A question about fixed points and non-expansive map The 2019 Stack Overflow Developer Survey Results Are InDiameters, distances and contraction mappings on a subset of $C_mathbbR[0,1]$Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ has no fixed pointsAbout fixed point and dampingConnection between codata and greatest fixed pointsNo fixed points imply no periodic pointsFixed points of complex rational functions dilemma.Contractions and Fixed PointsFind the Fixed points (Knaster-Tarski Theorem)On subgroups of isometries and their respective fixed-pointsWhy orientation-preserving self-homeomorphism of $mathbbCP^n$ when $n$ is odd must have fixed points?Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ has no fixed pointsTo Prove $T $ is a self map and $T$ have no fixed points

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A question about fixed points and non-expansive map



The 2019 Stack Overflow Developer Survey Results Are InDiameters, distances and contraction mappings on a subset of $C_mathbbR[0,1]$Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ has no fixed pointsAbout fixed point and dampingConnection between codata and greatest fixed pointsNo fixed points imply no periodic pointsFixed points of complex rational functions dilemma.Contractions and Fixed PointsFind the Fixed points (Knaster-Tarski Theorem)On subgroups of isometries and their respective fixed-pointsWhy orientation-preserving self-homeomorphism of $mathbbCP^n$ when $n$ is odd must have fixed points?Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ has no fixed pointsTo Prove $T $ is a self map and $T$ have no fixed points










2












$begingroup$


Let $$K=_2le 1 text and x(n)ge 0 text for all nin mathbbN $$ and define $T:Kto c_0$ by $T(x)=(1-|x|_2,x(1),x(2),ldots)$. Prove :



(1) $T$ is self map on $K$ and $|Tx-Ty|_2le sqrt2 |x-y|_2$



(2) $T $ does not have fixed points in $K$



my attempt



for (2):



suppose $T$ have fixed point i.e., $Tx=x$



then $(1-|x|_2, x(1),x(2),ldots)=(x(1),x(2),ldots)$



then $x(1)=1-|x|_2, x(2)=x(1), x(3)=x(2),ldots$



$$therefore |x|_2 =left(sum ^n_n=infty |x(n)|^2right)^frac12 = left(sum ^n_n=infty (1-|x|_2)^2right)^frac12$$



but how to prove this $x$ is not in $K$?



how to prove (1)










share|cite|improve this question











$endgroup$











  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    Mar 24 at 8:01











  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    Mar 24 at 8:13















2












$begingroup$


Let $$K=_2le 1 text and x(n)ge 0 text for all nin mathbbN $$ and define $T:Kto c_0$ by $T(x)=(1-|x|_2,x(1),x(2),ldots)$. Prove :



(1) $T$ is self map on $K$ and $|Tx-Ty|_2le sqrt2 |x-y|_2$



(2) $T $ does not have fixed points in $K$



my attempt



for (2):



suppose $T$ have fixed point i.e., $Tx=x$



then $(1-|x|_2, x(1),x(2),ldots)=(x(1),x(2),ldots)$



then $x(1)=1-|x|_2, x(2)=x(1), x(3)=x(2),ldots$



$$therefore |x|_2 =left(sum ^n_n=infty |x(n)|^2right)^frac12 = left(sum ^n_n=infty (1-|x|_2)^2right)^frac12$$



but how to prove this $x$ is not in $K$?



how to prove (1)










share|cite|improve this question











$endgroup$











  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    Mar 24 at 8:01











  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    Mar 24 at 8:13













2












2








2


1



$begingroup$


Let $$K=_2le 1 text and x(n)ge 0 text for all nin mathbbN $$ and define $T:Kto c_0$ by $T(x)=(1-|x|_2,x(1),x(2),ldots)$. Prove :



(1) $T$ is self map on $K$ and $|Tx-Ty|_2le sqrt2 |x-y|_2$



(2) $T $ does not have fixed points in $K$



my attempt



for (2):



suppose $T$ have fixed point i.e., $Tx=x$



then $(1-|x|_2, x(1),x(2),ldots)=(x(1),x(2),ldots)$



then $x(1)=1-|x|_2, x(2)=x(1), x(3)=x(2),ldots$



$$therefore |x|_2 =left(sum ^n_n=infty |x(n)|^2right)^frac12 = left(sum ^n_n=infty (1-|x|_2)^2right)^frac12$$



but how to prove this $x$ is not in $K$?



