A question about fixed points and non-expansive map












2












$begingroup$


Let $$K={x=(x(n))_nin l_2(mathbb{N}):|x|_2le 1 text{ and } x(n)ge 0 text{ for all } nin mathbb{N} }$$ 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 $|Tx-Ty|_2le sqrt{2} |x-y|_2$



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



$$therefore |x|_2 =left(sum ^n_{n=infty} |x(n)|^2right)^frac{1}{2} = left(sum ^n_{n=infty} (1-|x|_2)^2right)^frac{1}{2}$$



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



how to prove (1)






functional-analysis fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    yesterday












  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    yesterday
















2












$begingroup$


Let $$K={x=(x(n))_nin l_2(mathbb{N}):|x|_2le 1 text{ and } x(n)ge 0 text{ for all } nin mathbb{N} }$$ 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 $|Tx-Ty|_2le sqrt{2} |x-y|_2$



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



$$therefore |x|_2 =left(sum ^n_{n=infty} |x(n)|^2right)^frac{1}{2} = left(sum ^n_{n=infty} (1-|x|_2)^2right)^frac{1}{2}$$



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



how to prove (1)






functional-analysis fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    yesterday












  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    yesterday














2












2








2


1



$begingroup$


Let $$K={x=(x(n))_nin l_2(mathbb{N}):|x|_2le 1 text{ and } x(n)ge 0 text{ for all } nin mathbb{N} }$$ 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 $|Tx-Ty|_2le sqrt{2} |x-y|_2$



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



$$therefore |x|_2 =left(sum ^n_{n=infty} |x(n)|^2right)^frac{1}{2} = left(sum ^n_{n=infty} (1-|x|_2)^2right)^frac{1}{2}$$



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



how to prove (1)






functional-analysis fixed-point-theorems







share|cite|improve this question











$endgroup$




Let $$K={x=(x(n))_nin l_2(mathbb{N}):|x|_2le 1 text{ and } x(n)ge 0 text{ for all } nin mathbb{N} }$$ 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 $|Tx-Ty|_2le sqrt{2} |x-y|_2$



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



$$therefore |x|_2 =left(sum ^n_{n=infty} |x(n)|^2right)^frac{1}{2} = left(sum ^n_{n=infty} (1-|x|_2)^2right)^frac{1}{2}$$



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



how to prove (1)






functional-analysis fixed-point-theorems




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 yesterday









mechanodroid

29k62648




29k62648










asked yesterday









Inverse ProblemInverse Problem

1,028918




1,028918












  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    yesterday












  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    yesterday


















  • $begingroup$
    It seems that your calculation does not match the definition of $T$.
    $endgroup$
    – Song
    yesterday












  • $begingroup$
    @Song..now i edited correctly thank you
    $endgroup$
    – Inverse Problem
    yesterday
















$begingroup$
It seems that your calculation does not match the definition of $T$.
$endgroup$
– Song
yesterday






$begingroup$
It seems that your calculation does not match the definition of $T$.
$endgroup$
– Song
yesterday














$begingroup$
@Song..now i edited correctly thank you
$endgroup$
– Inverse Problem
yesterday




$begingroup$
@Song..now i edited correctly thank you
$endgroup$
– Inverse Problem
yesterday










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



begin{align}
|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
end{align}



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbb{N}$ 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
    yesterday










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    yesterday










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    yesterday












  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    yesterday





















2












$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_{n=1} |1-||x|||_2|^2)^frac{1}{2}$



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:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_{n=2} |x(n)-y(n)|^2)^frac{1}{2}\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_{n=1} |x(n)-y(n)|^2)^frac{1}{2}\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac{1}{2}\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac{1}{2}\ leq sqrt2||x-y||_2$



(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
    yesterday










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    yesterday










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    yesterday



















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
    yesterday










  • $begingroup$
    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
    yesterday












  • $begingroup$
    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
    yesterday













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160256%2fa-question-about-fixed-points-and-non-expansive-map%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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



begin{align}
|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
end{align}



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbb{N}$ 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
    yesterday










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    yesterday










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    yesterday












  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    yesterday


















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



begin{align}
|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
end{align}



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbb{N}$ 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
    yesterday










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    yesterday










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    yesterday












  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    yesterday
















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



begin{align}
|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
end{align}



which means $|Tx|_2 le 1$.



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



begin{align}
|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
end{align}



which means $|Tx|_2 le 1$.



Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbb{N}$ 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 yesterday

























answered yesterday









mechanodroidmechanodroid

29k62648




29k62648












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










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    yesterday










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    yesterday












  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    yesterday




















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










  • $begingroup$
    @InverseProblem Have a look.
    $endgroup$
    – mechanodroid
    yesterday










  • $begingroup$
    @mechanodroid......thank you so much ......for your help
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
    $endgroup$
    – Inverse Problem
    yesterday












  • $begingroup$
    @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
    $endgroup$
    – mechanodroid
    yesterday


















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




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












$begingroup$
@InverseProblem Have a look.
$endgroup$
– mechanodroid
yesterday




$begingroup$
@InverseProblem Have a look.
$endgroup$
– mechanodroid
yesterday












$begingroup$
@mechanodroid......thank you so much ......for your help
$endgroup$
– Inverse Problem
yesterday




$begingroup$
@mechanodroid......thank you so much ......for your help
$endgroup$
– Inverse Problem
yesterday












$begingroup$
@mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
$endgroup$
– Inverse Problem
yesterday






$begingroup$
@mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/…
$endgroup$
– Inverse Problem
yesterday














$begingroup$
@InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
$endgroup$
– mechanodroid
yesterday






$begingroup$
@InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$?
$endgroup$
– mechanodroid
yesterday













2












$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_{n=1} |1-||x|||_2|^2)^frac{1}{2}$



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:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_{n=2} |x(n)-y(n)|^2)^frac{1}{2}\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_{n=1} |x(n)-y(n)|^2)^frac{1}{2}\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac{1}{2}\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac{1}{2}\ leq sqrt2||x-y||_2$



(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
    yesterday










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    yesterday










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    yesterday
















2












$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_{n=1} |1-||x|||_2|^2)^frac{1}{2}$



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:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_{n=2} |x(n)-y(n)|^2)^frac{1}{2}\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_{n=1} |x(n)-y(n)|^2)^frac{1}{2}\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac{1}{2}\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac{1}{2}\ leq sqrt2||x-y||_2$



(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
    yesterday










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    yesterday










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    yesterday














2












2








2





$begingroup$

You're on the right track.



As you said - $||x||_2 =(sum ^infty_{n=1} |1-||x|||_2|^2)^frac{1}{2}$



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:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_{n=2} |x(n)-y(n)|^2)^frac{1}{2}\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_{n=1} |x(n)-y(n)|^2)^frac{1}{2}\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac{1}{2}\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac{1}{2}\ leq sqrt2||x-y||_2$



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






share|cite|improve this answer











$endgroup$



You're on the right track.



As you said - $||x||_2 =(sum ^infty_{n=1} |1-||x|||_2|^2)^frac{1}{2}$



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:



$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_{n=2} |x(n)-y(n)|^2)^frac{1}{2}\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_{n=1} |x(n)-y(n)|^2)^frac{1}{2}\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac{1}{2}\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac{1}{2}\ leq sqrt2||x-y||_2$



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









GSoferGSofer

781313




781313












  • $begingroup$
    ..can you tell me how to prove self map
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    yesterday










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    yesterday


















  • $begingroup$
    ..can you tell me how to prove self map
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    I'm not familiar with that term, what does it mean?
    $endgroup$
    – GSofer
    yesterday










  • $begingroup$
    it means that we have to prove $T$ map from $K$ to $K$
    $endgroup$
    – Inverse Problem
    yesterday
















$begingroup$
..can you tell me how to prove self map
$endgroup$
– Inverse Problem
yesterday




$begingroup$
..can you tell me how to prove self map
$endgroup$
– Inverse Problem
yesterday












$begingroup$
I'm not familiar with that term, what does it mean?
$endgroup$
– GSofer
yesterday




$begingroup$
I'm not familiar with that term, what does it mean?
$endgroup$
– GSofer
yesterday












$begingroup$
it means that we have to prove $T$ map from $K$ to $K$
$endgroup$
– Inverse Problem
yesterday




$begingroup$
it means that we have to prove $T$ map from $K$ to $K$
$endgroup$
– Inverse Problem
yesterday











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
    yesterday










  • $begingroup$
    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
    yesterday












  • $begingroup$
    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
    yesterday


















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
    yesterday










  • $begingroup$
    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
    yesterday












  • $begingroup$
    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
    yesterday
















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 yesterday









Yu DingYu Ding

5186




5186












  • $begingroup$
    Ding....can you tell how to prove this is self map
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    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
    yesterday












  • $begingroup$
    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
    yesterday




















  • $begingroup$
    Ding....can you tell how to prove this is self map
    $endgroup$
    – Inverse Problem
    yesterday










  • $begingroup$
    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
    yesterday












  • $begingroup$
    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
    yesterday


















$begingroup$
Ding....can you tell how to prove this is self map
$endgroup$
– Inverse Problem
yesterday




$begingroup$
Ding....can you tell how to prove this is self map
$endgroup$
– Inverse Problem
yesterday












$begingroup$
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
yesterday






$begingroup$
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
yesterday














$begingroup$
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
yesterday






$begingroup$
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
yesterday




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160256%2fa-question-about-fixed-points-and-non-expansive-map%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How did Captain America manage to do this?

Image
2 How does Captain America lift Thor's hammer? In Avengers: Age of Ultron , he is not worthy but how he is worthy in this film. plot-explanation marvel-cinematic-universe avengers-age-of-ultron avengers-endgame share | improve this question edited 19 mins ago A J ♦ 42.7k 16 229 246 asked 39 mins ago Andi AR Andi AR 325 1 4 14
Read more

迪纳利

Image
迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
Read more

南乌拉尔铁路局

Image
南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
Read more