Fixed point of a function defined in terms of a metric on a compact metric space
$begingroup$
Let $X$ be a compact metric space, and assume $f : X → X$ satisfies
$$d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$$
Define a function $g : X → mathbb{R}$ by $g(x) = d(x,f(x)), forall x ∈ X$. Prove the following statements.
a) $|g(x)-g(y)| leq 2d(x,y), forall x,y in X$.
b) $g$ is continuous and $g(x)=0$ for some $x in X$.
$text{Proof:}$
a) Observe that
begin{align*}|g(x)-g(y)|
&=|d(x,f(x))-d(y,f(y))| \
&=|d(x,f(x))-d(f(x),y)+d(f(x),y)-d(y,f(y))| \
&leq |d(x,f(x)-d(f(x),y)|+|d(f(x),y)-d(y,f(y))| \
&leq d(x,y)+d(f(x),f(y)) \
&leq 2d(x,y),
end{align*}
whre in the last step we used the hypothesis that $d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$
b) To show continuity we take $delta= largefrac{varepsilon}{2}$. Then if $d(x,y)< delta$, we have by part a) that $|g(x)-g(y)|< varepsilon$.
To show that there exists $xin X$ such that $g(x)=0$, we follow AlexL's answer and note that since $g$ is continuous and $X$ is compact, $g$ attains its minimum at some $x in X$. Suppose that for that for $g(x)neq 0$, and so $f(x)neq x$. Then
$$g(f(x))=d(f(x),f(f(x)))<d(x,f(x))=g(x),$$
which is a contraction to $g$ attaining its minimum at $x$. It follows that $g(x)=0$, and this completes the proof.
real-analysis metric-spaces fixed-point-theorems complete-spaces
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
$endgroup$
add a comment |
$begingroup$
Let $X$ be a compact metric space, and assume $f : X → X$ satisfies
$$d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$$
Define a function $g : X → mathbb{R}$ by $g(x) = d(x,f(x)), forall x ∈ X$. Prove the following statements.
a) $|g(x)-g(y)| leq 2d(x,y), forall x,y in X$.
b) $g$ is continuous and $g(x)=0$ for some $x in X$.
$text{Proof:}$
a) Observe that
begin{align*}|g(x)-g(y)|
&=|d(x,f(x))-d(y,f(y))| \
&=|d(x,f(x))-d(f(x),y)+d(f(x),y)-d(y,f(y))| \
&leq |d(x,f(x)-d(f(x),y)|+|d(f(x),y)-d(y,f(y))| \
&leq d(x,y)+d(f(x),f(y)) \
&leq 2d(x,y),
end{align*}
whre in the last step we used the hypothesis that $d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$
b) To show continuity we take $delta= largefrac{varepsilon}{2}$. Then if $d(x,y)< delta$, we have by part a) that $|g(x)-g(y)|< varepsilon$.
To show that there exists $xin X$ such that $g(x)=0$, we follow AlexL's answer and note that since $g$ is continuous and $X$ is compact, $g$ attains its minimum at some $x in X$. Suppose that for that for $g(x)neq 0$, and so $f(x)neq x$. Then
$$g(f(x))=d(f(x),f(f(x)))<d(x,f(x))=g(x),$$
which is a contraction to $g$ attaining its minimum at $x$. It follows that $g(x)=0$, and this completes the proof.
real-analysis metric-spaces fixed-point-theorems complete-spaces
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
$endgroup$
1
$begingroup$
You have a small error. In the statement of a) an in the last displayed line of he proof of a) the inequality $<$ should be $leq$ to include the case $x=y.$ It would be easier to make this change than to re-write it to cover only the case $xne y$.
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:47
1
$begingroup$
Another way to prove that $g$ is continuous: It suffices that if $ lim_{nto infty}d(x,y_n)=0$ then $lim_{nto infty}|g(x)-g(y_n)|=0.$ Use your proof of a), writing $y_n$ for $y.$ Then $|g(x)-g(y_n)| leq 2d(x,y_n)$..... So if $d(x,y_n)to 0$ then $|g(x)-g(y_n)|to 0 .$
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:59
add a comment |
$begingroup$
Let $X$ be a compact metric space, and assume $f : X → X$ satisfies
$$d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$$
Define a function $g : X → mathbb{R}$ by $g(x) = d(x,f(x)), forall x ∈ X$. Prove the following statements.
a) $|g(x)-g(y)| leq 2d(x,y), forall x,y in X$.
b) $g$ is continuous and $g(x)=0$ for some $x in X$.