how to prove (1)










share|cite|improve this question











$endgroup$




Let $$K=_2le 1 text and x(n)ge 0 text for all nin mathbbN $$ and define $T:Kto c_0$ by $T(x)=(1-|x|_2,x(1),x(2),ldots)$. Prove :



(1) $T$ is self map on $K$ and $|Tx-Ty|_2le sqrt2 |x-y|_2$



(2) $T $ does not have fixed points in $K$



my attempt



for (2):



suppose $T$ have fixed point i.e., $Tx=x$



then $(1-|x|_2, x(1),x(2),ldots)=(x(1),x(2),ldots)$



then $x(1)=1-|x|_2, x(2)=x(1), x(3)=x(2),ldots$



$$therefore |x|_2 =left(sum ^n_n=infty |x(n)|^2right)^frac12 = left(sum ^n_n=infty (1-|x|_2)^2right)^frac12$$



but how to prove this $x$ is not in $K$?



how to prove (1)







functional-analysis fixed-point-theorems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 24 at 10:46









mechanodroid

28.9k62648




28.9k62648










asked Mar 24 at 7:52









Inverse ProblemInverse Problem

1,032918




1,032918











  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    Mar 24 at 8:01











  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    Mar 24 at 8:13
















  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    Mar 24 at 8:01











  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    Mar 24 at 8:13















$begingroup$
It seems that your calculation does not match the definition of $T$.
$endgroup$
– Song
Mar 24 at 8:01





$begingroup$
It seems that your calculation does not match the definition of $T$.
$endgroup$
– Song
Mar 24 at 8:01













$begingroup$
@Song..now i edited correctly thank you
$endgroup$
– Inverse Problem
Mar 24 at 8:13




$begingroup$
@Song..now i edited correctly thank you
$endgroup$
– Inverse Problem
Mar 24 at 8:13










3 Answers
3






active

oldest

votes


















2












$begingroup$

For $(2)$ you got that if $Tx = x$ then $$x = (1-|x|_2, 1-|x|_2, ldots)$$



so $$+infty > |x|_2^2 = sum_n=1^infty (1-|x|_2)^2$$
The only way this series converges is if $1-|x|_2 = 0$, or $|x|_2 = 1$, so $x = (1,1,1ldots )$. But then clearly $|x|_2 = +infty$ and not $1$ so this is a contradiction.






To show that $T$ is actually a map $K to K$, take $x in K$ and we claim that $Tx in K$ as well.

Since $|x|_2 le 1$ we have $1-|x|_2 ge 0$ so



beginalign
|Tx|_2^2 &= (1-|x|_2)^2 + sum_n=1^infty |x_n|^2 \
&= (1-|x|_2)^2 + |x|_2^2 \
&le (1-|x|_2)^2 + 2(1-|x|_2)|x|_2 + |x|_2^2 \
&= (1-|x|_2+|x|_2)^2 \
&= 1
endalign



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbbN$ and $1-|x|_2 ge 0$.



Therefore $Tx in K$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    @mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1)
    $endgroup$
    – Inverse Problem
    Mar 24 at 10:27










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    Mar 24 at 10:42










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    Mar 24 at 12:51










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    Mar 24 at 13:09











  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    Mar 24 at 15:48



















2












$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_n=1 |1-||x|||_2|^2)^frac12$



There is only one possible way for this sum to converge - if and only if $||x||_2=1$. But in this case, we also get $||x||_2=0$ - a contradiction.



About (1) - let's try evaluating the required norm:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_n=2 |x(n)-y(n)|^2)^frac12\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_n=1 |x(n)-y(n)|^2)^frac12\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac12\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac12\ leq sqrt2||x-y||_2$



(Every $leq$ sign is due to triangle inequality)






share|cite|improve this answer











$endgroup$












  • $begingroup$
    ..can you tell me how to prove self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:09










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    Mar 24 at 9:11










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:13


















1












$begingroup$

It seems to me one may not need the assumption $x(n)geq 0$. In fact, if $Tx=x$,
then $|Tx|^2=|x|^2$ which says $(1-|x|)^2+x(1)^2+x(2)^2...=x(1)^2+x(2)^2+...$,
so we see $|x|=1$. Then $x(1)=0$. Let $n$ be the first integer with $x(n)neq 0$,
there exists such $n$ since $|x|=1$.
So $x(n-1)=0$, but the $n$-th component in $Tx$ is $x(n-1)=0$, contradicts
$Tx=x$ and $x(n)neq 0$.



Part 1: $|Tx-Ty|^2=[(1-|x|)-(1-|y|)]^2+[x(1)-y(1)]^2+...=(|x|-|y|)^2+|x-y|^2leq |x-y|^2+|x-y|^2=2|x-y|^2$ .






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Ding....can you tell how to prove this is self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:08










  • $begingroup$
    From $|x|leq 1$, the condition $x(n)geq 0$ is preserved by $T$. Next, $|Tx|^2=(1-|x|)^2+|x|^2leq 1$, here one uses the elementary inequality $(1-a)^2+a^2leq 1$ when $0leq aleq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$.
    $endgroup$
    – Yu Ding
    Mar 24 at 9:17











  • $begingroup$
    what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:19












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

For $(2)$ you got that if $Tx = x$ then $$x = (1-|x|_2, 1-|x|_2, ldots)$$



so $$+infty > |x|_2^2 = sum_n=1^infty (1-|x|_2)^2$$
The only way this series converges is if $1-|x|_2 = 0$, or $|x|_2 = 1$, so $x = (1,1,1ldots )$. But then clearly $|x|_2 = +infty$ and not $1$ so this is a contradiction.






To show that $T$ is actually a map $K to K$, take $x in K$ and we claim that $Tx in K$ as well.

Since $|x|_2 le 1$ we have $1-|x|_2 ge 0$ so



beginalign
|Tx|_2^2 &= (1-|x|_2)^2 + sum_n=1^infty |x_n|^2 \
&= (1-|x|_2)^2 + |x|_2^2 \
&le (1-|x|_2)^2 + 2(1-|x|_2)|x|_2 + |x|_2^2 \
&= (1-|x|_2+|x|_2)^2 \
&= 1
endalign



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbbN$ and $1-|x|_2 ge 0$.



Therefore $Tx in K$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    @mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1)
    $endgroup$
    – Inverse Problem
    Mar 24 at 10:27










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    Mar 24 at 10:42










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    Mar 24 at 12:51










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    Mar 24 at 13:09











  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    Mar 24 at 15:48
















2












$begingroup$

For $(2)$ you got that if $Tx = x$ then $$x = (1-|x|_2, 1-|x|_2, ldots)$$



so $$+infty > |x|_2^2 = sum_n=1^infty (1-|x|_2)^2$$
The only way this series converges is if $1-|x|_2 = 0$, or $|x|_2 = 1$, so $x = (1,1,1ldots )$. But then clearly $|x|_2 = +infty$ and not $1$ so this is a contradiction.






To show that $T$ is actually a map $K to K$, take $x in K$ and we claim that $Tx in K$ as well.

Since $|x|_2 le 1$ we have $1-|x|_2 ge 0$ so



beginalign
|Tx|_2^2 &= (1-|x|_2)^2 + sum_n=1^infty |x_n|^2 \
&= (1-|x|_2)^2 + |x|_2^2 \
&le (1-|x|_2)^2 + 2(1-|x|_2)|x|_2 + |x|_2^2 \
&= (1-|x|_2+|x|_2)^2 \
&= 1
endalign



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbbN$ and $1-|x|_2 ge 0$.



Therefore $Tx in K$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    @mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1)
    $endgroup$
    – Inverse Problem
    Mar 24 at 10:27










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    Mar 24 at 10:42










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    Mar 24 at 12:51










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    Mar 24 at 13:09











  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    Mar 24 at 15:48














2












2








2





$begingroup$

For $(2)$ you got that if $Tx = x$ then $$x = (1-|x|_2, 1-|x|_2, ldots)$$



so $$+infty > |x|_2^2 = sum_n=1^infty (1-|x|_2)^2$$
The only way this series converges is if $1-|x|_2 = 0$, or $|x|_2 = 1$, so $x = (1,1,1ldots )$. But then clearly $|x|_2 = +infty$ and not $1$ so this is a contradiction.






To show that $T$ is actually a map $K to K$, take $x in K$ and we claim that $Tx in K$ as well.

Since $|x|_2 le 1$ we have $1-|x|_2 ge 0$ so



beginalign
|Tx|_2^2 &= (1-|x|_2)^2 + sum_n=1^infty |x_n|^2 \
&= (1-|x|_2)^2 + |x|_2^2 \
&le (1-|x|_2)^2 + 2(1-|x|_2)|x|_2 + |x|_2^2 \
&= (1-|x|_2+|x|_2)^2 \
&= 1
endalign



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbbN$ and $1-|x|_2 ge 0$.



