This is a follow-up to my previous question.
Let $I \subseteq \mathbb{R}$, $D \subseteq I \times \mathbb{R}^n$ open, $(t_0,y_0) \in D$. Assume that $f: D \to \mathbb{R}^n$ is locally Lipschitz w.r.t. to $y$ on $D$.
Define $Q:=[t_0,t_0+h] \times \bar{B}_c(y_0) \subset U(t_0,y_0)$ where $U(t_0,y_0)$ is the neighbourhood of $(t_0,y_0)$ from the local Lipschitz condition.
I am trying to prove that the function
$\bar{f}(t,y)=f(t,\Psi(y))$ where $\Psi(y)=y_0+\frac{c}{max\{c,\lVert y-y_0\rVert\}}(y-y_0)$ is globally Lipschitz w.r.t. $y$ on $[t_0,t_0+h] \times \mathbb{R}^n$.
I have been able to prove this for the cases
i) $y_1,y_2 \in \bar{B}_c(y_0)$:
$\lVert \bar{f}(t,y_1)-\bar{f}(t,y_2)\rVert =\lVert f(t,y_1)-f(t,y_2)\rVert \leq K_{t_0,y_0} \lVert y_1-y_2 \rVert$ where $K_{t_0,y_0}$ is the local Lipschitz constant on $Q$.
ii) $y_1,y_2 \notin \bar{B}_c(y_0)$:
$\lVert \bar{f}(t,y_1)-\bar{f}(t,y_2)\rVert =\lVert f(t,y_0+\frac{c}{\lVert y_1-y_0 \rVert}(y_1-y_0))-f(t,y_0+\frac{c}{\lVert y_2-y_0 \rVert}(y_2-y_0))\rVert \leq c \lVert \frac{y_1-y_0}{\lVert y_1-y_0 \rVert} – \frac{y_2-y_0}{\lVert y_2-y_0 \rVert} \rVert \leq \frac{2c}{\lVert y_2-y_0\rVert} \lVert y_1-y_2\rVert < 2 \lVert y_1-y_2\rVert$
Note that I have used the inequality $\left\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\right\|
\le\left\|\frac{a}{\|a\|}-\frac{a}{\|b\|}\right\|+\left\|\frac{a}{\|b\|}-\frac{b}{\|b\|}\right\|
=\|a\|\,\frac{|\|b\|-\|a\||}{\|b\|\,\|a\|}+\frac{\|a-b\|}{\|b\|}
\le\frac{2}{\|b\|}\|a-b\|$.
Now I am stuck on case
iii) $y_1 \in \bar{B}_c(y_0)$, $y_2 \notin \bar{B}_c(y_0)$
$\lVert \bar{f}(t,y_1)-\bar{f}(t,y_2)\rVert =\lVert f(t,y_1)-f(t,y_0+\frac{c}{\lVert y_2-y_0 \rVert}(y_2-y_0))\rVert \leq \lVert y_1 – (y_0+\frac{c}{\lVert y_2-y_0\rVert}(y_2-y_0)) \rVert$
Intuitively, the function $\bar{f}$ should be globally Lipschitz since it extrapolates $f$ constantly outside of $\bar{B}_c(y_0)$, but I cannot seem to prove case iii) gives some upper bound in terms of a constant and $\lVert y_1-y_2 \rVert$ as in the cases i) and ii).
Can anyone please help me out here?
Thanks a lot!
Best Answer
To simplify the notation, let $y_0 = 0$. All in all, it is enough to prove that $$ |y_1| \leq c, \ |y_2| > c \quad\Longrightarrow\quad |y_1 - z| \leq |y_1-y_2|, \qquad z := c\, \frac{y_2}{|y_2|}\,. $$ It holds that $$ |y_1 - y_2|^2 = |(y_1 - z) + (z - y_2)|^2 = |y_1 - z|^2 + 2 (y_1 - z) \cdot (z - y_2) + |z-y_2|^2\,. $$ On the other hand, $(y_1 - z) \cdot (z - y_2) \geq 0$ (if you think the two vectors applied in $z$, then both are pointing inside the ball), so that the claim follows.