The distance is attained by a unique point

Question

Theorem:

Let $K$ be a convex and closed subset of a Hilbert space $X$ and $x \in X$. Then there is a unique $y_x \in K$ such that $$\|x-y_x\|=d(x,K):=\inf \{\|x-y\|: y \in K \}$$

Remarks:

• if $K$ is not closed then the distance $d(x,K)$ isn't attained in general.

If for example $X=\mathbb{R}^2$ and $D(0,1)=\{ y \in \mathbb{R}^2: \|y\|_2 <1\}$ and $x=(2,0)$ then $d(x,K)=1$ and is attained at the point $(1,0) \notin K$.

Could you explain me how we deduce that $d(x,K)=1$ ?

• if $K$ isn't convex then the uniqueness isn't satisfied.

  Could you explain me why?

• If $A \subset \mathbb{R} (\neq \varnothing)$ then there is a sequence $(a_n) \in A$ such that $a_n \to \inf A$ .

From the last remark we have that there is a sequence $(y_n) \in K$ such that $\|x-y_n\| \to d(x,K)$.

If we apply the last remark don't we get that there is a $y_n \in K$ such that $y_n \to \inf K$ ? How do we conclude that $\|x-y_n\| \to d(x,K)$ ?

Firstly we want to show that $(y_n)$ is Cauchy and since $X$ is a complete metric space it will converge to a $y \in X$, i.e. $\|y_n-y\| \to 0$ and since $K$ is closed we will have that $y \in K$.

In order to show this we will use the Parallelogram law $\|x-y\|^2+\|x+y\|^2=2 \|x\|^2+2 \|y\|^2$. But what $x$ and $y$ do we have to use?

Answer 1

Could you explain me how we deduce that $d(x,K)=1$?

$K$ is the interior of the unit circle about the origin. If $y \in K$, then $d(\mathbf 0,y) < 1$. Now for $x = (0,2), d(\mathbf 0, x) = 2$. By the triangle inequality $d(\mathbf 0, x) \le d(\mathbf 0,y) + d(x, y)$. So $$d(x, y) \ge d(\mathbf 0, x) - d(\mathbf 0,y) > 2 - 1 = 1.$$ Therefore $d(x, K) \ge 1$. On the other hand for $y = (0, 1 - h), 0 < h < 1$, we have $d(x, y) = 2 - (1 - h) = 1 + h$, so $d(x, K) \le 1 + h$ for all $h > 0$. Hence $d(x, K) \le 1$ as well.

if $K$ isn't convex then the uniqueness isn't satisfied.

Could you explain me why?

Tryss has already given an excellent example, but let me expand a little more on your question "how do we justify that the circle isn't convex?"

Please don't get so caught up in the statements of the definitions that you lose sight of what they mean: Convex means that the line segment connecting any two points in a set lies entirely within the set: I.e. if you have two points in the set, then every point between is also in the set. Look at the circle. This is clearly not true of it. the chord between two points on the circle is in the interior, not part of the circle itself.

From the last remark we have that there is a sequence $(y_n)\in K$ such that $\|x−y_n\| \to d(x,K)$.

If we apply the last remark don't we get that there is a $y_n\in K$ such that $y_n\to\inf K$ ? How do we conclude that $\|x−y_n\|\to d(x,K)$ ?

There is no such thing as $\inf K$. $K\subseteq X$, not of $\Bbb R$.

Unfortunately, just knowing $\|x - y_n\| \to d(x, K)$ does not even require $(y_n)$ to have a convergent subsequence. For example, in $\ell_2$ if we let $x = (1, 0, 0, ...)$ and let $K = \{ (\delta_{mn})_n\ |\ m \in \Bbb N, m > 1\}$. That is, $K$ is the remaining elements of the canonical basis of $\ell_2$ other than $x$. Then $d(x, y) = \sqrt 2$ for all $y \in K$, but no non-constant sequence in $K$ converges. (You will note that this $K$ is not convex, so it is not a counter-example to the theorem.)

Answer 2

If $K$ is not convex then we do not have in general a unique $y_x$: