Product Inequality for Schatten p-norm of Matrices

Question

Is it true that for all square, complex matrices A, B $$ \left\|AB\right\|_p\leq\left\|A\right\|\left\|B\right\|_p$$

where $\left\| .\right\|_p$ refers to the Schatten p-norm and $\left\| .\right\|$ refers to the spectral norm? How would I prove this?

Answer 1

One can use the minimax principle for singular values to prove that $\sigma_k(AB)\le\sigma_1(A)\sigma_k(B)$ for each $k$. The inequality in question now follows directly.

Answer 2

You can also prove this (and most Schatten norm results) via the theory of majorization

It's worth noting that p norms for real non-negative vectors and Schatten p norms for diagonal Positive (semi)definite matrices are essentially the same thing. In both cases the norms are homogenous with respect to re-scaling by positive numbers and they are sub-additive (i.e. as norms they must obey triangle inequality). Thus they are convex. They are also increasing functions in the sense that (when we restrict to real non-negative values) if we have the following component-wise inequalities
$\mathbf 0 \leq \mathbf x_1 \leq \mathbf x_2\implies \Big\Vert \mathbf 0\Big\Vert_p \leq \Big\Vert\mathbf x_1\Big\Vert_p \leq \Big\Vert\mathbf x_2\Big\Vert_p$

Letting $\Sigma_Z$ be the $n\times n$ diagonal matrix containing the singular values of $Z$ in the usual ordering of largest to smallest

$\Sigma_{AB} \preceq_w \Sigma_{A}\Sigma_{B}$
where $\preceq_w$ denotes weak majorization. (This takes some work to prove and e.g. one may find a proof in the book Inequalities: The Theory of Majorization by Olkin et. al)

Putting this all together
$\Big \Vert AB\Big\Vert_{S_p} $
$= \Big \Vert \Sigma_{AB}\Big\Vert_{S_p}$
$\leq \Big \Vert \Sigma_{A}\Sigma_{B}\Big\Vert_{S_p}$
$=\Big(\sum_{k=1}^n \big(\sigma_{k}^{(A)}\big)^p\cdot \big(\sigma_{k}^{(B)}\big)^p\Big)^\frac{1}{p}$
$\leq \Big(\sum_{k=1}^n \big(\sigma_{k}^{(A)}\big)^p\cdot \big(\sigma_{1}^{(B)}\big)^p\Big)^\frac{1}{p}$
$= \sigma_{1}^{(B)} \cdot \Big(\sum_{k=1}^n \big(\sigma_{k}^{(A)}\big)^p\Big)^\frac{1}{p}$
$= \Big \Vert\Sigma_{B}\Big\Vert_{S_\infty}\cdot \Big \Vert \Sigma_{A}\Big\Vert_{S_p}$

where the first inequality comes from applying a function that is symmetric, convex and increasing to a weak majorization relation, and the second inequality comes from the point-wise bound $\big(\sigma_{k}^{(A)}\big)^p\cdot \big(\sigma_{k}^{(B)}\big)^p \leq \big(\sigma_{k}^{(A)}\big)^p\cdot \big(\sigma_{1}^{(B)}\big)^p$. Finally note that when dealing with Schatten norms, that the "spectral norm" is more commonly known as the Schatten $\infty$ norm.