$text{Proof:}$
a) Observe that
begin{align*}|g(x)-g(y)|
&=|d(x,f(x))-d(y,f(y))| \
&=|d(x,f(x))-d(f(x),y)+d(f(x),y)-d(y,f(y))| \
&leq |d(x,f(x)-d(f(x),y)|+|d(f(x),y)-d(y,f(y))| \
&leq d(x,y)+d(f(x),f(y)) \
&leq 2d(x,y),
end{align*}
whre in the last step we used the hypothesis that $d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$
b) To show continuity we take $delta= largefrac{varepsilon}{2}$. Then if $d(x,y)< delta$, we have by part a) that $|g(x)-g(y)|< varepsilon$.
To show that there exists $xin X$ such that $g(x)=0$, we follow AlexL's answer and note that since $g$ is continuous and $X$ is compact, $g$ attains its minimum at some $x in X$. Suppose that for that for $g(x)neq 0$, and so $f(x)neq x$. Then
$$g(f(x))=d(f(x),f(f(x)))<d(x,f(x))=g(x),$$
which is a contraction to $g$ attaining its minimum at $x$. It follows that $g(x)=0$, and this completes the proof.
real-analysis metric-spaces fixed-point-theorems complete-spaces
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
$endgroup$
Let $X$ be a compact metric space, and assume $f : X → X$ satisfies
$$d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$$
Define a function $g : X → mathbb{R}$ by $g(x) = d(x,f(x)), forall x ∈ X$. Prove the following statements.
a) $|g(x)-g(y)| leq 2d(x,y), forall x,y in X$.
b) $g$ is continuous and $g(x)=0$ for some $x in X$.
$text{Proof:}$
a) Observe that
begin{align*}|g(x)-g(y)|
&=|d(x,f(x))-d(y,f(y))| \
&=|d(x,f(x))-d(f(x),y)+d(f(x),y)-d(y,f(y))| \
&leq |d(x,f(x)-d(f(x),y)|+|d(f(x),y)-d(y,f(y))| \
&leq d(x,y)+d(f(x),f(y)) \
&leq 2d(x,y),
end{align*}
whre in the last step we used the hypothesis that $d(f(x), f(y))< d(x, y), forall x neq y ∈ X.$
b) To show continuity we take $delta= largefrac{varepsilon}{2}$. Then if $d(x,y)< delta$, we have by part a) that $|g(x)-g(y)|< varepsilon$.
To show that there exists $xin X$ such that $g(x)=0$, we follow AlexL's answer and note that since $g$ is continuous and $X$ is compact, $g$ attains its minimum at some $x in X$. Suppose that for that for $g(x)neq 0$, and so $f(x)neq x$. Then
$$g(f(x))=d(f(x),f(f(x)))<d(x,f(x))=g(x),$$
which is a contraction to $g$ attaining its minimum at $x$. It follows that $g(x)=0$, and this completes the proof.
real-analysis metric-spaces fixed-point-theorems complete-spaces
real-analysis metric-spaces fixed-point-theorems complete-spaces
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
edited Dec 9 '18 at 7:48
Gaby Alfonso
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
asked Oct 1 '18 at 0:03
Gaby AlfonsoGaby Alfonso
839316
839316
1
$begingroup$
You have a small error. In the statement of a) an in the last displayed line of he proof of a) the inequality $<$ should be $leq$ to include the case $x=y.$ It would be easier to make this change than to re-write it to cover only the case $xne y$.
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:47
1
$begingroup$
Another way to prove that $g$ is continuous: It suffices that if $ lim_{nto infty}d(x,y_n)=0$ then $lim_{nto infty}|g(x)-g(y_n)|=0.$ Use your proof of a), writing $y_n$ for $y.$ Then $|g(x)-g(y_n)| leq 2d(x,y_n)$..... So if $d(x,y_n)to 0$ then $|g(x)-g(y_n)|to 0 .$
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:59
add a comment |
1
$begingroup$
You have a small error. In the statement of a) an in the last displayed line of he proof of a) the inequality $<$ should be $leq$ to include the case $x=y.$ It would be easier to make this change than to re-write it to cover only the case $xne y$.
