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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 $\mathcal{L}(\mathcal l^1, \mathcal l^\infty)$ .

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  • $\begingroup$ "$\sup_{n\in N}|nx_n|=n\|x\|_{\infty}$" is meaningless . What is the $n$ on the RHS? $\endgroup$ – DanielWainfleet Apr 21 '17 at 18:40
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If $$x_k=(0,\cdots, 0,\underbrace{1}_{k-th} ,0,\cdots)$$ then $\| Sx\|_\infty =k$ Hence $S$ is unbounded.

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  • $\begingroup$ 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

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