Sub-multiplicativity of the norm of Hadamard product?

Question

I am wondering if the following holds

$$ \lVert A \odot B \rVert_{p} \leq \lVert A \rVert_{p}\lVert B \rVert_{p} $$

where $\odot$ indicates Hadamard product for any $p$, and $A$ and $B$ are matrices (or vectors).

Answer 1

It does if the coefficients of $A$ and $B$ are all positive (but fails in some cases, see bellow):

Since norms are positive and $x \mapsto x^p$ is increasing, this amounts to prove that $\lVert A \odot B \rVert_{p}^{\ p} \leq \lVert A \rVert_{p}^{\ p}\lVert B \rVert_{p}^{\ p}$.

Say $A$ and $B$ are vectors of $\mathbb{R}^n$. Write $I := \{1,\dots, n\}$
We have $$\lVert A \lVert_p^{\ p} = \sum_{(i_1,\dots i_p) \in I^p} \prod_{k=1}^p a_{i_k}$$ Hence, $$\lVert A \lVert_p^{\ p} \lVert B \lVert_p^{\ p} = \sum_{\stackrel{(i_1,\dots i_p) \in I^p}{(j_1,\dots j_p) \in I^p}} \prod_{k=1}^p a_{i_k}\cdot \prod_{k=1}^p b_{j_k}$$

Whereas $$\lVert A \odot B \lVert_p^{\ p} = \sum_{i=1}^n a_i^p \cdot b_i^p$$

It is now clear that $\lVert A \odot B \lVert_p^{\ p}$ is a sub-sum of $\lVert A \lVert_p^{\ p} \lVert B \lVert_p^{\ p}$, where we take only the terms corresponding to $i_1 = \dots = i_p = j_1 = \dots j_p$. Whence the claim.

Important edit
I just realize that what I've written above shows that this fails if all coefficients of $A$ are negative, all coefficients of $B$ are positive, and $p$ is odd!