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I'm trying to follow the Nielsen and Chuang proof (equation 9.49 of Chapter 9, page 408). I reproduce it here for completeness.

With trace distance defined as $D(\rho, \sigma) = \frac{1}{2}tr(|\rho - \sigma|)$, it is known that there exists a projector $P$, such that $D(\rho, \sigma) = tr(P(\rho-\sigma))$. We wish to prove a result for a probabilistic mixture of states, $\sum_i p_i\rho_i$ and $\sum_i q_i\sigma_i$.

There exists a projector $P$ such that

$\begin{align} D\left(\sum_i p_i\rho, \sum_i q_i\sigma\right) &= \sum_i p_i tr(P\rho_i) - \sum_i q_i tr(P\sigma_i) \\ &=\sum_i p_i tr(P(\rho_i - \sigma_i)) + \sum_i (p_i - q_i) tr(P\sigma_i) \\ &\leq \sum_i p_i D(\rho_i,\sigma_i) + D(p_i,q_i) \end{align} $

Here, $D(p_i,q_i)$ is the classical probability distance given by $D(p_i,q_i) = \frac{1}{2}\sum_i |p_i - q_i|$ for any pair of probability vectors.

I don't understand the very last step, particularly the second term and how $tr(P\sigma_i)$ disappeared. Why is $D(p_i,q_i) = \frac{1}{2}\sum_i |p_i - q_i| \geq \sum_i (p_i - q_i) tr(P\sigma_i)$?

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What you are missing is that $$ D(X,Y)=\max_P \mathrm{tr}|P(X-Y)|\ , $$ where the maximum is over all projectors $P$.

Then, the inequality follows since you are replacing one maximization -- the whole expression is maximized over one P -- by many optimizations over independent $P$'s, which thus can take more optimal values (together with the triangle inequality): $$ \max_P \mathrm{tr}|P(\sum X_i)| \le \sum \max_P \mathrm{tr}|PX_i|\ , $$ with $X_i$ all the terms being summed in the second line.

In addition, it is used that $\mathrm{tr}[P\sigma_i]\le \mathrm{tr}[\sigma_i]=1$.


Let me add that this variational characterization of the trace norm -- $$ \|X\|_1=\mathrm{tr}|X| = \max_P \mathrm{tr}|PX| $$ over all projectors $P$ -- is extremely useful, so it is good to keep it in mind. It is a special case of the also very useful min-max-theorem.

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