MHB The distance is attained by a unique point

[evinda]

Hello! (Wave)

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? (Thinking)

 

[Ackbach]

I don't have time to answer all your questions just now, but I can answer a few, and perhaps nudge you on the others.

Hello! (Wave)

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 $d(x,K)=1$ ?

So $K$ is the open circle of radius $1$, centered at the origin, right? If you visualize the point $(2,0)$ relative to the circle, then the point on the boundary of the circle closest to $x$ is $(1,0)$. However, this point is not in the open circle. Consider the sequence $\{(1-1/n,0)\} \subset K$, and the distance from each of these points to $(2,0)$ in the Euclidean metric. Then those distances are $2-(1-1/n)=1+1/n$. If we take the infimum of these distances, we will see that it is $1$, and hence $d(x,K)=1$.

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

Could you explain me why?

See the following figure:

View attachment 4982

The point on the right is equidistant from the two different lobes on the left. The two circles on the left make a decidedly non-convex shape, and so you can see why the distance from a point to a set might not be unique if the set isn't convex.

The rest of your post will take considerably more thought. I can't promise a time when I can look at it.

 

[Opalg]

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 $$||x-y_x||=d(x,K):=inf \{||x-y||: y \in K \}$$
.
.
.

- 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 $||x-y_n|| \to d(x,K)$.

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? (Thinking)

In the parallelogram law, replace "$x$" by $x-y_n$ and "$y$" by $x-y_m$. Then the law says that $$\|(x-y_n) - (x-y_m)\|^2 + \|(x-y_n) + (x-y_m)\|^2 = 2\|x-y_n\|^2 + 2\|x-y_m\|^2.$$ On the left side of that equation, the first term is $\|y_m-y_n\|^2$. The second term can be written as $\|2x - (y_m + y_n)\|^2 = 4\|x - \frac12(y_m+y_n)\|^2.$ But $\frac12(y_m+y_n) \in K$ because $K$ is convex. Therefore $\|x - \frac12(y_m+y_n)\| \geqslant d = d(x,K)$. It follows that $$\|y_m-y_n\|^2 \leqslant 2\|x-y_n\|^2 + 2\|x-y_m\|^2 - 4d^2.$$ When $m$ and $n$ are large, $\|x-y_n\|$ and $\|x-y_m\|$ are both close to $d$, and it follows that the right side of that inequality will be close to zero. It follows that $\|y_m-y_n\|^2$ on the left side of the inequality will also be close to zero. This shows that the sequence $(y_n)$ is Cauchy.

 

[evinda]

So $K$ is the open circle of radius $1$, centered at the origin, right? If you visualize the point $(2,0)$ relative to the circle, then the point on the boundary of the circle closest to $x$ is $(1,0)$. However, this point is not in the open circle. Consider the sequence $\{(1-1/n,0)\} \subset K$, and the distance from each of these points to $(2,0)$ in the Euclidean metric. Then those distances are $2-(1-1/n)=1+1/n$. If we take the infimum of these distances, we will see that it is $1$, and hence $d(x,K)=1$.

Could you explain it further to me? (Sweating)

See the following figure:
The point on the right is equidistant from the two different lobes on the left. The two circles on the left make a decidedly non-convex shape, and so you can see why the distance from a point to a set might not be unique if the set isn't convex.

Ah, I see... (Nod)

 

[Ackbach]

Could you explain it further to me? (Sweating)

Sure! Let me show you a picture:

View attachment 4990

Here I've included the Mathematica code I used to generate this plot. Now, the circle is shown filled, with a dotted line, to show that the boundary itself is not included in the circle (it's an open circle). The point is shown on the right with a blue dot. We're trying to find the distance from the point to the circle. Now, imagine any sequence of points converging to the right-most point of the circle, which is at $(1,0)$, and is clearly the point on the (closed) circle closest to the point $(2,0)$. The infimum of the distances from the sequence points to the point $(2,0)$ will be $1$, which is simply the Euclidean distance from $(1,0)$ to $(2,0)$.

Does that answer your question?

 

[evinda]

Sure! Let me show you a picture:
Here I've included the Mathematica code I used to generate this plot. Now, the circle is shown filled, with a dotted line, to show that the boundary itself is not included in the circle (it's an open circle). The point is shown on the right with a blue dot. We're trying to find the distance from the point to the circle. Now, imagine any sequence of points converging to the right-most point of the circle, which is at $(1,0)$, and is clearly the point on the (closed) circle closest to the point $(2,0)$. The infimum of the distances from the sequence points to the point $(2,0)$ will be $1$, which is simply the Euclidean distance from $(1,0)$ to $(2,0)$.

Does that answer your question?

Yes, I got it... Thanks a lot! (Smirk)