TITLE: Kind of submultiplicativity of the Frobenius norm: $\|AB\|_F \leq \|A\|_2\|B\|_F$?
QUESTION [10 upvotes]: Let $\|\cdot\|_F$ and $\|\cdot\|_2$ be the Frobenius norm and the spectral norm, respectively.
I'm reading Ji-Guang Sun's paper 'Perturbation Bounds for the Cholesky and QR Factorizations' from BIT 31 in 1991.
While deriving the perturbation bound on Cholesky factorization $A=LL^T$ with $A\in\mathbb{R}^{n\times k}$ of rank $k$, he begins with the perturbed matrix $A+E$ and its Cholesky factorization:
$$A+E=(L+G)(L+G)^T.$$
Expanding the RHS and subtracting $A$ from both sides, we have $E = GL^T + LG^T + GG^T$. Then he claims in (2.12) of his paper that
$$\|E\|_F\leq 2\|L\|_2\|G\|_F + \|G\|_F^2\,.$$
If I read $\|L\|_F$ instead of $\|L\|_2$, everything seems perfectly normal to me by the following three properties of any matrix norm:

 $\|X^T\|=\|X\|$ for any matrix norm
 subadditivity: $\|X+Y\|_p\leq\|X\|_p + \|Y\|_p$
 submultiplicativity: $\|XY\|_p\leq \|X\|_p\cdot\|Y\|_p$ for $p=2$ or $F$

But I'm not sure why $\|L\|$ is a spectral norm in this equation even though all other norms are Frobenius norms. Does $\|XY\|_F\leq \|X\|_2\|Y\|_F$ always hold?

REPLY [11 votes]: A simpler, more direct proof that requires no SVD: let $Y_j$ be the $j$th column of $Y$ and $Z_j$ that of $Z=XY$. Then,
$$\|Z\|_F^2 = \sum_j \|Z_j\|_2^2 = \sum_j \|XY_j\|_2^2 \leq \sum_j \|X\|_2^2\|Y_j\|_2^2 = \|X\|_2^2\|Y\|_F^2.$$