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







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  • $begingroup$
    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 $Tx in K$.






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  • $begingroup$
    @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)






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






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  • $begingroup$
    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













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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



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




















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