Prove that $\|\mathrm{Tr}_B(T)\|_1\le\|T\|_1$ for bounded operators $T$ While reading Holevo's book on Quantum information, I saw the following result: For any bounded operator $T \in \mathcal{B}(H_A \otimes H_B)$, where $H_A$ and $H_B$ are Hilbert spaces with finite dimension, we have
$$\|\mathrm{Tr}_{B}(T)\|_1 \leq \|T\|_1$$
where $\mathrm{Tr}_B(T)$ is the partial trace (see this for the definition in Holevo) of $T$ with respect to $H_B$. The norm here is the Trace norm defined as $\|T\|_1 = \mathrm{Tr}(|T|) = \mathrm{Tr}(\sqrt{T^*T})$.
My approach was to first show that for any operator $T$, $\mathrm{Tr}(T) = \mathrm{Tr}(\mathrm{Tr}_B(T))$ and then to show $\mathrm{Tr}_B(|T|)\geq |\mathrm{Tr}_B(T)|$ (operator inequality). From the definition, I was able to show the first part easily. But the second part, which is like a Jensen inequality was elusive. Also, I'm not sure if it is true but it does hold for regular trace i.e., $\mathrm{Tr}(|T|)\geq |\mathrm{Tr}(T)|$.
Starting with the definition does not seem to help as I do not know how to work with $|Tr_B(T)|$. Could someone throw some light on this? (I appreciate hints as well as a proof if possible) Also is there an alternative, more useful way to define partial trace?
 A: The proof I know uses the partial trace definition via the trace, e.g. for $T\in\mathcal B_1(\mathcal H_A\otimes\mathcal H_B)$ now $\operatorname{tr}_B(T)$ is the unique operator such that $$\operatorname{tr}(\operatorname{tr}_B(T)X)=\operatorname{tr}(T(X\otimes\mathbb 1_B))$$ for all $X\in\mathcal B(\mathcal H_A)$. ($\mathcal B_1$ is the trace class but no worries, in finite dimensions this is the same as $\mathcal B$)
 The key observations are the following:


*

*Every operator admits a polar decomposition $T=U|T|$ with $U$ unitary (or in infinite dimensions: $U$ partial isometry), so in particular $\|U\|=1\,$.

*For any $X\in\mathcal B_1(\mathcal H),Y\in\mathcal B(\mathcal H)$ one has $|\operatorname{tr}(XY)|\leq \|X\|_1\|Y\|\,$.
The rest is simple as now there exists a unitary matrix $U$, such that
$$
\|\operatorname{tr}_B(T)\|_1=\operatorname{tr}(|\operatorname{tr}_B(T)|)=\operatorname{tr}(U\operatorname{tr}_B(T))=\operatorname{tr}((U\otimes\mathbb 1_B)T)\leq \|T\|_1\underbrace{\|U\otimes\mathbb 1_B\|}_{=\|U\|\|\mathbb 1_B\|=1}=\|T\|_1\,.
$$
(Side note that $\operatorname{tr}((U\otimes\mathbb 1_B)T)=|\operatorname{tr}((U\otimes\mathbb 1_B)T)|$ (non-negative) as we showed that it equals $\|\operatorname{tr}_B(T)\|_1$, so we can actually use the above trace inequality.)
