# Is the distance between two elements of a normed vector space at least the distance between their projections on the unit ball?

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?

• Why WLOG we can take $||x||=1$ ? – Jakobian Aug 29 '18 at 13:39
• @Rumpelstiltskin Otherwise as $||y|| \geq ||x||$ we can scale down both $x,y$ which reduces the R.H.S but keeps the L.H.S the same so if the inequality did not hold before the rescaling it would not hold after it as well. So we just need to prove the result for the rescaled version. – Abdul Hadi Khan Aug 29 '18 at 13:44
• By scaling, do you mean dividing the whole inequality by $||x||$ ? But then LHS will change – Jakobian Aug 29 '18 at 13:49
• @Rumpelstiltskin We can replace $x$ with $x' = x/||x||$ and $y$ with $y' = y/||x||$. Then as $||y|| \geq ||x|| \implies ||y'|| \geq 1$ and $||x'|| = 1$. However we still have $\pi(x') = \pi(x)$ and $\pi(y') = \pi(y)$ but $||x'-y'|| = ||x/||x||-y/||x|||| = ||x-y||/||x|| \leq ||x-y||$. So if $||\pi(x)-\pi(y)|| \leq ||x-y||$ did not hold neither would $||\pi(x')-\pi(y')|| \leq ||y'-x'||$. So a proof of $||\pi(x')-\pi(y')|| \leq ||y'-x'||$ would also get that $||\pi(x)-\pi(y)|| \leq ||y-x||$. Hence we can reduce the problem to the $x',y'$ case or equivalently rename to $x,y$ and take $||x|| = 1$. – Abdul Hadi Khan Aug 29 '18 at 13:58
• You can get proper norm bars by using \| instead of ||. – joriki Aug 29 '18 at 15:24

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 $$\|x-y\|=3/4,\ \ \ \|x-y/\|y\|\|=4/5$$
The answer is negative. The operator norms of the two matrices below are equal to $1$: $$A=\pmatrix{1&0\\ 0&0},\ B=(\sqrt{2}-1)\pmatrix{2&1\\ 1&0}.$$ However, numerically we have $\|A-B\|=0.50879>0.50406=\|A-\frac{17}{16}B\|$. So, a counterexample is exhibited by putting $(x,y)=(A,\frac{17}{16}B)$.