Skip to main content
additional results
Source Link

Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

  1. which metric spaces are central?
  2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty totally bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

A partial positive answer, when $\ S_1(\{a\ b\})\ $ always consists of a single point, was obvious from the moment the strongly convex spaces were introduced by Karol Borsuk:

    Operation $\ s\ $ is continuous for every compact strongly convex space.

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

A more specific fixed point theorem is a corollary of the above theorem, together with the easier part (I'd be willing to provide more material2) of the theorem below:

THEOREM

  1. Every injective (i.e. hyperconvex) metric space is absolutely central.
  2. The central point operation $\ s\ $, defined for all non-empty bounded subsets, is continuous.

Corollary   Let $\ f:X\rightarrow X\ $ be an isometry of an arbitrary injective metric space such that $\ f(A)=A\ $ for a certain non-empty bounded $\ A\subseteq X.\ $ Then there exists $\ p\in X\ $ such that $\ f(x)=x$.

(I'd be willing to provide more material).

Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

  1. which metric spaces are central?
  2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty totally bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

(I'd be willing to provide more material).

Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

  1. which metric spaces are central?
  2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty totally bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

A partial positive answer, when $\ S_1(\{a\ b\})\ $ always consists of a single point, was obvious from the moment the strongly convex spaces were introduced by Karol Borsuk:

    Operation $\ s\ $ is continuous for every compact strongly convex space.

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

A more specific fixed point theorem is a corollary of the above theorem, together with the easier part (2) of the theorem below:

THEOREM

  1. Every injective (i.e. hyperconvex) metric space is absolutely central.
  2. The central point operation $\ s\ $, defined for all non-empty bounded subsets, is continuous.

Corollary   Let $\ f:X\rightarrow X\ $ be an isometry of an arbitrary injective metric space such that $\ f(A)=A\ $ for a certain non-empty bounded $\ A\subseteq X.\ $ Then there exists $\ p\in X\ $ such that $\ f(x)=x$.

(I'd be willing to provide more material).

tag fixed-point-theorems
Link
typo (a word ommision)
Source Link

Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

  1. which metric spaces are central?
  2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty totally bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

(I'd be willing to provide more material).

Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

  1. which metric spaces are central?
  2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

(I'd be willing to provide more material).

Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

  1. which metric spaces are central?
  2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty totally bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

(I'd be willing to provide more material).

Source Link
Loading