Upper bound for the trace of the product of a matrix and a positive matrix Let $A \in M_N(\mathbb C)$ be positive semi-definite (i.e. $A = C^*C$ for some $C \in M_N(\mathbb C)$) and let $B \in M_N(\mathbb C)$ be arbitrary. Is it then true that
$$\tag{1}
\lvert Tr(AB) \rvert \leq \lVert B\rVert Tr(A)
$$
where $\lVert B \rVert$ denotes the operator norm of $B$?
I have shown the above inequality with the Frobenius norm instead of the operator norm and I have shown that
$$
\lvert Tr(AB) \rvert \leq \lvert Tr(B) \rvert \lVert A \rVert
$$
but I'm unable to prove of disprove (1).
 A: Yes this is true. This is a special case of Hölder's inequality. For an inner product space, Hölder's inequality states that
$$
|\langle x, y\rangle | \leq \|x\|_p \|y\|_q
$$
for all $p,q \geq 1$ with $1/p + 1/q = 1$.
Thinking of the space $M_N(\mathbb{C})$ as a vector space we can equip it with the following inner product $\langle A, B \rangle = \mathrm{Tr}[A^* B]$ where $X^*$ denotes the conjugate transpose of $X$. Our $p$-norm is now the Schatten $p$ norm defined as $\|X\|_p := \mathrm{Tr}[(X^*X)^{p/2}]^{1/p}$ note that $\|X\|_\infty$ is the operator norm. Applying Hölder's inequality in the limiting case $p \to 1$ and $q \to \infty$ we have
$$
\mathrm{Tr}[AB] \leq \|B\| \|A\|_1 = \|B\|\mathrm{Tr}[A].
$$
where for the final equality we used the fact that $A$ is positive semidefinite to write $\|A\|_1 = \mathrm{Tr}[A]$.
A: By von Neumann's trace inequality, $|\operatorname{tr}(AB)|\le\sum_i\sigma_i(A)\sigma_i(B)$. Since the singular values of $A$ are the eigenvalues of $A$ and $\sigma_i(B)\le\|B\|_2$ for each $i$, the result follows.
By the way, your claim that $|\operatorname{tr}(AB)|\leq|\operatorname{tr}(B)|\|A\|_2$ is false. Consider $A=\pmatrix{1&1\\ 1&1}$ and $B=\pmatrix{0&1\\ 0&0}$ for instance.
