# Thread: Closed subset of R^n has an element of minimal norm

1. ## Closed subset of R^n has an element of minimal norm

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 $R^n$ 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.)
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I proved part a, but I really have no idea how to do part b.

I hope someone can help me out! Thank you!

[also under discussion in math links forum]

2. Pass to $\{||y||:y\in F\}$ which is bounded below. Then chuse a sequence of points converging to the lower bound of this set and use its closedness. Alternatively, show that you can make this set compact by lopping off the its "top" (the points where it is greater than N).

3. Originally Posted by kingwinner
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[B]b) Show that a closed subset F of $R^n$ contains an element of minimal norm, that is, there is an x E F such that ||x||≤||y|| for all y E F.
What's to stop you from doing this:

Let $M=d(\bold{0},F)$ and let $K=B_{M+1}[0]\cap F$ (where $B_{M+1}(0)$ is the closed ball). Then, $K$ is closed since it is the intersection of two closed sets and it is bounded since $\text{diam }K\leqslant 2(M+1)$, thus compact. Also, it is not hard to prove that $\inf_{x\in F}\|x\|\geqslant \inf_{x\in K}\|x\|$. But, $\|\cdot\\:\mathbb{R}^n\to\mathbb{R}:\bold{x}\mapst o\|\bold{x}\|$ is continuous and since $K$ is compact we have that $\|\cdot\|:K\to\mathbb{R}$ assumes a minimum on $K$. The conclusion follows.

4. Originally Posted by Drexel28
What's to stop you from doing this:

Let $M=d(\bold{0},F)$ and let $K=B_{M+1}[0]\cap F$ (where $B_{M+1}(0)$ is the closed ball). Then, $K$ is closed since it is the intersection of two closed sets and it is bounded since $\text{diam }K\leqslant 2(M+1)$, thus compact. Also, it is not hard to prove that $\inf_{x\in F}\|x\|\geqslant \inf_{x\in K}\|x\|$. But, $\|\cdot\\:\mathbb{R}^n\to\mathbb{R}:\bold{x}\mapst o\|\bold{x}\|$ is continuous and since $K$ is compact we have that $\|\cdot\|:K\to\mathbb{R}$ assumes a minimum on $K$. The conclusion follows.
Why $\inf_{x\in F}\|x\|\geqslant \inf_{x\in K}\|x\|$? Shouldn't it be the other way around since F has more elements?

Also, K is a subset of F. If ||.|| attains a minimum on K, why does it follow that it attains a minimum on F??

Thanks for explaining!

5. Originally Posted by kingwinner
Why $\inf_{x\in F}\|x\|\geqslant \inf_{x\in K}\|x\|$? Shouldn't it be the other way around since F has more elements?

Also, K is a subset of F. If ||.|| attains a minimum on K, why does it follow that it attains a minimum on F??

Thanks for explaining!
It should be equality not inequality. Think about your second question again now.