Local conditioning times backward error controls forward error to first order

By [L2] applied to the set $S=\{S_{f,x}(\delta ):\delta >0\}$ whose infimum is $\kappa$ by [L1], every $c>\kappa$ admits some ${\delta }_0>0$ with $S_{f,x}({\delta }_0)<c$.

By the definition of $S_{f,x}({\delta }_0)$ as a supremum, every admissible $h$ with $0<\parallel h\parallel <{\delta }_0$ satisfies $\parallel f(x+h)-f(x)\parallel \mathrm{/}\parallel h\parallel \le S_{f,x}({\delta }_0)<c$, hence $\parallel f(x+h)-f(x)\parallel \le c\parallel h\parallel$, which is claim 1 with $\delta :={\delta }_0$.

For every $\epsilon >0$ the number $c:=\kappa +\epsilon$ is strictly larger than $\kappa$, so claim 1 supplies $\delta >0$ with $\parallel f(x+h)-f(x)\parallel \le (\kappa +\epsilon )\parallel h\parallel$ for all admissible $h$ of norm below $\delta$; this is exactly the stated bound, which is claim 2.

For the relative form, divide the inequality of claim 2 by the fixed positive number $\parallel f(x)\parallel$ and multiply by the fixed positive number $\parallel x\parallel$: $\parallel f(x+h)-f(x)\parallel \mathrm{/}\parallel f(x)\parallel \le (\kappa +\epsilon )\parallel h\parallel \mathrm{/}\parallel f(x)\parallel =(\kappa \parallel x\parallel \mathrm{/}\parallel f(x)\parallel +\epsilon \parallel x\parallel \mathrm{/}\parallel f(x)\parallel )\parallel h\parallel \mathrm{/}\parallel x\parallel$, and the error term is $o(1)\parallel h\parallel \mathrm{/}\parallel x\parallel$ because the positive constants $\parallel x\parallel ,\parallel f(x)\parallel$ are fixed, which is claim 3.

Claims 1, 2 and 3 are steps 2.1, 3.1 and 4.1.