Does elementwise matrix inequality extend to norms? The elements of $A$ and $B$ are non-negative and $0 \le A_{ij} \le B_{ij}$ $\forall i,j$.
Is it true that $\Vert A \Vert_p \leq \Vert B \Vert_p$ ? The norm is the operator norm induced by the usual vector $p$-norm.
This is an exercise problem from A Brief Introduction to Numerical Analysis by E. Tyrtyshnikov and am using the book for self-study. My intuition is that it is true, when I think of vectors with all non-negative components, but am unable to give a proof in general.
 A: The operator norm is given by $$||A||_p = \sup_{x\in\Bbb R^n : ||x||_p\leq 1} ||Ax||_p = \sup_{x\in\Bbb R^n:||x||\leq 1}\left(\sum_i \left|\sum_j A_{ij}x_j\right|^p\right)^{1/p}.$$
First, note that since $A$ has non-negative entries the supremum is attained for vectors with nonnegative entries: indeed, if $x$ has a negative entry $x_i$ and norm at most $1$, then the vector $x'=(x_1,\dots,x_{i-1},-x_i,x_{i+1},\dots,x_n)$ has $p$-norm at most $1$ and $||Ax'||\geq ||Ax||$.
If now you only consider the term $|\sum_j A_{ij}x_j|^p$ and expand it we obtain $$\sum_{k_1+\dots+k_m=p} \binom{p}{k_1,\dots,k_m} A_{i1}^{k_1}x_1^{k_1}\dots A_{im}^{k_m}x_m^{k_m}$$
and $(A_{ij}x_j)^{k_j}\leq (B_{ij}x_j)^{k_j}$ by non-negativity. This means that $|\sum_j A_{ij}x_j|^p\leq |\sum_j B_{ij}x_j|^p$. Rewind back to the original formula and we obtain $$||A||_p=\sup_{x\in\Bbb K^n : ||x||_p\leq 1} ||Ax||_p = \sup_{x\in\Bbb R^n:||x||\leq 1}\left(\sum_i \left|\sum_j A_{ij}x_j\right|^p\right)^{1/p} \leq \sup_{x\in\Bbb R^n:||x||\leq 1}\left(\sum_i \left|\sum_j B_{ij}x_j\right|^p\right)^{1/p} = ||B||_p$$
as desired.
