# Exercise 9

## from Example Sheet 4

### Exercise 9.

Continuity of the von Neumann entropy (Fannes’ inequality): Suppose $$\rho, \sigma \in \mathcal{D}( \mathcal{H})$$ are states such that their trace distance $$D(\rho, \sigma)$$ satisfies the bound $2 D(\rho, \sigma) = \|\rho - \sigma\|_1 \le 1/e.$ Then $|S(\rho) - S(\sigma)| \le \|\rho - \sigma\|_1 \log d + \eta\left( \|\rho - \sigma\|_1\right),$(1) where $$d = \dim \mathcal{H}$$, and $$\eta(x) := - x \log x.$$

Let us prove this theorem in steps:

1. Let $$r_1 \ge r_2 \ge \ldots \ge r_d$$ be the eigenvalues of $$\rho$$ arranged in non-increasing order, and let $$s_1 \ge s_2 \ge \ldots \ge s_d$$ be the eigenvalues of $$\sigma$$ arranged in non-increasing order. Then prove that: $\|\rho - \sigma\|_1 \ge \sum_{j=1}^d |r_j - s_j|$(2)
2. Check (using elementary calculus) that if $$|r-s| \le 1/2$$, then $|\eta(r) - \eta(s)| \le \eta(|r-s|),$ where $$\eta(x) := - x \log x.$$
3. Use Step 2 and the triangle inequality to prove that $|S(\rho) - S(\sigma)|\le \sum_j \eta(|r_j - s_j|).$
4. Let $$\varepsilon_j := |r_j - s_j|$$, $$\forall\, j=1,2,\ldots, d$$, and $$\varepsilon:= \sum_j \varepsilon_j$$. Let $$\lambda_j := \varepsilon_j/\varepsilon$$ and note that $$\{\lambda_j\}$$ forms a probability distribution. Use this fact and Step 3 to prove that $|S(\rho) - S(\sigma)|\le \varepsilon\log d + \eta(\varepsilon).$
5. Note that $$\eta(\varepsilon)$$ is a monotonically increasing function of $$\varepsilon$$ for $$0 \le \varepsilon\le 1/e$$. Use this to finally arrive at the statement (1).
1. I could not find a short proof of this fact (without assuming something that directly implies it). This subquestion is then nonexaminable; see e.g. the excellent book Matrix Analysis by Bhatia1 for several (more involved) proofs.

2. Without loss of generality, take $$r > s$$ and set $$\delta := r-s$$. Then \begin{aligned} \eta(s+\delta) - \eta(s) - \eta(\delta) &= - (s+\delta)\log (s+\delta) + s \log s + \delta\log(\delta) \\ &= - s\log (s+\delta) - \delta \log(s+\delta) + s \log s + \delta\log(\delta)\\ &= s\log\frac{s}{s+\delta} + \delta \log \frac{\delta}{s+\delta} \\ &\leq 0 \end{aligned} since both $$\frac{s}{s+\delta}\leq 1$$ and $$\frac{\delta}{s+\delta} \leq 1$$. Thus, $\eta(r) - \eta(s) \leq \eta(r-s).$ On the other hand, set $f(\delta) := \eta(s) - \eta(s+\delta) - \eta(\delta).$ It remains to show that $$f(\delta)\leq 0$$ for $$\delta \leq \frac{1}{2}$$ (and $$r=s+\delta\leq 1$$), which implies $\eta(s) - \eta(s+\delta) \leq \eta(\delta)$ meaning that we have proven $|\eta(r) - \eta(s)|\leq \eta(|r-s|).$

