Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ is non-expansive












2












$begingroup$


Let $X$ be a normed linear space and $X=c_0$ (the space of sequences of real numbers which converge to $0$). Define
begin{align} T:,&c_0to c_0 \ &xmapsto T(x)=(1,x_1,x_2,cdots), end{align}
for arbitrary $x=(1,x_1,x_2,cdots)in c_0.$ I want to show that $T$ is non-expansive.



Remark: We have that



$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$



MY TRIAL



Let $x,yin c_0.$ WTS: there exists $0leq kleq 1,$ such that $Vert T(x)-T(y) Vert leq k Vert x-y Vert . $



Now,
begin{align} Vert T(x)-T(y) Vert &=Vert (1,x_1,x_2,cdots)-(1,y_1,y_2,cdots) Vert\&=Vert ( 0,x_i-y_i)Vert\&= suplimits_{iin Bbb{N}} {0,|x_i-y_i|}\&= suplimits_{iin Bbb{N}} {|x_i-y_i|}\&=Vert x-y Vert end{align}
Taking $k=1,$ (based on mathworker21's advice and BigbearZzz's help), then



begin{align} Vert T(x)-T(y) Vert leq k Vert x-y Vert . end{align}
I'll be interested to see if you have a different kind of proof.






functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    in your 'WTS', it's equivalent to take $k=1$. also do you know what the norm $||cdot||$ is here? this problem is pretty trivial
    $endgroup$
    – mathworker21
    Dec 14 '18 at 8:16








  • 1




    $begingroup$
    Yes, your proof is correct.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 8:21










  • $begingroup$
    @Kavi Rama Murthy: Thank you very much Sir.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:24
















2












$begingroup$


Let $X$ be a normed linear space and $X=c_0$ (the space of sequences of real numbers which converge to $0$). Define
begin{align} T:,&c_0to c_0 \ &xmapsto T(x)=(1,x_1,x_2,cdots), end{align}
for arbitrary $x=(1,x_1,x_2,cdots)in c_0.$ I want to show that $T$ is non-expansive.



Remark: We have that



$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$



MY TRIAL



Let $x,yin c_0.$ WTS: there exists $0leq kleq 1,$ such that $Vert T(x)-T(y) Vert leq k Vert x-y Vert . $



Now,
begin{align} Vert T(x)-T(y) Vert &=Vert (1,x_1,x_2,cdots)-(1,y_1,y_2,cdots) Vert\&=Vert ( 0,x_i-y_i)Vert\&= suplimits_{iin Bbb{N}} {0,|x_i-y_i|}\&= suplimits_{iin Bbb{N}} {|x_i-y_i|}\&=Vert x-y Vert end{align}
Taking $k=1,$ (based on mathworker21's advice and BigbearZzz's help), then



begin{align} Vert T(x)-T(y) Vert leq k Vert x-y Vert . end{align}
I'll be interested to see if you have a different kind of proof.






functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    in your 'WTS', it's equivalent to take $k=1$. also do you know what the norm $||cdot||$ is here? this problem is pretty trivial
    $endgroup$
    – mathworker21
    Dec 14 '18 at 8:16








  • 1




    $begingroup$
    Yes, your proof is correct.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 8:21










  • $begingroup$
    @Kavi Rama Murthy: Thank you very much Sir.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:24














2












2








2


1



$begingroup$


Let $X$ be a normed linear space and $X=c_0$ (the space of sequences of real numbers which converge to $0$). Define
begin{align} T:,&c_0to c_0 \ &xmapsto T(x)=(1,x_1,x_2,cdots), end{align}
for arbitrary $x=(1,x_1,x_2,cdots)in c_0.$ I want to show that $T$ is non-expansive.



Remark: We have that



$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$



MY TRIAL



Let $x,yin c_0.$ WTS: there exists $0leq kleq 1,$ such that $Vert T(x)-T(y) Vert leq k Vert x-y Vert . $



Now,
begin{align} Vert T(x)-T(y) Vert &=Vert (1,x_1,x_2,cdots)-(1,y_1,y_2,cdots) Vert\&=Vert ( 0,x_i-y_i)Vert\&= suplimits_{iin Bbb{N}} {0,|x_i-y_i|}\&= suplimits_{iin Bbb{N}} {|x_i-y_i|}\&=Vert x-y Vert end{align}
Taking $k=1,$ (based on mathworker21's advice and BigbearZzz's help), then



begin{align} Vert T(x)-T(y) Vert leq k Vert x-y Vert . end{align}
I'll be interested to see if you have a different kind of proof.






functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$




Let $X$ be a normed linear space and $X=c_0$ (the space of sequences of real numbers which converge to $0$). Define
begin{align} T:,&c_0to c_0 \ &xmapsto T(x)=(1,x_1,x_2,cdots), end{align}
for arbitrary $x=(1,x_1,x_2,cdots)in c_0.$ I want to show that $T$ is non-expansive.



