Ordered eigenvalues of Hermitian matrices are operator-norm Lipschitz

Write $A=R+iS$ with $R,S$ real. Hermitian symmetry gives $R^{\mathsf{T}}=R$ and $S^{\mathsf{T}}=-S$. Define $\mathcal{R}(A)=\left(\begin{array}{cc}R& -S\\ S& R\end{array}\right)$ and $\mathrm{\Phi }(x+iy)=(x,y)$. Direct multiplication gives $\mathrm{\Phi }(Az)=\mathcal{R}(A)\mathrm{\Phi }(z)$ and $\mathrm{\parallel }\mathrm{\Phi }(z){\mathrm{\parallel }}_2=\mathrm{\parallel }z{\mathrm{\parallel }}_2$; the transpose identities show $\mathcal{R}(A)$ is real symmetric. Applying the same construction to $A-B$ yields $\mathrm{\parallel }\mathcal{R}(A)-\mathcal{R}(B){\mathrm{\parallel }}_2=\mathrm{\parallel }A-B{\mathrm{\parallel }}_2$ by the supremum definition in [F4].

By [F1], $A=U\mathrm{diag}({\lambda }_1(A),\dots ,{\lambda }_n(A))U^{\ast }$ for a unitary $U$. Its realification is orthogonal, and the realification of the diagonal matrix has each ${\lambda }_j(A)$ twice on its diagonal. Therefore the descending eigenvalues of $\mathcal{R}(A)$ are ${\lambda }_1(A),{\lambda }_1(A),\dots ,{\lambda }_n(A),{\lambda }_n(A)$; the same holds for $B$.

Put $M=\mathcal{R}(B)$, $N=\mathcal{R}(A-B)$, and $\delta =\mathrm{\parallel }A-B{\mathrm{\parallel }}_2$. By [F3] and step 1.1, every eigenvalue of the real symmetric $N$ lies in $[-\delta ,\delta ]$ using [F4]. Applying [F2] with the perturbation's largest eigenvalue for the upper bound and smallest eigenvalue for the lower bound gives $\mathrm{\mid }{\lambda }_k(M+N)-{\lambda }_k(M)\mathrm{\mid }\le \delta$ for every $k$. Taking $k=2j-1$ and using step 2.1 proves the claimed complex-Hermitian eigenvalue bound.