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:47
1
$begingroup$
Another way to prove that $g$ is continuous: It suffices that if $ lim_{nto infty}d(x,y_n)=0$ then $lim_{nto infty}|g(x)-g(y_n)|=0.$ Use your proof of a), writing $y_n$ for $y.$ Then $|g(x)-g(y_n)| leq 2d(x,y_n)$..... So if $d(x,y_n)to 0$ then $|g(x)-g(y_n)|to 0 .$
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:59
1
1
$begingroup$
You have a small error. In the statement of a) an in the last displayed line of he proof of a) the inequality $<$ should be $leq$ to include the case $x=y.$ It would be easier to make this change than to re-write it to cover only the case $xne y$.
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:47
$begingroup$
You have a small error. In the statement of a) an in the last displayed line of he proof of a) the inequality $<$ should be $leq$ to include the case $x=y.$ It would be easier to make this change than to re-write it to cover only the case $xne y$.
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:47
1
1
$begingroup$
Another way to prove that $g$ is continuous: It suffices that if $ lim_{nto infty}d(x,y_n)=0$ then $lim_{nto infty}|g(x)-g(y_n)|=0.$ Use your proof of a), writing $y_n$ for $y.$ Then $|g(x)-g(y_n)| leq 2d(x,y_n)$..... So if $d(x,y_n)to 0$ then $|g(x)-g(y_n)|to 0 .$
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:59
$begingroup$
Another way to prove that $g$ is continuous: It suffices that if $ lim_{nto infty}d(x,y_n)=0$ then $lim_{nto infty}|g(x)-g(y_n)|=0.$ Use your proof of a), writing $y_n$ for $y.$ Then $|g(x)-g(y_n)| leq 2d(x,y_n)$..... So if $d(x,y_n)to 0$ then $|g(x)-g(y_n)|to 0 .$
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:59
add a comment |
1 Answer
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Since $X$ is compact and $g$ continuous, there exists $x_0 in X$ such that $g(x_0)$ is minimal. I let you show that we must have $g(x_0)=0$.
answered Oct 1 '18 at 0:11
AlexLAlexL
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add a comment |
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Since $X$ is compact and $g$ continuous, there exists $x_0 in X$ such that $g(x_0)$ is minimal. I let you show that we must have $g(x_0)=0$.
answered Oct 1 '18 at 0:11
AlexLAlexL
1,24416
$endgroup$
add a comment |
$begingroup$
Since $X$ is compact and $g$ continuous, there exists $x_0 in X$ such that $g(x_0)$ is minimal. I let you show that we must have $g(x_0)=0$.
answered Oct 1 '18 at 0:11
AlexLAlexL
1,24416
$endgroup$
add a comment |
$begingroup$
Since $X$ is compact and $g$ continuous, there exists $x_0 in X$ such that $g(x_0)$ is minimal. I let you show that we must have $g(x_0)=0$.
answered Oct 1 '18 at 0:11
AlexLAlexL
1,24416
$endgroup$
Since $X$ is compact and $g$ continuous, there exists $x_0 in X$ such that $g(x_0)$ is minimal. I let you show that we must have $g(x_0)=0$.
answered Oct 1 '18 at 0:11
AlexLAlexL
1,24416
answered Oct 1 '18 at 0:11
AlexLAlexL
1,24416
answered Oct 1 '18 at 0:11
AlexLAlexL
1,24416
answered Oct 1 '18 at 0:11
AlexLAlexL
1,24416
1,24416
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1
$begingroup$
You have a small error. In the statement of a) an in the last displayed line of he proof of a) the inequality $<$ should be $leq$ to include the case $x=y.$ It would be easier to make this change than to re-write it to cover only the case $xne y$.
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:47
1
$begingroup$
Another way to prove that $g$ is continuous: It suffices that if $ lim_{nto infty}d(x,y_n)=0$ then $lim_{nto infty}|g(x)-g(y_n)|=0.$ Use your proof of a), writing $y_n$ for $y.$ Then $|g(x)-g(y_n)| leq 2d(x,y_n)$..... So if $d(x,y_n)to 0$ then $|g(x)-g(y_n)|to 0 .$
$endgroup$
– DanielWainfleet
Nov 1 '18 at 7:59