Show that $Vert Tx-Tp Vert leq Vert x-p Vert $, where $p=0.$
$begingroup$
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
$Tp=p$ if and only if $p=0;$
$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}
- 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
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
$endgroup$
add a comment |
$begingroup$
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
$Tp=p$ if and only if $p=0;$
$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}
- 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
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$Tp=p$ if and only if $p=0;$
$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}
- 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
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
$endgroup$
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
$Tp=p$ if and only if $p=0;$
$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}
- 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
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
edited Dec 14 '18 at 17:16
Omojola Micheal
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
asked Dec 14 '18 at 16:46
Omojola MichealOmojola Micheal
1,875324
1,875324
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.
Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $|x|_infty$?
answered Dec 14 '18 at 16:52
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.
Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.
Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.
Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
$endgroup$
The norm on $l^infty$ is $||x||:=sup_{ninBbb N} |x_n|$.
Hint: If $|lambda|le 1$, then $|lambda|^2le |lambda|$.
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
answered Dec 14 '18 at 16:53
BigbearZzzBigbearZzz
8,75121652
8,75121652
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $|x|_infty$?
answered Dec 14 '18 at 16:52
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $|x|_infty$?
answered Dec 14 '18 at 16:52
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $|x|_infty$?
answered Dec 14 '18 at 16:52
SmileyCraftSmileyCraft
3,591517
$endgroup$
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 $|x|_infty$?
answered Dec 14 '18 at 16:52
SmileyCraftSmileyCraft
3,591517
answered Dec 14 '18 at 16:52
SmileyCraftSmileyCraft
3,591517
answered Dec 14 '18 at 16:52
SmileyCraftSmileyCraft
3,591517
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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