A common separating resolvent contour certifies a Lipschitz response map

For $X,Y\in K$, set $H_X=F(X)$, $H_Y=F(Y)$, $\mathrm{\Delta }=H_X-H_Y$, and $H(t)=(1-t)H_Y+t{H}_X$. For $0\le t\le 1$, the operator-norm triangle inequality gives $\mathrm{\parallel }H(t)-H_0{\mathrm{\parallel }}_{op}\le (1-t)\mathrm{\parallel }{H}_Y-H_0{\mathrm{\parallel }}_{op}+t\mathrm{\parallel }{H}_X-H_0{\mathrm{\parallel }}_{op}<g\mathrm{/}4$. Thus the whole matrix segment lies in the open ball of [F3], even when $K$ is not convex. By [F2] and [L1] its occupied projector is well-defined, and by [F1] the path $h(t)={\mathrm{\Pi }}_r(H(t))$ has endpoints $h(0)=\mathrm{\Phi }(Y)$ and $h(1)=\mathrm{\Phi }(X)$.

The path $h$ is continuous on $[0,1]$ by [F3]. At each $t\in (0,1)$, the Frechet expansion in the direction $\mathrm{\Delta }$ gives $h(t+s)-h(t)=sD{\mathrm{\Pi }}_r(H(t))[\mathrm{\Delta }]+o(\mathrm{\mid }s\mathrm{\mid })$, so $h^{\mathrm{\prime }}(t)=D{\mathrm{\Pi }}_r(H(t))[\mathrm{\Delta }]$; if $\mathrm{\Delta }=0$, the path is constant and this identity still holds. By [F3], $\mathrm{\parallel }{h}^{\mathrm{\prime }}(t){\mathrm{\parallel }}_F\le (16R\mathrm{/}{g}^2)\mathrm{\parallel }F(X)-F(Y){\mathrm{\parallel }}_F$. Listing the real and imaginary parts of all matrix entries identifies the Frobenius norm with the Euclidean norm on ${\mathbb{R}}^{2n^2}$ by [F4], so [F5] applies to this path with precisely that bound.

By [F5] and the endpoint identities from step 1.1, $\mathrm{\parallel }\mathrm{\Phi }(X)-\mathrm{\Phi }(Y){\mathrm{\parallel }}_F\le (16R\mathrm{/}{g}^2)\mathrm{\parallel }F(X)-F(Y){\mathrm{\parallel }}_F\le (16R{L}_F\mathrm{/}{g}^2)\mathrm{\parallel }X-Y{\mathrm{\parallel }}_F$. This is the asserted estimate with $q=16R{L}_F\mathrm{/}{g}^2$. If $q<1$ and $\mathrm{\Phi }(K)\subseteq K$, it is exactly the contraction condition on $K$.