If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.
$begingroup$
Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that
$d(f(x),f(y)) < d(x,y)$ for all $xneq y$
a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.
b) Prove that $f$ has a unique fixed point in X.
My attempt:
a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$
Choose $epsilon>0$. Set $delta=epsilon/2$.
Then, let $delta>0$ such that $d(x,y)<delta$,
we have
|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$
=$||x-f(x)|-|y-f(y)||$
$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$
$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$
However, I'm stuck in this step,
what I need to do is to somehow make the following inequality
|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$
which would lead to
$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$
However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.
For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$
I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?
Any help would be greatly appreciated.
real-analysis uniform-continuity fixed-point-theorems fixedpoints
edited Dec 8 '18 at 12:21
amWhy
1
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
$endgroup$
add a comment |
$begingroup$
Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that
$d(f(x),f(y)) < d(x,y)$ for all $xneq y$
a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.
b) Prove that $f$ has a unique fixed point in X.
My attempt:
a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$
Choose $epsilon>0$. Set $delta=epsilon/2$.
Then, let $delta>0$ such that $d(x,y)<delta$,
we have
|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$
=$||x-f(x)|-|y-f(y)||$
$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$
$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$
However, I'm stuck in this step,
what I need to do is to somehow make the following inequality
|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$
which would lead to
$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$
However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.
For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$
I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?
Any help would be greatly appreciated.
real-analysis uniform-continuity fixed-point-theorems fixedpoints
edited Dec 8 '18 at 12:21
amWhy
1
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
$endgroup$
1
$begingroup$
$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37
1
$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41
add a comment |
$begingroup$
Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that
$d(f(x),f(y)) < d(x,y)$ for all $xneq y$
a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.
b) Prove that $f$ has a unique fixed point in X.
My attempt:
a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$
Choose $epsilon>0$. Set $delta=epsilon/2$.
Then, let $delta>0$ such that $d(x,y)<delta$,
we have
|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$
=$||x-f(x)|-|y-f(y)||$
$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$
$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$
However, I'm stuck in this step,
what I need to do is to somehow make the following inequality
|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$
which would lead to
$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$
However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.
For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$
I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?
Any help would be greatly appreciated.
real-analysis uniform-continuity fixed-point-theorems fixedpoints
edited Dec 8 '18 at 12:21
amWhy
1
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
$endgroup$
Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that
$d(f(x),f(y)) < d(x,y)$ for all $xneq y$
a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.
b) Prove that $f$ has a unique fixed point in X.
My attempt:
a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$
Choose $epsilon>0$. Set $delta=epsilon/2$.
Then, let $delta>0$ such that $d(x,y)<delta$,
we have
|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$
=$||x-f(x)|-|y-f(y)||$
$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$
$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$
However, I'm stuck in this step,
what I need to do is to somehow make the following inequality
|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$
which would lead to
$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$
However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.
For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$
I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?
Any help would be greatly appreciated.
real-analysis uniform-continuity fixed-point-theorems fixedpoints
real-analysis uniform-continuity fixed-point-theorems fixedpoints
edited Dec 8 '18 at 12:21
amWhy
1
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
edited Dec 8 '18 at 12:21
amWhy
1
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
edited Dec 8 '18 at 12:21
amWhy
1
edited Dec 8 '18 at 12:21
amWhy
1
edited Dec 8 '18 at 12:21
amWhy
1
1
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
asked Dec 8 '18 at 11:31
lastgunslingerlastgunslinger
628
628
1
$begingroup$
$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37
1
$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41
add a comment |
1
$begingroup$
$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37
1
$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41
1
1
$begingroup$
$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37
$begingroup$
$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37
1
1
$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41
$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.
Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.
Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.
Another way of proving uniform continuity of $g$ is given below :--
$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
$endgroup$
$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02
add a comment |
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$begingroup$
Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.
Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.
Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.
Another way of proving uniform continuity of $g$ is given below :--
$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
$endgroup$
$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02
add a comment |
$begingroup$
Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.
Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.
Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.
Another way of proving uniform continuity of $g$ is given below :--
$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
$endgroup$
$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02
add a comment |
$begingroup$
Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.
Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.
Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.
Another way of proving uniform continuity of $g$ is given below :--
$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
$endgroup$
Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.
Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.
Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.
Another way of proving uniform continuity of $g$ is given below :--
$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
edited Dec 8 '18 at 13:36
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
answered Dec 8 '18 at 13:19
UserSUserS
1,5391112
1,5391112
$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02
add a comment |
$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02
$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02
$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02
add a comment |
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$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37
1
$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41