Therefore $Tx in K$.






share|cite|improve this answer











$endgroup$



For $(2)$ you got that if $Tx = x$ then $$x = (1-|x|_2, 1-|x|_2, ldots)$$



so $$+infty > |x|_2^2 = sum_n=1^infty (1-|x|_2)^2$$
The only way this series converges is if $1-|x|_2 = 0$, or $|x|_2 = 1$, so $x = (1,1,1ldots )$. But then clearly $|x|_2 = +infty$ and not $1$ so this is a contradiction.






To show that $T$ is actually a map $K to K$, take $x in K$ and we claim that $Tx in K$ as well.

Since $|x|_2 le 1$ we have $1-|x|_2 ge 0$ so



beginalign
|Tx|_2^2 &= (1-|x|_2)^2 + sum_n=1^infty |x_n|^2 \
&= (1-|x|_2)^2 + |x|_2^2 \
&le (1-|x|_2)^2 + 2(1-|x|_2)|x|_2 + |x|_2^2 \
&= (1-|x|_2+|x|_2)^2 \
&= 1
endalign



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbbN$ and $1-|x|_2 ge 0$.



Therefore $Tx in K$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 24 at 10:42

























answered Mar 24 at 10:15









mechanodroidmechanodroid

28.9k62648




28.9k62648











  • $begingroup$
    @mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1)
    $endgroup$
    – Inverse Problem
    Mar 24 at 10:27










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    Mar 24 at 10:42










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    Mar 24 at 12:51










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    Mar 24 at 13:09











  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    Mar 24 at 15:48

















  • $begingroup$
    @mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1)
    $endgroup$
    – Inverse Problem
    Mar 24 at 10:27










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    Mar 24 at 10:42










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    Mar 24 at 12:51










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    Mar 24 at 13:09











  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    Mar 24 at 15:48
















$begingroup$
@mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1)
$endgroup$
– Inverse Problem
Mar 24 at 10:27




$begingroup$
@mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1)
$endgroup$
– Inverse Problem
Mar 24 at 10:27












$begingroup$
@InverseProblem Have a look.
$endgroup$
– mechanodroid
Mar 24 at 10:42




$begingroup$
@InverseProblem Have a look.
$endgroup$
– mechanodroid
Mar 24 at 10:42












$begingroup$
@mechanodroid......thank you so much ......for your help
$endgroup$
– Inverse Problem
Mar 24 at 12:51




$begingroup$
@mechanodroid......thank you so much ......for your help
$endgroup$
– Inverse Problem
Mar 24 at 12:51












$begingroup$
@mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
$endgroup$
– Inverse Problem
Mar 24 at 13:09





$begingroup$
@mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
$endgroup$
– Inverse Problem
Mar 24 at 13:09













$begingroup$
@InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
$endgroup$
– mechanodroid
Mar 24 at 15:48





$begingroup$
@InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
$endgroup$
– mechanodroid
Mar 24 at 15:48












2












$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_n=1 |1-||x|||_2|^2)^frac12$



There is only one possible way for this sum to converge - if and only if $||x||_2=1$. But in this case, we also get $||x||_2=0$ - a contradiction.



About (1) - let's try evaluating the required norm:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_n=2 |x(n)-y(n)|^2)^frac12\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_n=1 |x(n)-y(n)|^2)^frac12\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac12\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac12\ leq sqrt2||x-y||_2$



(Every $leq$ sign is due to triangle inequality)






share|cite|improve this answer











$endgroup$












  • $begingroup$
    ..can you tell me how to prove self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:09










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    Mar 24 at 9:11










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:13















2












$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_n=1 |1-||x|||_2|^2)^frac12$



There is only one possible way for this sum to converge - if and only if $||x||_2=1$. But in this case, we also get $||x||_2=0$ - a contradiction.



About (1) - let's try evaluating the required norm:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_n=2 |x(n)-y(n)|^2)^frac12\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_n=1 |x(n)-y(n)|^2)^frac12\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac12\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac12\ leq sqrt2||x-y||_2$



(Every $leq$ sign is due to triangle inequality)






share|cite|improve this answer











$endgroup$












  • $begingroup$
    ..can you tell me how to prove self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:09










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    Mar 24 at 9:11










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:13













2












2








2





$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_n=1 |1-||x|||_2|^2)^frac12$



There is only one possible way for this sum to converge - if and only if $||x||_2=1$. But in this case, we also get $||x||_2=0$ - a contradiction.



