Injective mapping theorem proof discrepancy.

Question

I do not understand from where the 2 comes in the proof of letter(a) of the statement of the theorem:

And this is proposition 12.2.4

I think the author has used the mean value inequality, but I do not know how, could anyone explain this for me please?

Answer 1

From Proposition 12.2.4, since $Df(c)$ is injective the exists a constant $\gamma'$ such that $\|Df(c)(u)\| \geqslant \gamma'\|u\|$. This is easily proved by using the fact that a linear operator is continuous and attains a maximum and minimum on the (compact) unit sphere.

We can define $\gamma = \frac{1}{2}\gamma'$ so that $\|Df(c)(u)\| \geqslant \gamma'\|u\| \geqslant 2\gamma\|u\|$.

There is a lemma that states if $f$ is continuously differentiable, then given $\gamma > 0$, for all points $u$ and $v$ sufficiently close to $c$, i.e. in the ball $B$ centered at $c$, we have

$$\|f(u) - f(v) - Df(c)(u-v)\| \leqslant \gamma \|u-v\|$$

The proof does, in fact, use the mean value theorem.

By the reverse triangle inequality,

$$\|Df(c)(u-v)\| - \|f(u) - f(v)\|\leqslant\|f(u) - f(v) - Df(c)(u-v)\| \leqslant \gamma \|u-v\| $$

Hence,

$$\|f(u) - f(v)\| \geqslant \|Df(c)(u-v)\| - \gamma \|u-v\| \geqslant 2\gamma\|u-v\| - \gamma \|u-v\| = \gamma \|u-v\|$$