Let $n\geq 2$, $C>0$, $i=1,…,n$ and define $f: \mathbb{R}^n \rightarrow \mathbb{R}, f(x)=\left\{\begin{matrix}
C^{-n}x_i & |x|\leq C\\
|x|^{-n}x_i& |x|\geq C
\end{matrix}\right.$.
Show that $f$ is Lipschitz
Both the functions $x\mapsto x_i, x\mapsto \left\{\begin{matrix}
C^{-n}& |x|\leq C\\
|x|^{-n}& |x|\geq C
\end{matrix}\right.$ are Lipschitz and the second one is also bounded.$f$ is bounded too as $x_i|x|^{-n}$ vanishes as $|x|\rightarrow \infty$, being $n\geq 2$. These observations make $f$ a bounded locally Lipschitz function, however I don't think this fact will help.
I guess I have to resort to the definition of Lipschitz continuity, but I couldn't come up with anything clever. The problems arise when computing $|f(x)-f(y)|$ for $|x|,|y|$ not both $\leq C$. Do you have any suggestion?
Best Answer
Suppose $f$ is separately Lipschitz on $B(0,C)$ and its complement. You might as well choose the same constant $L$
Pick $x,y$. If $x,y$ are both in $B(0,C)$ or the complement then clearly $|f(x)-f(y)| \le L\|x-y\|$.
If $x \in B(0,C) $ and $y$ in the complement then let $t \in [0,1]$ be such that $z=tx+(1-t)y $ is on the boundary.
Then $|f(x)-f(y)| \le |f(x)-f(z)| + |f(z)-f(y)| \le L(\|x-z\|+\|z-y\|) = L\|x-y\|$.
Note:
If $\phi(x) = \|x\|^n$, then $D\phi(x)h = n \|x\|^{n-1} x^T h$ and so with $\eta(x) = {1 \over \|x\|^n}$ we have $D \eta(x)h = -n{ 1 \over \|x\|^{n+2}} x^Th$. In particular, $\|D\eta(x)\| \le n { 1 \over \|x\|^{n+1}}$, so we can see that with $\|x\|>C$ we have $\|Df(x)\| \le K {1 \over \|x\|^{n}}$ for some $K$. Hence $f$ is Lipschitz with rank $K$ for $\|x\| > C$.