Baby Rudin Theorem 9.17 In proof of Theorem 9.17 Rudin writes:
$$
|f(x+h)-f(x)-Ah|\leq \frac{1}{2}|Ah| \tag{23}
$$
It follows that
$$
|f(x+h)-f(x)|\geq \frac{1}{2}|Ah| \tag{24}
$$
I note that $f$ is a $C'$ mapping of a subset $E$ of $R^n$ into $R^m$ and $A = f'(A)$. I don't see how (24) follows from (23).
:

Prove: |u-v|<=1/2|v| -> |u|>=1/2|v|.
|u-v|>=||u|-|v||
Assume |u|<1/2|v|
|u|-|v|<-1/2|v| => |v|-|u|>1/2|v|
||u|-|v||=||v|-|u||=|v|-|u|
|u-v|>1/2v contradiction. so |u|>=1/2|v|
