On a local existence theorem for ODEs I have a question about the proof of Theorem 3.10 in Applied Analysis by Hunter and Nachtergaele. The theorem is stated as follows:
Theorem.
Let $f: I \times \overline{B}_R(u_0) \to \mathbb{R}^n,$ where
$$
I = \{t \in \mathbb{R}: |t- t_0| \leq T \}
$$
is an interval in $\mathbb{R}$ and
$$
\overline{B}_R(u_0) = \{u \in \mathbb{R}^n : ||u - u_0|| \leq R \}
$$
is the closed ball of radius $R > 0$ centered at $u_0 \in \mathbb{R}^n.$ Suppose $f(t,u)$ is continuous on $I \times \overline{B}_R(u_0)$ and Lipschitz continuous with respect to $u$ uniformly in $t.$ Set
$$
M = \sup \{||f(t,u)|| : t \in I \text{ and } u \in \overline{B}_R(u_0) \} < \infty.
$$
Then the initial value problem
\begin{align*}
    u'(t) &= f(u)\\
    u(t_0) &= u_0.
\end{align*}
has a unique continuously differentiable local solution $u(t),$ defined in the time interval $|t-t_0| < \delta,$ where
$$
\delta = \min(T, R/M).
$$
The proof involves showing that the map defined by
$$
Tu(t) = u_0 + \int_{t_0}^t f(s, u(s)) ds
$$
is a contraction that maps $X \to X$, where
$$
X = \{ u : \lbrack t_0 - \eta, t_0 + \eta \rbrack \to \overline{B}_R(u_0) : u \text{ is continuous}\}
$$
for some $\eta \in (0, \delta).$ I understand how to show that $T$ maps $X \to X$, but I don't understand the last parts of the proof that $T$ contracts. I understand how we obtain the estimate
$$
|| Tu - Tv ||_\infty \leq C \eta ||u-v||_\infty,
$$
where $C$ is the Lipschitz constant for $f$. Next, the authors write that we should choose $\eta = C/2,$ and this proves $T$ is a contraction. How does this show $T$ contracts? So far as I know, the only information I have about about $C$, being the Lipschitz constant for $f$, is that $C \geq 0.$ How does it follow that $C^2/2 < 1?$ (which is what I believe we need for $T$ to be a contraction).
Lastly, the authors write that "since $\eta$ depends only on the Lipschitz constant of $f$ and on the distance $R$ of the initial data from the boundary of $\overline{B}_R(u_0)$, repeated application of this result gives a unique local solution defined for $|t-t_0| < \delta.$" How am I to understand what is meant by this last part, the "repeated application of this result?"
Thanks in advance for any and all replies.
 A: Obviously, for a contraction you want
$$
Cη\le\frac12
$$
(or any other value smaller $1$). Transcription of manuscripts and editing of books may be carried out by non-mathematicians, so such a mangling of small details in formulas is not uncommon.
As to the second part, you got a solution $u$ on $(t_0-η,t_0+η)$. Now consider the IVP at the point $t_1=t_0+\frac12η$ with value $u(t_1)$. By using the same sets and constants, you can conclude that a local solution on $(t_1-η,t_1+η)=(t_0-\frac12η,t_0+\frac32η)$ exists and can be used to extend the first solution. Now apply the same centered on $t_{-1}=t_0-\frac12η$, $t_2=t_0+η$ etc.
If you treat the end point $η$ dynamically, you might conclude that for any $t\in[t_0-δ,t_0+δ]$ you get
\begin{align}
\|(T^2u)(t)-(T^2v)(t)\|&\le\int_{t_0}^tC\|(Tu)(s)-(Tv)(s)\|\,ds
\\
&\le\int_{t_0}^tC^2|s-t_0|\|u(s)-v(s)\|\,ds
\\
&\le \frac{|C(t-t_0)|^2}2\|u-v\|_\infty
\\&\vdots\\
\|(T^nu)(t)-(T^nv)(t)\|&\le \frac{|C(t-t_0)|^n}{n!}\|u-v\|_\infty
\end{align}
This allows to get contractivity for some $T^n$ on the full interval $[t_0-δ,t_0+δ]$ at once, as $\frac{(Cδ)^n}{n!}$ converges to zero. Some minimal manipulation is required to show that the fixed point of $T^n$ is indeed also a fixed point of $T$, essentially if $u_*$ is the fixed point of $T^n$ then $Tu_*$ has exactly the same defining properties, uniqueness then requires $Tu_*=u_*$.
A: Appears that the book has a typo: a newer version of this same chapter has $\eta = 1/2C$, which makes much more sense. https://www.math.ucdavis.edu/~hunter/book/ch3.pdf
