Kind of submultiplicativity of the Frobenius norm: $\|AB\|_F \leq \|A\|_2\|B\|_F$? 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? 
 A: This inequality is true. But let me first make a comment. When you say that any matrix norm is submultiplicative ($\|XY\|\le\|X\|\cdot\|Y\|$), you understate that a matrix norm over $M_n(\mathbb C)$ is subordinated to a norm of $\mathbb C^n$:
$$\|A\|:=\sup_{x\ne0}\frac{\|Ax\|}{\|x\|}.$$
But the Frobenius norm is not subordinated, for instance because $\|I_n\|_F=\sqrt{n}$, whereas $\|I_n\|=1$ for any matrix norm. The reason for which the Frobenius norm is submultiplicative is therefore specific; it is more or less a consequence of Cauchy-Schwarz inequality.
Now, back to your question. Both norms are unitarily invariant, in the sense that $\|UAV\|=\|A\|$ whenever $U$ and $V$ are unitary matrices. Therefore we may assume that $A$ is diagonal, with diagonal entries $a_1,\ldots,a_n$. Now, we have $\|A\|_2=\max_i|a_i|$ and therefore
$$\|AB\|_F^2 = \sum |a_i b_{ij}|^2\le \|A\|_2^2 \sum |b_{ij}|^2 =\|A\|_2^2\|B\|_F^2.$$
A: 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,
$$\|Z\|_F^2 = \sum_j \|Z_j\|_2^2 = \sum_j \|XY_j\|_2^2 \leq \sum_j \|X\|_2^2\|Y_j\|_2^2 = \|X\|_2^2\|Y\|_F^2.$$
A: The general result is equivalent to the result for diagonal matrices (with positive entries), since this is what maximizes the LHS, given specified singular values for $X, Y.$ I leave that case up to OP.
