1
$\begingroup$

Let $X$ be a normed vector space and let $S_X=\{x\in X:\|x\|=1\}$ (i.e. the unit sphere). Also, given a set $A\subset X$ and any $x\in X$, define dist$(x,A)=\inf\{\|x-a\|:a\in A\}$. Given any $x\in X$, is it necessarily true that

dist$(x,S_X)=\|x-\frac{x}{\|x\|}\|$?

Obviously we have dist$(x,S_X)\leq\|x-\frac{x}{\|x\|}\|$, but the other inequality is not so apparent. Proving that dist$(x,S_X)\geq\|x-\frac{x}{\|x\|}\|$ would be equivalent to showing that for all $y\in S_X$, $\|x-y\|\geq\|x-\frac{x}{\|x\|}\|$, but I can't really think of where to go with this.

Any hints on proving this would be appreciated, or perhaps a counterexample if it is not true.

$\endgroup$
1
  • $\begingroup$ Draw a picture!! $\endgroup$
    – bubba
    Mar 14, 2015 at 2:28

1 Answer 1

0
$\begingroup$

Yes, it is true.

Assume that there is $y$ with $\|y\|=1$, such that $\|x-y\|<\|x-\frac{x}{\|x\|}\|$. It follows that$$\|x\|\leq\|y\|+\|x-y\|<1+\|x-\frac{x}{\|x\|}\|=1+\|(1-\frac{1}{\|x\|})x\|=1+\|x\|-1=\|x\|,$$which is a contradiction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .