Inequality for trace of product of matrices given norms of the matrices I have to show that $\sup_{n \in \mathbb{N}}|Tr(A_nB_n)| < \infty$. I know that $\sup_{n \in \mathbb{N}}||A_n||, \sup_{n \in \mathbb{N}}||B_n|| < \infty$, where $A_n,B_n$ are $2 \times 2$ random matrices and $ ||X|| = Tr(XX^T)^{1/2}$. Is there any inequality which I can use?
 A: Following @whuber's suggestion, starting with
$$\begin{eqnarray*}
A &=& \left(\begin{array}{rrrr} a & b \\ c & d \end{array}\right)\\
P &=& \left(\begin{array}{rrrr} p & q \\ r & s \end{array}\right)\\
\end{eqnarray*}_.$$
then the matrix product is
$$\begin{eqnarray*}
AP^T
&=& \left(\begin{array}{rrrr} a & b \\ c & d \end{array}\right)
    \left(\begin{array}{rrrr} p & r \\ q & s \end{array}\right)\\
&=& \left(\begin{array}{llll} ap+bq & \ldots \\ \ldots & cr + ds \end{array}\right)\\
\end{eqnarray*}_.$$
The norm of a matrix
$$\begin{eqnarray*}
||A||
&=& Tr(AA^T)^{1/2} \\
&=& (a^2 + b^2 + c^2 + d^2)^{1/2}
\end{eqnarray*}_,$$
is just the Euclidian norm of the matrix treated as a vector of its coefficients $\mathbf{a} = (a, b, c, d)^T$.
The trace of the product
$$\begin{eqnarray*}
Tr(AP^T)
&=& ap + bq + cr + ds
\end{eqnarray*},$$
is just the dot product (inner product) of the matrices treated as vectors of their coefficients.
The Cauchy-Schwarz inequality states that 
$$ |\mathbf{x}\cdot\mathbf{y}|  \le ||\mathbf{x}||\,||\mathbf{y}||,$$
where the vector norms on the right are Euclidian. This inequality can then be applied to the vectors $\mathbf{a}$ and $\mathbf{p}$ of coefficients of the matrices $A$ and $P$.
$$ |Tr(AP^T)| \le ||A||\,||P|| $$
We are given $\sup_{n\in N}||A_n|| < \infty$, and $\sup_{n\in N}||B_n|| < \infty$. Hence 
$$ \sup_{n\in N}|Tr(AP^T)| \le \sup_{n\in N}(||A||\,||P||) < \infty, \;\;\mbox{and}$$
$$\sup_{n\in N}|Tr(AP^T)| < \infty .$$
However this is not the form required in the question: it gives an inequality for the trace of the product $AP^T$ rather than $AP$. But $||X||=||X^T||$, which follows from the fact that for Euclidian vector norms $||\mathbf{x}^T|| = ||\mathbf{x}||$, or equivalently, for any matrix $X$, $Tr(XX^T)=Tr(X^TX)$. Setting $B=P^T$ 
$$\sup_{n\in N}|Tr(AB)| < \infty .$$
