For every infinite sequence x=(x1,x2,x3,...) of complex numbers define S(x) by S(x1,x2,x3,...)=(x1,2x2,3x3,...). Is S in L(l1,l∞)?
I argue that S is unbounded and hence not in L(l1,l∞).
Proof: Firstly, using ||x||∞≥||x||1, we have that ||S(x)||∞=sup
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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