdot||x||_1.$$

This means that

$$\frac{||S(x)||_\infty}{|| x||_1} \geq n \rightarrow\infty$$

and hence, $S$ is unbounded with $||\cdot||_\infty$ norm and not a member of $\mathcal{L}(\mathcal l^1, \mathcal l^\infty)$ .

  • $\begingroup$ "$\sup_{n\in N}|nx_n|=n\|x\|_{\infty}$" is meaningless . What is the $n$ on the RHS? $\endgroup$ – DanielWainfleet Apr 21 '17 at 18:40

If $$x_k=(0,\cdots, 0,\underbrace{1}_{k-th} ,0,\cdots)$$ then $\| Sx\|_\infty =k$ Hence $S$ is unbounded.

  • $\begingroup$ Thanks for this. But I was looking for a comment on my proof, although yours is of course more elegant and simple. $\endgroup$ – Vladimir Nabokov Apr 21 '17 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.