Trace Norm properties

Here are some (edit: more) ideas: First, it seems useful to restrict oneself to square matrices by "squaring" A as in this reference (p. 2 bottom of http://www.drhea.net/wp-content/uploads/2011/11/vonNeumann.pdf - just add zeros to $A$ to make it square which does not affect the SVD except some diagonal ones to $U$ or $V$ and some zeros to $\Sigma$).

3. (I think this should anyway only hold if $A \in M_{n\times n}(\mathbb{C})$ is a square matrix.) In that case, I believe you can prove this by considering $$\mathrm{tr}(A) = \mathrm{tr}(U\Sigma V^*) = \mathrm{tr}(\Sigma V^*U) = \sum_i(\Sigma e_i)\cdot(V^*U e_i) \le \sum_i \| \Sigma e_i \| \| V^*U e_i \|\\ \le \sum_i \sigma_i.$$ (Note $U,V,\Sigma$ are all square now.)

4. First of all, let us assume von Neumann's trace inequality ( http://en.wikipedia.org/wiki/Von_Neumann%27s_trace_inequality ). This inequality implies $|\mathrm{tr}(BA)| \le \|B\| \|A\|_1$, i.e.\ $\sup_{\|B\|\le 1} |\mathrm{tr}(BA)| \le \|B\| \|A\|_1$. The other direction follows with the choice $B=VU^*$, where $U,V$ unitary are such that $A=U\Sigma V^*$, i.e. $\sqrt{A^*A} = V \Sigma V^*$, because then $\mathrm{tr}(BA) = \mathrm{tr}(VU^* U\Sigma V^*) = \mathrm{tr}(V\Sigma V^*) = \|A\|_1$.

The statement $|\mathrm{tr}(AB)|\le ||B||\cdot |\mathrm{tr}(A)|$ is untrue in general, consider the matrix $A=B=\begin{bmatrix} 0 & 1\\ 1 & 0\\ \end{bmatrix}$. $|\mathrm{tr}(AB)|=2$ while $|\mathrm{tr}(A)| = 0$.

4. Now we deduce 4) from 5) and von Neumann's trace inequality: $\|BA\|_1 = \sup_{\|C\|\le 1}|\mathrm{tr}(CBA)| \le \sup_{\|C\|\le 1}|\|B\|\mathrm{tr}(CA)| = |\|B\| \|A\|_1$