Given $x_{0}$, let $f(x) = \|x-x_{0}\|$. Show tha $f$ has a minumum on any closed, nonempty set $A \subset \mathbb{R^{n}}$

Question

Given $x_{0}$, let $f(x) = \|x-x_{0}\|$. Show tha $f$ has a minumum on any closed, nonempty set $A \subset \mathbb{R^{n}}$.

I tried to do a test for reduction to the absurd, but it was a little difficult to get the result. Perhaps you find a simpler way to demonstrate this exercise using elements of real analysis.

Any bounded closed subset of $\mathbb{R}^n$ is compact. Show that $f$ is continuous. — Dzoooks, Oct 07 '19 at 23:09
So $f$ is continuous and, as $f$ wanders off to $\infty$ in a very nice way as $|x|\to\infty$, you can always look for the minimum of $f$ on a bounded closed subset of $A$. Can you take it from here? — Maximilian Janisch, Oct 07 '19 at 23:10

score 0 · Answer 1 · answered Oct 07 '19 at 23:09

0

Did you tried use the following fact?

All limited sequence in a closed set has a convergent subsequence to an element of this set.

Try construct a sequence in $A$ s.t. the distances to $x_0$ are decreasing (at least non-increasing).

answered Oct 07 '19 at 23:09

Quiet_waters

1,475

score 0 · Answer 2 · answered Oct 07 '19 at 23:12

Examine $\ell=\inf_{x\in A} f(x)$. We know $\ell \ge 0$, because $f(x)\ge 0$ always. We need to show that $f$ attains its infimum. Here's a sketch of how: let $(x_n)$ denote a sequence of elements in $A$ so that $f(x_n)\to \ell$ (why does such a sequence exist?). In particular, we see that $\lVert x_n-x_0\rVert \le\ell+\varepsilon$ for $n$ sufficiently large, so that the sequence is eventually contained in a compact neighborhood of $x_0$, and so up to passing to a subsequence, we may assume it converges.

So, $(x_n)\to x_\infty$, a limit point. Now, $x_\infty\in A$, because $A$ is closed. By continuity of $f$ (justify!), we see that $f(x_\infty)=\ell$ (justify!), and so $f$ attains its minimum on $A$.

score 0 · Accepted Answer · answered Oct 07 '19 at 23:47

Let $C$ be a closed subset and $x\in C$. We denote by $r=\|x-x_0\|$. Write $L=B(x_0,2r)\cap C$ is compact, since $f$ is continuous, the restriction of $f$ to $L$ attains its minimum at $y$. $f(y)$ is the minimum of the restriction of $f$ to $C$. Suppose that $f(z)<f(y), z\in C$ it is equivalent to saying that $\|z-x_0\|<\|y-x_0\|<r$. This implies that $z\in B(x_0,2r)\cap C$, contradiction.

score 0 · Answer 4 · answered Oct 08 '19 at 00:32

Without loss of generality, $x_0=0.$ If $0\in A$, the minimum is $0.$ If not, let $N$ be the first integer such that $A\cap \overline B_N(0)\neq \emptyset.$ Now, $A\cap \overline B_N(0)$ is compact, and $f$ is continuous, so there is a $y\in A\cap \overline B_N(0)$ such that $f(x)=\|x\|\ge \|y\|$ for all $x\in A\cap \overline B_N(0).$ Now let $x\in A.$ There is a first integer $n$ such that $x\in \overline B_n(0).$ If $n\le N$, then $f(x)=\|x\|\ge \|y\|$. On the other hand, if $n>N,$ then again, $\|x\|>\|y\|$ because $x\notin \overline B_N(0).$ So, if fact, $y$ is a global minimum for $f$.

score 0 · Answer 5 · answered Oct 08 '19 at 00:37

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Hint Pick some $a \in A$. Let $\|x_0-a\|=R$.

Denote by $K:= \{ x \in \mathbb R^n : \| x -x_0\| \leq R$.

Show that $A \cap K$ is closed in $K$, and hence compact and non-empty.

Deduce that $f$ attains its minimum on $A \cap K$.

answered Oct 08 '19 at 00:37

N. S.

134,609

Given $x_{0}$, let $f(x) = \|x-x_{0}\|$. Show tha $f$ has a minumum on any closed, nonempty set $A \subset \mathbb{R^{n}}$

5 Answers5