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).
:

show |u+2v|<|v|-> |u|>|v|
Played with this in 2 dimensions and its true.
Assume |u|<|v|
Draw the vector 2v and from its tip draw a circle of radius |u|. Then u+2v is a vector from the base of 2v to a point on the circle whose length is always greater than |v|. Contradiction. Therefore |u|>|v|.
In n dimensions you would draw a ball of radius |u| at the tip of 2v and tnen calculate distances using the distance formula.
Note. |u+2v| is the same as |u-2v| because v can be + or -. It's just easier to think about.