About (1) - let's try evaluating the required norm:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_n=2 |x(n)-y(n)|^2)^frac12\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_n=1 |x(n)-y(n)|^2)^frac12\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac12\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac12\ leq sqrt2||x-y||_2$



(Every $leq$ sign is due to triangle inequality)






share|cite|improve this answer











$endgroup$



You're on the right track.



As you said - $||x||_2 =(sum ^infty_n=1 |1-||x|||_2|^2)^frac12$



There is only one possible way for this sum to converge - if and only if $||x||_2=1$. But in this case, we also get $||x||_2=0$ - a contradiction.



About (1) - let's try evaluating the required norm:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_n=2 |x(n)-y(n)|^2)^frac12\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_n=1 |x(n)-y(n)|^2)^frac12\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac12\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac12\ leq sqrt2||x-y||_2$



(Every $leq$ sign is due to triangle inequality)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 24 at 8:55

























answered Mar 24 at 8:16









GSoferGSofer

7801315




7801315











  • $begingroup$
    ..can you tell me how to prove self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:09










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    Mar 24 at 9:11










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:13
















  • $begingroup$
    ..can you tell me how to prove self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:09










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    Mar 24 at 9:11










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:13















$begingroup$
..can you tell me how to prove self map
$endgroup$
– Inverse Problem
Mar 24 at 9:09




$begingroup$
..can you tell me how to prove self map
$endgroup$
– Inverse Problem
Mar 24 at 9:09












$begingroup$
I'm not familiar with that term, what does it mean?
$endgroup$
– GSofer
Mar 24 at 9:11




$begingroup$
I'm not familiar with that term, what does it mean?
$endgroup$
– GSofer
Mar 24 at 9:11












$begingroup$
it means that we have to prove $T$ map from $K$ to $K$
$endgroup$
– Inverse Problem
Mar 24 at 9:13




$begingroup$
it means that we have to prove $T$ map from $K$ to $K$
$endgroup$
– Inverse Problem
Mar 24 at 9:13











1












$begingroup$

It seems to me one may not need the assumption $x(n)geq 0$. In fact, if $Tx=x$,
then $|Tx|^2=|x|^2$ which says $(1-|x|)^2+x(1)^2+x(2)^2...=x(1)^2+x(2)^2+...$,
so we see $|x|=1$. Then $x(1)=0$. Let $n$ be the first integer with $x(n)neq 0$,
there exists such $n$ since $|x|=1$.
So $x(n-1)=0$, but the $n$-th component in $Tx$ is $x(n-1)=0$, contradicts
$Tx=x$ and $x(n)neq 0$.



Part 1: $|Tx-Ty|^2=[(1-|x|)-(1-|y|)]^2+[x(1)-y(1)]^2+...=(|x|-|y|)^2+|x-y|^2leq |x-y|^2+|x-y|^2=2|x-y|^2$ .






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Ding....can you tell how to prove this is self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:08










  • $begingroup$
    From $|x|leq 1$, the condition $x(n)geq 0$ is preserved by $T$. Next, $|Tx|^2=(1-|x|)^2+|x|^2leq 1$, here one uses the elementary inequality $(1-a)^2+a^2leq 1$ when $0leq aleq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$.
    $endgroup$
    – Yu Ding
    Mar 24 at 9:17











  • $begingroup$
    what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:19
















1












$begingroup$

It seems to me one may not need the assumption $x(n)geq 0$. In fact, if $Tx=x$,
then $|Tx|^2=|x|^2$ which says $(1-|x|)^2+x(1)^2+x(2)^2...=x(1)^2+x(2)^2+...$,
so we see $|x|=1$. Then $x(1)=0$. Let $n$ be the first integer with $x(n)neq 0$,
there exists such $n$ since $|x|=1$.
So $x(n-1)=0$, but the $n$-th component in $Tx$ is $x(n-1)=0$, contradicts
$Tx=x$ and $x(n)neq 0$.



Part 1: $|Tx-Ty|^2=[(1-|x|)-(1-|y|)]^2+[x(1)-y(1)]^2+...=(|x|-|y|)^2+|x-y|^2leq |x-y|^2+|x-y|^2=2|x-y|^2$ .






