Skip to main content
cap, not cap (typo)
Source Link

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $$\ \mu\in m(P)\cap[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|g_\nu(x)-F_\nu(x)|\ =\ |(1-\nu)\cdot c_\mu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|g_\nu(x)-F_\nu(x)|\ =\ |(1-\nu)\cdot c_\mu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cap[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|g_\nu(x)-F_\nu(x)|\ =\ |(1-\nu)\cdot c_\mu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

"old" typo fixed
Source Link

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|g_\nu(x)-F_\nu(x)|\ =\ |q_\nu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$$$|g_\nu(x)-F_\nu(x)|\ =\ |(1-\nu)\cdot c_\mu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|g_\nu(x)-F_\nu(x)|\ =\ |q_\nu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|g_\nu(x)-F_\nu(x)|\ =\ |(1-\nu)\cdot c_\mu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

a silly error + typos
Source Link

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  heThe following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition of $\ f_\mu\ $ satisfies$\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point $\ f_\nu=\bigcirc^k f_\mu$.(I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof; then of course $\ F_\kappa = f_\kappa,\ $ but never mind :-proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\lambda-\mu|<\epsilon\ $$\ |\nu-\kappa|<\epsilon\ $ for $\ \lambda:=\mu^k,\ $$\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |f_\nu(x)-F_\kappa(x)|\ \le\ |f_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|f_\nu(x)-F_\nu(x)|\ =\ |q_\nu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$$$|g_\nu(x)-F_\nu(x)|\ =\ |q_\nu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |f_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |f_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  he following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition of $\ f_\mu\ $ satisfies $\ f_\nu=\bigcirc^k f_\mu$.

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof; then of course $\ F_\kappa = f_\kappa,\ $ but never mind :-)

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\lambda-\mu|<\epsilon\ $ for $\ \lambda:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |f_\nu(x)-F_\kappa(x)|\ \le\ |f_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|f_\nu(x)-F_\nu(x)|\ =\ |q_\nu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |f_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |f_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

Let $\ L\ $ be a Hilbert space. Let $\ P\subseteq L\ $ be a non-empty compact subset. Then $\ P\ $ is called $\ \mu$-shrinkable $\ \Leftarrow:\Rightarrow$

$$\exists_{q\in L}\ \ \mu\cdot P\ +\ q\ \subseteq\ P$$

for arbitrary $\ \mu\ge 0\ $ (thus $\ \mu \le 1\ $ when $\ |P|>1$).

Let $\ m(P)\ $ be the set of all $\ \mu\ge 0\ $ such that $\ P\ $ is $\ \mu$-shrinkable. Following @E.S.Halevi, let $\ P\ $ be called shrinkable $\ \Leftarrow:\Rightarrow\ \ m(P) = [0;1].\ $ Then

THEOREM  The following three properties of a non-empty compact $\ P\subseteq L\ $are equivalent:

  1. P is a star set;
  2. P is shrinkable;
  3. $\ \sup (\ m(P)\cap[0;1)\ )\ =\ 1$

PROOF   Implications $\ 1\Rightarrow 2\Rightarrow 3\ $ are trivial. we need only $\ 3\Rightarrow 1.\ $ Thus assume condition $3$.

Consider map $\ f_\mu : x\mapsto \mu\cdot x + q_\mu,\ $ of $\ P\ $ into itself, for every $\ \mu\in m(P)\cup[0;1).\ $ Then by Banach's fpp there exists a unique $c_\mu\in P\ $ such that $\ c_\mu = \mu\cdot c_\mu + q_\mu,\ $ so that $\ q_\mu = (1-\mu)\cdot c_\mu.\ $ Thus there is a limit point $\ c_1\in P\ $ of a certain sequence of points $\ c_\mu\ $ for which $ \lim \mu = 1$.

Observe that for $\ \nu:=\mu^k\ $ the composition $\ g_\nu:=\bigcirc^k f_\mu\ $ has the same fixed point (I am going to choose at the most one $\ k\ $ for each $\ \mu;\ $ also

$$\forall_{x\in L}\ \ g_\nu(x)\ =\ \nu\cdot x\ +\ (1-\nu)\cdot c_\mu$$

Now let's consider an arbirary $\ \kappa\in[0;1).\ $ I'll show that function

$$\ F_\kappa\ :\ x\ \mapsto\ \kappa\cdot (x-c_1)+c_1\ \ =\ \ \kappa\cdot x\ +\ (1-\kappa)\cdot c_1$$

maps $\ P\ $ into itself (for every such $\kappa,\ $ so that will be the end of the proof).

Thus let $\ \epsilon > 0.\ $ Then there exist $\ \mu\in[0;1)\ $ and natural $\ k,\ $ such that $\ |c_\mu-c_1|<\epsilon\ $ and $\ |\nu-\kappa|<\epsilon\ $ for $\ \nu:=\mu^k,\ $ hence for arbitrary $\ x\in P$:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ |g_\nu(x)-F_\nu(x)|\ +\ |F_\nu(x)-F_\kappa(x)|$$

where

$$\ |F_\nu(x)-F_\kappa(x)|\ =\ |(\nu-\kappa)\cdot x + (\kappa-\nu)\cdot c_1|\ =\ |\nu-\kappa|\cdot|x-c_1|$$

henceforth

$$|F_\nu(x)-F_\kappa(x)|\ \le\ \epsilon\cdot |x-c_1|$$

Next

$$|g_\nu(x)-F_\nu(x)|\ =\ |q_\nu - (1-\nu)\cdot c_1|\ =\ |1-\nu|\cdot|c_\mu-c_1|\ \le\ \epsilon$$

These inequalities imply:

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ (|x-c_1|+1)\cdot\epsilon$$

or

$$ |g_\nu(x)-F_\kappa(x)|\ \le\ D\cdot\epsilon$$

where  $\ D := diam(P)$

Thus for every $\ \delta > 0\ $ let $\ \epsilon:=\frac\delta D\ $ such that... OK, enough of this $\delta$-$\epsilon$ business, $\ F_\kappa(x)\in P$.

END of PROOF

REMARK The theorem holds not just for the Hilbert spaces but also for Banach spaces. One should be also able to replace translations by arbitrary linear isometries. I am even curious and hopeful about considering this kind of theorems for the locally convex linear spaces.

Source Link
Loading