Let T be an trace class operator and A be bounded. I found on wiki: https://en.wikipedia.org/wiki/Trace_class that $$ |Tr(TA)|\leq || T|| ||A|| $$ Does anyone know how to start the proof? And more important, does anyone know if this implies that
$$ |Tr(TA)|\leq |Tr(T)| ||A|| $$
On
With the norms being given as is, this is false, as the other answer says. However, it is true that if $T$ is trace-class and $A$ is bounded, $$|tr(AT)| \leq \|A\| \|T\|_1,$$ where $\|\cdot\|_1$ denotes the trace-class norm. To see this, consider the polar decomposition of $T$ given by $T = W|T|$, and an orthonormal basis $E$ of your Hilbert space $\mathcal{H}$. Then, we get \begin{align} |tr(AT)| &= |\sum_{x \in E} \langle ATx, x \rangle | \\ &= |\sum_{x \in E} \langle|T|^{1/2}x, |T|^{1/2} W^*A^*x\rangle \\ &\leq \sum_{x \in E} \||T|^{1/2}x\| \||T|^{1/2}W^*A^*x\| \\ &\leq (\sum_{x \in E}\||T|^{1/2}x\|^2)^{1/2}(\sum_{x \in E}\||T|^{1/2}W^*A^*x\|^2)^{1/2} \\ &= \|T\|_1^{1/2} \||T|^{1/2}W^*A^*\|_2 \\ &\leq \|T\|_1^{1/2} \||T|^{1/2}\|_2 \|A\| \\ &= \|T\|_1 \|A\|. \end{align} This gives us the result.
If $\|A\|$ denotes the operator norm, this is false. Take for example $A=T$ an orthogonal projection of rank $N$. Then $\mathrm{Tr}(TA) = N$ but $\|A\|\,\|T\| =1$.
Your second inequality is also false. Think to what happens if $\mathrm{Tr}(T)=0$ and $A=T^*$ ... For example if $\pi_1$ and $\pi_2$ are two rank $1$ orthogonal projections on orthogonal subspaces and $A=T= \pi_1-\pi_2$, then $\mathrm{Tr}(T)=0$ but $\mathrm{Tr}(TA)=\mathrm{Tr}(\pi_1+\pi_2) = 2$.