Let $f \in L^\infty[0,1].$ It is clear that the norm of the multiplication operator $M_f : g \mapsto fg$ on $L^p[0,1]$ is $\|f\|_\infty.$
What happens in the noncommutative situation? Let us consider the full matrix algebra $M_n(\mathbb{C})$ endowed with the Schatten $p$-norm: $\|T\|_p^p = {\mbox{ Tr }} |T|^p$. Denote $M_f$ the diagonal matrix diag$(f_1,..., f_n).$ Does it follow that $$ \|M_fT\|_p \leq \max_i |f_i| \|T\|_p ?$$
Okay, I got it. The minimax theorem for the $k$th singular value $\sigma_k(T)$ of $T \in M_n(\mathbb{C})$ says that $$\sigma_k(T) = \min \: \{\|T(I-P) \| : P \mbox{ is a projection of rank } k-1 \},$$ where $\|\cdot\|$ denotes the operator norm. Then the inequality $$ \sigma_k (M_fT) \leq \|M_f\| \sigma_k(T) = \max_i|f_i| \sigma_k(T)$$ holds. Hence $$ \mbox{ Tr } |M_fT|^p = \sum_{k=1}^n \sigma_k(M_fT)^p \leq \sum_{k=1}^n \|M_f\|^p \sigma_k(T)^p; $$ that is, $$ \|M_fT\|_p \leq \max_i|f_i| \|T\|_p.$$