Frobenius norm of product of matrix

Actually there is $$||FG||^2_F \leqslant||F||^2_F||G||^2_F$$ The proof is as follows. \begin{align} \|FG\|^2_F&=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{p}\left|\sum\limits_{k=1}^nf_{ik}g_{kj}\right|^2 \\ &\leqslant\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{p}\left(\sum\limits_{k=1}^n|f_{ik}|^2\sum\limits_{k=1}^n|g_{kj}|^2\right)\tag{Cauchy-Schwarz} \\ &=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{p}\left(\sum\limits_{k,l=1}^n|f_{ik}|^2|g_{lj}|^2\right) \\ &=\sum\limits_{i=1}^{m}\sum\limits_{k=1}^{n}|f_{ik}|^2\sum\limits_{l=1}^{n}\sum\limits_{j=1}^{p}|g_{lj}|^2 \\ &=\|F\|^2_F\|G\|^2_F \end{align} Frobenius norm is like vector norm and similar to $l_2$.

Let $A$ be $m \times r$ and $B$ be $r \times n$. A better bound here is $$ \| A B \|_F \le \|A\| \|B\|_F \quad (*) $$ where $\|A\|$ is the $\ell_2$ operator norm: $$ \|A\| = \max_{\|x\|_2\, \le\, 1} \|A x\|_2. $$ It is also equal to the largest singular value of A. From this definition $\|A x\|_2 \le \|A\| \|x\|_2$ for any vector $x$ in $\mathbb R^r.$

Since $\|A\| \le \|A\|_F$, inequality (*) is a strictly better inequality than the sub-multiplicative inequality for the Frobenius norm.

To see the inequality, let $B = [b_1 \mid b_2 \mid \cdots \mid b_n]$ be the column decomposition of $B$. Then, $A B = [Ab_1 \mid A b_2 \mid \dots \mid Ab_n]$ is the column decomposition of $AB$. It follows that \begin{align*} \| A B \|_F^2 = \sum_{j=1}^n \|A b_j\|^2 \le \|A\|^2 \sum_j \|b_j\|^2 = \|A\|^2 \|B\|_F^2. \end{align*}

EDIT in response to the question in the comments, ``Is there a lower bound for the Frobenius norm of the product of two matrices?''. In general, no, except for the obvious lower bound of zero. Consider the following two matrices \begin{align} A = \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} -b & 0 \\ a & 0 \end{pmatrix}. \end{align} Then $\|A\|_F = \|B\|_F = \sqrt{a^2 + b^2}$, while $\|AB\|_F = 0$.

What if the two matrices are symmetric? Consider \begin{align} A = \begin{pmatrix} a & b \\ b & a \end{pmatrix}, \quad B = \begin{pmatrix} -b & a \\ a & -b \end{pmatrix}, \quad A B = \begin{pmatrix} 0 & a^2-b^2 \\ a^2-b^2 & 0 \end{pmatrix}. \end{align} Then, $\|A\|_F^2 = \|B\|_F^2 = 2(a^2 + b^2)$ while $\|AB\|_F^2 = 2(a^2 - b^2)^2$ which can be made arbitrarily smaller than either of $\|A\|_F^2$ or $\|B\|_F^2$. For example, take $a=b$.

We will prove that $$ \Vert FG \Vert_f^2 \leq \Vert F \Vert_f^2 \cdot \Vert G \Vert_f^2.$$ We have $$\Vert FG \Vert_f^2 = \mathsf{Tr}(FG G^TF^T) = \mathsf{Tr} (F^TFGG^T),$$ by the cyclic property of trace function. To prove the theorem, it is enough if we show that $$\mathsf{Tr}(F^TFGG^T) \leq \mathsf{Tr}(F^TF) \mathsf{Tr}(GG^T).$$ Observe that $F^TF$ and $GG^T$ are positive semidefinite and symmetric matrices. Hence, they are diagonalizable and can be written as $$ F^TF = U\Sigma_F U^\dagger$$ $$GG^T = V\Sigma_G V^\dagger,$$ where $U,V$ are unitary matrices and $\Sigma_F,\Sigma_G$ are diagonal matrices with eigenvalues (non-negative) of $F^TF$ and $GG^T$ respectively as diagonal entries. Again, by using the cyclic property of trace function, we can write the left hand side as $$\mathsf{Tr}(F^TFGG^T) = \mathsf{Tr}(U\Sigma_F U^\dagger V\Sigma_G V^\dagger) = \mathsf{Tr}(V^\dagger U\Sigma_F U^\dagger V\Sigma_G ).$$

It is easy to show the following properties of diagonal matrices: Let $D$ be a diagonal matrix with non-negative diagonal entries.

1) Under any unitary transformation of $D$, the resulting matrix has non-negative diagonal entries.

2) If $M$ is a square matrix with non-negative diagonal entries, then $\mathsf{Tr}(MD)\leq \mathsf{Tr}(M) \cdot \mathsf{Tr} (D)$.

The above properties directly imply that $$\mathsf{Tr}(V^\dagger U\Sigma_F U^\dagger V\Sigma_G ) \leq \mathsf{Tr}(V^\dagger U\Sigma_F U^\dagger V) \cdot \mathsf{Tr}( \Sigma_G ) =\mathsf{Tr}( \Sigma_F ) \cdot \mathsf{Tr}( \Sigma_G ) = \mathsf{Tr}(F^TF) \mathsf{Tr}(GG^T),$$ where the last two equalities follow from the fact that trace is preserved under unitary transformation.