How can I get an expression for the operator norm?

Question

Suppose $X = \ell^p(\mathbb{N})$ and suppose $\lambda \in \ell^\infty(\mathbb{N})$. Define $M_\lambda \in B(X)$ by $M_\lambda(x)(n) = \lambda(n)x(n)$. My lecturer says that the operator norm of $M_\lambda$ is $$ \|M_\lambda\| = \|\lambda\|_\infty $$ Does he have some formula or theorem for the operator norm to claim this directly without doing a calculation?

Answer 1

The "calculation" is trivial. First, $|\lambda(n)|\le||\lambda||_\infty$ for every $n$, hence $||M_\lambda f||_p\le||\lambda||_\infty||f||_p$, so $||M_\lambda||\le||\lambda||_\infty$.

Since $M_\lambda e_n=\lambda(n)e_n$ for all $n$, this shows that $||M_\lambda||\ge|\lambda(n)|$ for all $n$, and so by taking sups, $||M_\lambda||\ge||\lambda||_\infty$.