I am stuck on the following question which I am having some difficulty with. Any help is much appreciated.
Let $X$ be a normed vector space and define the projection onto the unit ball as $\pi(x)$ = $x/||x||$ for $x \in X \setminus \{0\}$. Let $x,y \in X \setminus \{0\}$ with $||x||,||y|| \geq 1$. Must it be the case that we have $||\pi(x) − \pi(y)|| \leq ||x − y||$?
I can't seem to find a counterexample so am trying to go for a proof. So far the only observation I have been able to make is this:
WLOG we can take $||x|| \leq ||y||$. We may also WLOG take $||x|| = 1$ as otherwise we can replace $x$ with $x' = x/||x||$ and $y$ with $y' = y/||x||$ which does not affect the L.H.S of our inequality which reducing the R.H.S which only makes the inequality more strict. Hence the problem reduces to showing $||x-\pi(y)|| \leq ||x-y||$ with $||x|| = 1$ and $||y|| \geq 1$.
If we have $||y|| \geq 3$ then we can use the triangle inequality twice to show this. However I can't find a proof using just $||y|| \geq 1$. How should I proceed?
Best Answer
Not true in general, as balls are not necessarily round. Take for instance $X=\mathbb R^2 $ with $$\|x\|_\infty=\max\{|x_1|,|x_2|\}. $$ Choose $$x=(1,3/4), \ \ \ y =(1/4,5/4) $$ Then $\|x\|=1$, and $$\|x-y\|=3/4,\ \ \ \|x-y/\|y\|\|=4/5 $$