Background
If $\rho$ and $\sigma$ are density matrices such that the trace distance between them satisfies $T(\rho,\sigma)\leq1/e$. Then the Fannes' inequality states that $$|S(\rho)-S(\sigma)|\leq T(\rho,\sigma)\log d+\eta(T(\rho,\sigma))$$
Proof of Fannes' inequality is given in Box 11.2, Page 512, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, as
Let $r_1\geq r_2\geq\cdots\geq r_d$ be the eigenvalues of $\rho$, and Let $s_1\geq s_2\geq\cdots\geq s_d$ be the eigenvalues of $\sigma$, in descending order, where $d$ is the dimension of the Hilbert space.
Let $H=\rho-\sigma$, which is hermitian since both $\rho$ and $\sigma$ are positive semidefinite.
The spectral decomposition of $H$ can be written as, $H=\rho-\sigma=VD_HV^\dagger$.
Let $D_Q$ be the diagonal matrix that has all the positive eigenvalues of $H$, and $D_R$ contains the negation of all the negative eigenvalues of $H$, such that we can define $VD_QV^\dagger=Q$ and $VD_RV^\dagger=R$. Therefore, $$ D_H=D_Q-D_R\\ H=\rho-\sigma=VD_HV^\dagger=V(D_Q-D_R)V^\dagger=VD_QV^\dagger-VD_RV^\dagger=Q-R $$ $\implies D_Q,D_R$, and thereby $Q,R$ have orthogonal supports, since $Q$ only supports the positive eigenspace and $R$ only supports the negative eigenspace.
$\therefore D_QD_R=D_RD_Q=0\implies QR=RQ=0$
The trace distance between $\rho$ and $\sigma$ is defined as, $T(\rho,\sigma)=\dfrac{1}{2}tr|\rho-\sigma|=\dfrac{1}{2}tr\sqrt{(\rho-\sigma)^\dagger(\rho-\sigma)}$ \begin{align} (\rho-\sigma)^\dagger(\rho-\sigma)&=(Q-R)^\dagger(Q-R)=(Q^\dagger-R^\dagger)(Q-R)\\ &=Q^\dagger Q+R^\dagger R-Q^\dagger R-R^\dagger Q\\ &=QQ+RR-QR-RQ=Q^2+R^2\text{ , since }QR=RQ=0\\ &=QQ+RR+QR+RQ=(Q+R)^2\\ \sqrt{(\rho-\sigma)^\dagger(\rho-\sigma)}&=Q+R \end{align} The trace distance becomes $T(\rho,\sigma)=\dfrac{1}{2}tr(Q+R)=\dfrac{1}{2}\Big(tr(Q)+tr(R)\Big)$
So we have $\rho-\sigma=Q-R$ with $|\rho-\sigma|=Q+R$, where $Q$ and $R$ are positive operators with orthogonal supports.
Defining $V\equiv R+\rho=Q+\sigma$ obtains $2T(\rho,\sigma)=tr(Q)+tr(R)=tr(2V)-tr(\rho)-tr(\sigma)$
Applying the min-max theorem, as discussed in How to show $T(\rho,\sigma)≥\sum_i|r_i − s_i|$ with $r_i,s_i$ eigenvalues of $\rho,\sigma$?, we can prove $2t_k\ge2\max(r_k, s_k)=r_k+s_k+|r_k-s_k|$, which upper bounds the trace distance as
$$2T(\rho,\sigma)\ge\sum_k|r_k-s_k|$$
This much is clear!
The remaining part of the proof is as follows :
By calculus whenever $|r − s| ≤ 1/2$ it follows that $|η(r) − η(s)| ≤ η(|r − s|)$. A moment’s thought shows that $|r_i − s_i | ≤ 1/2$ for all $i$, so $$ |S(ρ) − S(σ)|=|\sum_{i}(η(r_i)−η(s_i))|≤η(|r_i−s_i|) $$ Setting $Δ≡\sum_i|r_i − s_i|$ and observing that $η(|r_i−s_i|)=Δη(|r_i−s_i|/Δ)− |r_i−s_i|\log(Δ)$, we see that $$ |S(ρ)−S(σ)|≤Δη(|r_i−s_i|/Δ)+η(Δ)≤Δ\log d+η(Δ) $$ where we applied Theorem 11.2 to obtain the second inequality. But $Δ≤T(ρ, σ)$ by $(11.46)$, so by the monotonicity of $η(·)$ on the interval $[0, 1/e]$, $$ |S(ρ) − S(σ)|≤T(ρ,σ)\log d + η(T (ρ, σ)) $$ whenever $T (ρ, σ) ≤ 1/e$, which is Fannes’ inequality
Question
How do I prove that, whenever $|r − s| ≤ 1/2$, it follows that $|η(r) − η(s)| ≤ η(|r − s|)$?
The graph of $\eta(x)=-x\log x$ function is
In the domain $0\leq x\leq y\leq 1$ we have
\begin{align} |\eta(x)-\eta(y)|&=|-x\log x+y\log y|\\ &=|-x\log x+\frac{x}{\ln 2}+y\log y -\frac{y}{\ln 2}+\frac{y}{\ln 2}-\frac{x}{\ln 2}|\\ &\leq |-(x\log x-\frac{x}{\ln 2})+(y\log y-\frac{y}{\ln 2})|+\frac{|y-x|}{\ln 2}\\ &=|\int_x^y \log tdt|+\frac{|y-x|}{\ln 2}=-\int_x^y \log tdt+\frac{|y-x|}{\ln 2}\\ &=-\int_x^{x+(y-x)} \log tdt+\frac{|y-x|}{\ln 2}\\ &\leq -\int_0^{y-x} \log tdt+\frac{|y-x|}{\ln 2}\\ &=-\Big(t\log t-t\Big)_0^{y-x}+\frac{|y-x|}{\ln 2}\\ &=-(y-x)\log(y-x)+(y-x)+\frac{|y-x|}{\ln 2}\\ &=\eta(|y-x|)+|y-x|+\frac{|y-x|}{\ln 2} \end{align}
I am not able to find the intended inequality $|η(r) − η(s)| ≤ η(|r − s|)$.
It would be helpful if it is clarified so that the rest of the proof can be understood.
