Show that an operator is bounded.

Question

Let $\{\alpha_{mn} ;m,n\geq 1\}$ be scalars satisfying

a- $M=\sup_n\sum_{m\geq 1}|\alpha_{mn}|<\infty $ , and

b- $\sup_n|\alpha_{mn}|<\infty$,

then $(Af)(n) = \sum_{m\geq 1}\alpha_{mn} f(m)$ defines a bounded operator $A$ on $\ell^1$ and $||A|| =M$

Answer 1

By the Holder inequality, $\lVert Af \rVert \leq ||\sum_{m \geq 1} \alpha_{mn} ||_{\infty} ||f||_1$. By (a) and the definition of sup norm, $M=||\sum_{m \geq 1} \alpha_{mn} ||_{\infty}$.

Thus, $A$ is a bounded operator with operator norm at most $M$. Can you get the second part now?