A resolvent contour bound transfers scalar rational error to matrices

Since $f$ and $r$ are holomorphic on $\mathrm{\Omega }$, [F1] applied to each gives $f(A)-r(A)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}(f(z)-r(z))(zI-A{)}^{-1}dz\mathrm{.}$ The cycle is disjoint from $\sigma (A)$, so every resolvent in the integrand exists.

Write the finite chain integral as ${\sum }_{\nu }{m}_{\nu }{\int }_{{\gamma }_{\nu }}G$. Parametrize each rectifiable contour piece by arc length. Multiplying its estimate by $\mathrm{\mid }{m}_{\nu }\mathrm{\mid }$ and applying the triangle inequality to the finite chain sum gives the coefficient-weighted total ${\sum }_{\nu }\mathrm{\mid }{m}_{\nu }\mathrm{\mid }L({\gamma }_{\nu })=L_{\mathrm{\Gamma }}$, including zero or negative coefficients. For every row, the absolute value of the integral is at most the integral of the absolute value; summing the row bounds and using [F2] gives ${\parallel {\int }_{\mathrm{\Gamma }}G(z)dz\parallel }_{\mathrm{\infty },\mathbb{C}}\le L_{\mathrm{\Gamma }}{\sup }_{z\in \mathrm{\Gamma }}\mathrm{\parallel }G(z){\mathrm{\parallel }}_{\mathrm{\infty },\mathbb{C}}\mathrm{.}$ For $G(z)=(f(z)-r(z))(zI-A{)}^{-1}$, the hypotheses give ${\sup }_{\mathrm{\Gamma }}\mathrm{\parallel }G(z){\mathrm{\parallel }}_{\mathrm{\infty },\mathbb{C}}\le \epsilon M$, with the resolvent bound certified by [F3].

Taking norms in step 1.1, applying step 1.2, and using $\mathrm{\mid }1\mathrm{/}(2\pi i)\mathrm{\mid }=1\mathrm{/}(2\pi )$ proves the claimed estimate.