First, $$f(0) = 0$$, and $f'(\delta) = - \eta'(s+\delta) -\eta'(\delta)$ while $\eta'(x) = \frac{\mathrm{d}}{\mathrm{d}x}(- x \log x) = - \log x - \frac{1}{\ln 2},$ so $f'(\delta) = \log ( s + \delta) + \frac{1}{\ln 2} + \log (\delta) + \frac{1}{\ln 2}$ and $f''(\delta) = \frac{1}{\ln 2} \left[ \frac{1}{s+\delta} + \frac{1}{\delta}\right] > 0$ for $$\delta > 0$$. Thus, $$f$$ is convex. We wish to prove $$f(\delta)\leq 0$$ for $$\delta \in [0, \min(1-s, \frac{1}{2})]$$, and by convexity, we only need to check the boundaries (since $$f$$ will be less than either boundary point between the two). We have immediately $$f(0)=0$$. Next, consider $$s \geq \frac{1}{2}$$. Then let’s check $$g(s) := f(1-s) = \eta(s) - \eta(1-s) \leq 0$$ for $$s\geq \frac{1}{2}$$. We can see at $$s= \frac{1}{2}$$, $$g(s)=0$$, and $$g'(s) = -\log(s)+\log(1-s) \leq 0$$ since $$1-s \leq s$$. Thus, $$g$$ is decreasing for all $$s\in [ \frac{1}{2},1]$$, proving indeed, $$g(s)\leq 0$$ on that interval.

Lastly, we need to check when $$s < \frac{1}{2}$$ that $$f(\frac{1}{2}) \leq 0$$. In this case, $h(s) := f(\frac{1}{2} ) = \eta(s) - \eta(s+\tfrac{1}{2}) - \eta(\tfrac{1}{2}).$ Then $$h(0) = - 2 \eta(\frac{1}{2})\leq 0$$, and $$h'(s) = - \log(s) + \log(s + \frac{1}{2}) > 0$$, so $$h$$ is increasing. Thus, $$h$$ takes a maximum at the end of the interval we wish to test, at $$s= \frac{1}{2}$$. But here, $$h(\tfrac{1}{2}) = \eta(\tfrac{1}{2} ) - \eta(1) - \eta(\tfrac{1}{2}) = 0$$. Thus, $$f(\tfrac{1}{2}) \leq 0$$ for $$s\in [0, \tfrac{1}{2}]$$.

3. We have that $$S(\rho) = \sum_j \eta(r_j)$$ and $$S(\sigma) = \sum_j \eta(s_j)$$. Therefore, \begin{aligned} |S(\rho) - S(\sigma)| &= \left| \sum_j \eta(r_j) - \eta(s_j)\right| \\ &\leq \sum_j |\eta(r_j) - \eta(s_j)| \\ &\leq \sum_j \eta(|r_j-s_j|). \end{aligned}
4. Note that $\eta(xy) = - xy \log (xy) = - xy (\log x + \log y) = - xy \log x - xy \log y = y \eta(x) + x \eta(y)$ Thus, $$\eta(\varepsilon_j) = \eta(\varepsilon\lambda_j) = \varepsilon\eta(\lambda_j) + \lambda_j\eta(\varepsilon)$$, and \begin{aligned} |S(\rho) - S(\sigma)| &\leq \sum_j \eta(\varepsilon_j) \\ &= \varepsilon\sum_j \eta(\lambda_j) + \eta(\varepsilon) \\ &= \eta(\varepsilon) + H(\{ \lambda_j\}_{j})\\ &\leq \eta(\varepsilon)+ \varepsilon\log d \end{aligned} using that $$\{ \lambda_j\}_{j}$$ is a probability distribution, and the entropy of any probability distribution with $$d$$ elements is bounded by $$\log d$$.
5. By step 1, $$\varepsilon\leq \|\rho-\sigma\|_1$$. Since $$\varepsilon\mapsto \eta(\varepsilon)$$ is monotonically increasing on the range $$[0,1/e]$$, if $$\|\rho-\sigma\|_1\leq 1/e$$, then \begin{aligned} |S(\rho) - S(\sigma)| &\leq\eta(\varepsilon) + \varepsilon\log d \\ &\leq \eta(\|\rho-\sigma\|_1) + \|\rho-\sigma\|_1 \log d \end{aligned} as desired.
1. DOI: 10.1007/978-1-4612-0653-8. In particular, equation IV.62 of that book is a restatement of this question.