# How to prove the strong convexity of the trace distance?

On page $$408$$ of Nielsen & Chuang in the step going from equation $$(9.48)$$ to $$(9.49)$$, I don't see how:

$$\sum\limits_i (p_i - q_i)tr(P \sigma_i) \leq D(p_i, q_i)$$

I proceed as follows:

$$\sum\limits_i (p_i - q_i)tr(P \sigma_i)\leq \sum\limits_i |p_i - q_i| tr(P \sigma_i)$$

How is this $$\leq \frac{1}{2}\sum\limits_i |p_i - q_i|$$?

Let $$A$$ be the set of indices $$i$$ such that $$p_i\ge q_i$$. We have \begin{align} \sum_i(p_i-q_i)\mathrm{tr}(P\sigma_i)\le&\,\sum_{i\in A}(p_i-q_i)\mathrm{tr}(P\sigma_i)\tag1\\ \le&\,\sum_{i\in A}(p_i-q_i)\tag2\\ \le&\,\max_S\sum_{i\in S}(p_i-q_i)\tag3\\ =&\,D(p,q)\tag4 \end{align} where in the first step we drop non-positive terms, in the second we use $$\mathrm{tr}(P\sigma_i)\le 1$$, in the third we take maximum over subsets $$S$$ of the index set and in the final step we use $$(9.4)$$ on page $$401$$.
• Thank you Adam. How do I prove tr$(P\sigma_i) \leq 1$? It seems intuitive as P is a projection operator and tr$(\sigma_i) = 1$.
• Choose a basis $\psi_i$ with $i=1,\dots,k$ of the image of $P$ and extend it to a basis $\psi_i$ with $i=1,\dots,n$ of the full Hilbert space . Then $\mathrm{tr}(P\sigma)=\sum_{i=1}^k\langle\psi_i|\sigma|\psi_i\rangle\le\sum_{i=1}^n\langle\psi_i|\sigma|\psi_i\rangle=\mathrm{tr}(\sigma)=1$. Nov 28, 2022 at 16:21