Show that $Vert Tx-Tp Vert leq Vert x-p Vert $, where $p=0.$












1












$begingroup$


Let $X = l_infty$ (the space of sequences of real numbers which are bounded). Let $K={xin l_infty:Vert x Vert_inftyleq 1}.$ Defined begin{align} T:& Kto K \&xmapsto Tx=(0,x^2_1,x^2_2,x^2_3,cdots)end{align}



I want to show that





  1. $Tp=p$ if and only if $p=0;$


  2. $Vert Tx-Tp Vert leq Vert x-p Vert $, where $p=0.$


MY TRIAL



1.



begin{align} Tp=p&iff(0,p^2_1,p^2_2,p^2_3,cdots)=(p_1,p_2,p_3,cdots)\ &iff p_1=0,;p_2=p^2_1,;p_3=p^2_2,;cdots\ &iff p_n=0,;forall nin Bbb{N}\ &iff p=0end{align}




  1. Let $x,pin K$ s.t. $p=0,$ then
    begin{align} Vert Tx-Tp Vert=Vert (0,x^2_1,x^2_2,x^2_3,cdots)-(0,0,0,0,cdots)Vertend{align}
    Honestly, I don't know what to do from here. Any help please?






functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||le ||x||$?
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 16:51










  • $begingroup$
    @BigbearZzz: Yes, that's it!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:17
















1












$begingroup$


Let $X = l_infty$ (the space of sequences of real numbers which are bounded). Let $K={xin l_infty:Vert x Vert_inftyleq 1}.$ Defined begin{align} T:& Kto K \&xmapsto Tx=(0,x^2_1,x^2_2,x^2_3,cdots)end{align}



I want to show that





  1. $Tp=p$ if and only if $p=0;$


  2. $Vert Tx-Tp Vert leq Vert x-p Vert $, where $p=0.$


MY TRIAL



1.



begin{align} Tp=p&iff(0,p^2_1,p^2_2,p^2_3,cdots)=(p_1,p_2,p_3,cdots)\ &iff p_1=0,;p_2=p^2_1,;p_3=p^2_2,;cdots\ &iff p_n=0,;forall nin Bbb{N}\ &iff p=0end{align}




  1. Let $x,pin K$ s.t. $p=0,$ then
    begin{align} Vert Tx-Tp Vert=Vert (0,x^2_1,x^2_2,x^2_3,cdots)-(0,0,0,0,cdots)Vertend{align}
    Honestly, I don't know what to do from here. Any help please?






functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||le ||x||$?
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 16:51










  • $begingroup$
    @BigbearZzz: Yes, that's it!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:17














1












1








1


1



$begingroup$


Let $X = l_infty$ (the space of sequences of real numbers which are bounded). Let $K={xin l_infty:Vert x Vert_inftyleq 1}.$ Defined begin{align} T:& Kto K \&xmapsto Tx=(0,x^2_1,x^2_2,x^2_3,cdots)end{align}



I want to show that





  1. $Tp=p$ if and only if $p=0;$


  2. $Vert Tx-Tp Vert leq Vert x-p Vert $, where $p=0.$


MY TRIAL



1.



begin{align} Tp=p&iff(0,p^2_1,p^2_2,p^2_3,cdots)=(p_1,p_2,p_3,cdots)\ &iff p_1=0,;p_2=p^2_1,;p_3=p^2_2,;cdots\ &iff p_n=0,;forall nin Bbb{N}\ &iff p=0end{align}




  1. Let $x,pin K$ s.t. $p=0,$ then
    begin{align} Vert Tx-Tp Vert=Vert (0,x^2_1,x^2_2,x^2_3,cdots)-(0,0,0,0,cdots)Vertend{align}
    Honestly, I don't know what to do from here. Any help please?






functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$




Let $X = l_infty$ (the space of sequences of real numbers which are bounded). Let $K={xin l_infty:Vert x Vert_inftyleq 1}.$ Defined begin{align} T:& Kto K \&xmapsto Tx=(0,x^2_1,x^2_2,x^2_3,cdots)end{align}



