Fixed point of a function defined in terms of a metric on a compact metric space












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













share|cite|improve this question




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edited Dec 9 '18 at 7:48







Gaby Alfonso

















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
















  • 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












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$begingroup$

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 $g(x_0)=0$.






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    $begingroup$

    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 $g(x_0)=0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $g(x_0)=0$.






      share|cite|improve this answer









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        $begingroup$

        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 $g(x_0)=0$.






        share|cite|improve this answer









        $endgroup$



        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 $g(x_0)=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 1 '18 at 0:11









        AlexLAlexL

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