Closed subset of R^n has an element of minimal norm

1. Apr 12, 2010

kingwinner

1. The problem statement, all variables and given/known data
a) Let V be a normed vector space. Then show that (by the triangle inequality) the function f(x)=||x|| is a Lipschitz function from V into [0,∞). In particular, f is uniformly continuous on V.

b) Show that a closed subset F of contains an element of minimal norm, that is, there is an x E F such that ||x||≤||y|| for all y E F. (here ||x|| refers to the usual Euclidean norm).
(hint: F may not be compact, so work on a suitable compact subset of F.)

2. Relevant equations
3. The attempt at a solution
I proved part a, but I really have no idea how to do part b. Why do we need compactness, and what is the suitable compact subset of F?

I hope someone can help me out! Thank you!

2. Apr 13, 2010

lanedance

just ideas that will need some working but hope they help

you want to take a compact subset of F, is a closed an bounded subset enough to be compact (it is in Rn - Hiene Borel)?

you could pick any element of F, then use it to set your bound, and show elements outside the bounded set have a larger norm

now you have a compact subset of a metric space and i seem to remember something along the lines of a continuous real valued function always takes a maximal & minimal value on a compact set...

Last edited: Apr 13, 2010
3. Apr 13, 2010

kingwinner

If we intersect F with a compact set K, will F intersect K stilll be compact? Why or why not?
Also, should we choose K to be very small? How can we formalize this choice of K? (i.e. what is the so-called suitable compact subset of F mentioned in the hint?)

"...and show elements outside the bounded set have a larger norm"
How??

4. Apr 13, 2010

JSuarez

Consider the closed balls of radius r, centred at the origin:

$$B\left(0,r\right)=\left\{x \in V:\left\|x\right\|\leq r\right\}$$

There must be one that intersects F. How do you go on from there?

5. Apr 13, 2010

kingwinner

Why does it have to intersect F?

And how can we show elements outside the bounded set have a larger norm? I don't get this idea at all. How is this possible? Can someone please explain this in greater detail?

Thanks a lot!

6. Apr 13, 2010

Office_Shredder

Staff Emeritus
By definition, elements outside the ball of radius r have a larger norm than elements inside a ball of radius r. Look at how Suarez described the set

7. Apr 14, 2010

JSuarez

For sufficiently large r, it must intersect F; otherwise, F would be empty.

8. Apr 17, 2010

kingwinner

What would go wrong if F were empty??

9. Apr 17, 2010

JSuarez

It wouldn't have an element with minimum norm; in fact, the original statement of your question should be "a closed nonempty subset F contains an element of minumum norm". As the empty set is closed, the statement is false without that condition.