I want to show that





  1. $Tp=p$ if and only if $p=0;$


  2. $Vert Tx-Tp Vert leq Vert x-p Vert $, where $p=0.$


MY TRIAL



1.



begin{align} Tp=p&iff(0,p^2_1,p^2_2,p^2_3,cdots)=(p_1,p_2,p_3,cdots)\ &iff p_1=0,;p_2=p^2_1,;p_3=p^2_2,;cdots\ &iff p_n=0,;forall nin Bbb{N}\ &iff p=0end{align}




  1. Let $x,pin K$ s.t. $p=0,$ then
    begin{align} Vert Tx-Tp Vert=Vert (0,x^2_1,x^2_2,x^2_3,cdots)-(0,0,0,0,cdots)Vertend{align}
    Honestly, I don't know what to do from here. Any help please?






functional-analysis analysis normed-spaces




functional-analysis analysis normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 17:16







Omojola Micheal

















asked Dec 14 '18 at 16:46









Omojola MichealOmojola Micheal

1,853324




1,853324












  • $begingroup$
    If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||le ||x||$?
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 16:51










  • $begingroup$
    @BigbearZzz: Yes, that's it!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:17


















  • $begingroup$
    If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||le ||x||$?
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 16:51










  • $begingroup$
    @BigbearZzz: Yes, that's it!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:17
















$begingroup$
If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||le ||x||$?
$endgroup$
– BigbearZzz
Dec 14 '18 at 16:51




$begingroup$
If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||le ||x||$?
$endgroup$
– BigbearZzz
Dec 14 '18 at 16:51












$begingroup$
@BigbearZzz: Yes, that's it!
$endgroup$
– Omojola Micheal
Dec 14 '18 at 17:17




$begingroup$
@BigbearZzz: Yes, that's it!
$endgroup$
– Omojola Micheal
Dec 14 '18 at 17:17










2 Answers
2






active

oldest

votes


















1












$begingroup$

The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.



Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I don't quite get the hint. Can you please, break it down?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:18










  • $begingroup$
    Your set $K$ consists of elements $x=(x_1,x_2,dots)$ such that $|x_i|le 1$ (see the definition of the norm on $l^infty$).
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:22



















1












$begingroup$

Your answer for 1 is maybe not very well written, but it looks correct to me.



For 2 notice that $Tp=p=0$, so you just need to show $|Tx|_inftyleq|x|_infty$. Maybe you can even find a closed form of $|Tx|_infty$ in terms of $|x|_infty$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for number one but how do I find the closed form of $Vert Tx Vert_infty$ in terms of $Vert x Vert_infty$? I'm new into Functional Analysis.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:13












  • $begingroup$
    Maybe first try to find $|Tx|_infty$ for $x=(t,0,0,...)$ in terms of $t$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 17:16











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.



Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I don't quite get the hint. Can you please, break it down?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:18










  • $begingroup$
    Your set $K$ consists of elements $x=(x_1,x_2,dots)$ such that $|x_i|le 1$ (see the definition of the norm on $l^infty$).
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:22
















1












$begingroup$

The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.



Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I don't quite get the hint. Can you please, break it down?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:18










  • $begingroup$
    Your set $K$ consists of elements $x=(x_1,x_2,dots)$ such that $|x_i|le 1$ (see the definition of the norm on $l^infty$).
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:22














1












1








1





$begingroup$

The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.



Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.






share|cite|improve this answer









$endgroup$



The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.



Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 16:53









BigbearZzzBigbearZzz

8,69121652




8,69121652












  • $begingroup$
    Sorry, I don't quite get the hint. Can you please, break it down?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:18










  • $begingroup$
    Your set $K$ consists of elements $x=(x_1,x_2,dots)$ such that $|x_i|le 1$ (see the definition of the norm on $l^infty$).
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:22


















  • $begingroup$
    Sorry, I don't quite get the hint. Can you please, break it down?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:18










  • $begingroup$
    Your set $K$ consists of elements $x=(x_1,x_2,dots)$ such that $|x_i|le 1$ (see the definition of the norm on $l^infty$).
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 17:22
















$begingroup$
Sorry, I don't quite get the hint. Can you please, break it down?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 17:18




$begingroup$
Sorry, I don't quite get the hint. Can you please, break it down?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 17:18












$begingroup$
Your set $K$ consists of elements $x=(x_1,x_2,dots)$ such that $|x_i|le 1$ (see the definition of the norm on $l^infty$).
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:22




$begingroup$
Your set $K$ consists of elements $x=(x_1,x_2,dots)$ such that $|x_i|le 1$ (see the definition of the norm on $l^infty$).
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:22











1












$begingroup$

Your answer for 1 is maybe not very well written, but it looks correct to me.



For 2 notice that $Tp=p=0$, so you just need to show $|Tx|_inftyleq|x|_infty$. Maybe you can even find a closed form of $|Tx|_infty$ in terms of $|x|_infty$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for number one but how do I find the closed form of $Vert Tx Vert_infty$ in terms of $Vert x Vert_infty$? I'm new into Functional Analysis.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:13












  • $begingroup$
    Maybe first try to find $|Tx|_infty$ for $x=(t,0,0,...)$ in terms of $t$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 17:16
















1












$begingroup$

Your answer for 1 is maybe not very well written, but it looks correct to me.



For 2 notice that $Tp=p=0$, so you just need to show $|Tx|_inftyleq|x|_infty$. Maybe you can even find a closed form of $|Tx|_infty$ in terms of $|x|_infty$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for number one but how do I find the closed form of $Vert Tx Vert_infty$ in terms of $Vert x Vert_infty$? I'm new into Functional Analysis.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:13












  • $begingroup$
    Maybe first try to find $|Tx|_infty$ for $x=(t,0,0,...)$ in terms of $t$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 17:16














1












1








1





$begingroup$

Your answer for 1 is maybe not very well written, but it looks correct to me.



For 2 notice that $Tp=p=0$, so you just need to show $|Tx|_inftyleq|x|_infty$. Maybe you can even find a closed form of $|Tx|_infty$ in terms of $|x|_infty$?






share|cite|improve this answer









$endgroup$



Your answer for 1 is maybe not very well written, but it looks correct to me.



For 2 notice that $Tp=p=0$, so you just need to show $|Tx|_inftyleq|x|_infty$. Maybe you can even find a closed form of $|Tx|_infty$ in terms of $|x|_infty$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 16:52









SmileyCraftSmileyCraft

3,591517




3,591517












  • $begingroup$
    Thanks for number one but how do I find the closed form of $Vert Tx Vert_infty$ in terms of $Vert x Vert_infty$? I'm new into Functional Analysis.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:13












  • $begingroup$
    Maybe first try to find $|Tx|_infty$ for $x=(t,0,0,...)$ in terms of $t$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 17:16


















  • $begingroup$
    Thanks for number one but how do I find the closed form of $Vert Tx Vert_infty$ in terms of $Vert x Vert_infty$? I'm new into Functional Analysis.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 17:13












  • $begingroup$
    Maybe first try to find $|Tx|_infty$ for $x=(t,0,0,...)$ in terms of $t$.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 17:16
















$begingroup$
Thanks for number one but how do I find the closed form of $Vert Tx Vert_infty$ in terms of $Vert x Vert_infty$? I'm new into Functional Analysis.
$endgroup$
– Omojola Micheal
Dec 14 '18 at 17:13






$begingroup$
Thanks for number one but how do I find the closed form of $Vert Tx Vert_infty$ in terms of $Vert x Vert_infty$? I'm new into Functional Analysis.
$endgroup$
– Omojola Micheal
Dec 14 '18 at 17:13














$begingroup$
Maybe first try to find $|Tx|_infty$ for $x=(t,0,0,...)$ in terms of $t$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:16




$begingroup$
Maybe first try to find $|Tx|_infty$ for $x=(t,0,0,...)$ in terms of $t$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:16


















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