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Ding....can you tell how to prove this is self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:08










  • $begingroup$
    From $|x|leq 1$, the condition $x(n)geq 0$ is preserved by $T$. Next, $|Tx|^2=(1-|x|)^2+|x|^2leq 1$, here one uses the elementary inequality $(1-a)^2+a^2leq 1$ when $0leq aleq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$.
    $endgroup$
    – Yu Ding
    Mar 24 at 9:17











  • $begingroup$
    what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:19














1












1








1





$begingroup$

It seems to me one may not need the assumption $x(n)geq 0$. In fact, if $Tx=x$,
then $|Tx|^2=|x|^2$ which says $(1-|x|)^2+x(1)^2+x(2)^2...=x(1)^2+x(2)^2+...$,
so we see $|x|=1$. Then $x(1)=0$. Let $n$ be the first integer with $x(n)neq 0$,
there exists such $n$ since $|x|=1$.
So $x(n-1)=0$, but the $n$-th component in $Tx$ is $x(n-1)=0$, contradicts
$Tx=x$ and $x(n)neq 0$.



Part 1: $|Tx-Ty|^2=[(1-|x|)-(1-|y|)]^2+[x(1)-y(1)]^2+...=(|x|-|y|)^2+|x-y|^2leq |x-y|^2+|x-y|^2=2|x-y|^2$ .






share|cite|improve this answer









$endgroup$



It seems to me one may not need the assumption $x(n)geq 0$. In fact, if $Tx=x$,
then $|Tx|^2=|x|^2$ which says $(1-|x|)^2+x(1)^2+x(2)^2...=x(1)^2+x(2)^2+...$,
so we see $|x|=1$. Then $x(1)=0$. Let $n$ be the first integer with $x(n)neq 0$,
there exists such $n$ since $|x|=1$.
So $x(n-1)=0$, but the $n$-th component in $Tx$ is $x(n-1)=0$, contradicts
$Tx=x$ and $x(n)neq 0$.



Part 1: $|Tx-Ty|^2=[(1-|x|)-(1-|y|)]^2+[x(1)-y(1)]^2+...=(|x|-|y|)^2+|x-y|^2leq |x-y|^2+|x-y|^2=2|x-y|^2$ .







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 24 at 8:48









Yu DingYu Ding

7187




7187











  • $begingroup$
    Ding....can you tell how to prove this is self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:08










  • $begingroup$
    From $|x|leq 1$, the condition $x(n)geq 0$ is preserved by $T$. Next, $|Tx|^2=(1-|x|)^2+|x|^2leq 1$, here one uses the elementary inequality $(1-a)^2+a^2leq 1$ when $0leq aleq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$.
    $endgroup$
    – Yu Ding
    Mar 24 at 9:17











  • $begingroup$
    what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:19

















  • $begingroup$
    Ding....can you tell how to prove this is self map
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:08










  • $begingroup$
    From $|x|leq 1$, the condition $x(n)geq 0$ is preserved by $T$. Next, $|Tx|^2=(1-|x|)^2+|x|^2leq 1$, here one uses the elementary inequality $(1-a)^2+a^2leq 1$ when $0leq aleq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$.
    $endgroup$
    – Yu Ding
    Mar 24 at 9:17











  • $begingroup$
    what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set
    $endgroup$
    – Inverse Problem
    Mar 24 at 9:19
















$begingroup$
Ding....can you tell how to prove this is self map
$endgroup$
– Inverse Problem
Mar 24 at 9:08




$begingroup$
Ding....can you tell how to prove this is self map
$endgroup$
– Inverse Problem
Mar 24 at 9:08












$begingroup$
From $|x|leq 1$, the condition $x(n)geq 0$ is preserved by $T$. Next, $|Tx|^2=(1-|x|)^2+|x|^2leq 1$, here one uses the elementary inequality $(1-a)^2+a^2leq 1$ when $0leq aleq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$.
$endgroup$
– Yu Ding
Mar 24 at 9:17





$begingroup$
From $|x|leq 1$, the condition $x(n)geq 0$ is preserved by $T$. Next, $|Tx|^2=(1-|x|)^2+|x|^2leq 1$, here one uses the elementary inequality $(1-a)^2+a^2leq 1$ when $0leq aleq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$.
$endgroup$
– Yu Ding
Mar 24 at 9:17













$begingroup$
what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set
$endgroup$
– Inverse Problem
Mar 24 at 9:19





$begingroup$
what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set
$endgroup$
– Inverse Problem
Mar 24 at 9:19


















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