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](http://dx.doi.org/10.1007%2FBF01931293)' 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? 



Kind of submultiplicativity of the Frobenius norm: $\|AB\|_F \leq \|A\|_2\|B\|_F$?
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
$$\|AB\|\_F^2 = \sum |a\_i b\_{ij}|^2\le \|A\|\_2^2 \sum |b\_{ij}|^2 =\|A\|\_2^2\|B\|\_F^2.$$