Remark: We have that



$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$



MY TRIAL



Let $x,yin c_0.$ WTS: there exists $0leq kleq 1,$ such that $Vert T(x)-T(y) Vert leq k Vert x-y Vert . $



Now,
begin{align} Vert T(x)-T(y) Vert &=Vert (1,x_1,x_2,cdots)-(1,y_1,y_2,cdots) Vert\&=Vert ( 0,x_i-y_i)Vert\&= suplimits_{iin Bbb{N}} {0,|x_i-y_i|}\&= suplimits_{iin Bbb{N}} {|x_i-y_i|}\&=Vert x-y Vert end{align}
Taking $k=1,$ (based on mathworker21's advice and BigbearZzz's help), then



begin{align} Vert T(x)-T(y) Vert leq k Vert x-y Vert . end{align}
I'll be interested to see if you have a different kind of proof.






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 8:51







Omojola Micheal

















asked Dec 14 '18 at 8:12









Omojola MichealOmojola Micheal

1,920324




1,920324








  • 1




    $begingroup$
    in your 'WTS', it's equivalent to take $k=1$. also do you know what the norm $||cdot||$ is here? this problem is pretty trivial
    $endgroup$
    – mathworker21
    Dec 14 '18 at 8:16








  • 1




    $begingroup$
    Yes, your proof is correct.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 8:21










  • $begingroup$
    @Kavi Rama Murthy: Thank you very much Sir.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:24














  • 1




    $begingroup$
    in your 'WTS', it's equivalent to take $k=1$. also do you know what the norm $||cdot||$ is here? this problem is pretty trivial
    $endgroup$
    – mathworker21
    Dec 14 '18 at 8:16








  • 1




    $begingroup$
    Yes, your proof is correct.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 8:21










  • $begingroup$
    @Kavi Rama Murthy: Thank you very much Sir.
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:24








1




1




$begingroup$
in your 'WTS', it's equivalent to take $k=1$. also do you know what the norm $||cdot||$ is here? this problem is pretty trivial
$endgroup$
– mathworker21
Dec 14 '18 at 8:16






$begingroup$
in your 'WTS', it's equivalent to take $k=1$. also do you know what the norm $||cdot||$ is here? this problem is pretty trivial
$endgroup$
– mathworker21
Dec 14 '18 at 8:16






1




1




$begingroup$
Yes, your proof is correct.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:21




$begingroup$
Yes, your proof is correct.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:21












$begingroup$
@Kavi Rama Murthy: Thank you very much Sir.
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:24




$begingroup$
@Kavi Rama Murthy: Thank you very much Sir.
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:24










1 Answer
1






active

oldest

votes


















1












$begingroup$

You've computed that
$$begin{align}
||Tx-Ty|| &= sup {0,|x_1-y_1|,|x_2-y_2|,dots},
end{align}$$
whereas we know that
$$
||x-y|| = sup {|x_1-y_1|,|x_2-y_2|,dots}.
$$
All you need to know to solve the problem is that norm for $c_0$ is the supremum norm.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do we need to have $sup?$, seeing that $c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:23










  • $begingroup$
    @Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:24










  • $begingroup$
    The norm on $c_0$ is defined to be $||x|| = sup_{iinBbb N} |x_i|$.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:25










  • $begingroup$
    Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:26










  • $begingroup$
    Thanks about the norm info. That is an addition. Thanks!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:27











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You've computed that
$$begin{align}
||Tx-Ty|| &= sup {0,|x_1-y_1|,|x_2-y_2|,dots},
end{align}$$
whereas we know that
$$
||x-y|| = sup {|x_1-y_1|,|x_2-y_2|,dots}.
$$
All you need to know to solve the problem is that norm for $c_0$ is the supremum norm.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do we need to have $sup?$, seeing that $c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:23










  • $begingroup$
    @Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:24










  • $begingroup$
    The norm on $c_0$ is defined to be $||x|| = sup_{iinBbb N} |x_i|$.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:25










  • $begingroup$
    Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:26










  • $begingroup$
    Thanks about the norm info. That is an addition. Thanks!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:27
















1












$begingroup$

You've computed that
$$begin{align}
||Tx-Ty|| &= sup {0,|x_1-y_1|,|x_2-y_2|,dots},
end{align}$$
whereas we know that
$$
||x-y|| = sup {|x_1-y_1|,|x_2-y_2|,dots}.
$$
All you need to know to solve the problem is that norm for $c_0$ is the supremum norm.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do we need to have $sup?$, seeing that $c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:23










  • $begingroup$
    @Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:24










  • $begingroup$
    The norm on $c_0$ is defined to be $||x|| = sup_{iinBbb N} |x_i|$.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:25










  • $begingroup$
    Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:26










  • $begingroup$
    Thanks about the norm info. That is an addition. Thanks!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:27














1












1








1





$begingroup$

You've computed that
$$begin{align}
||Tx-Ty|| &= sup {0,|x_1-y_1|,|x_2-y_2|,dots},
end{align}$$
whereas we know that
$$
||x-y|| = sup {|x_1-y_1|,|x_2-y_2|,dots}.
$$
All you need to know to solve the problem is that norm for $c_0$ is the supremum norm.






share|cite|improve this answer









$endgroup$



You've computed that
$$begin{align}
||Tx-Ty|| &= sup {0,|x_1-y_1|,|x_2-y_2|,dots},
end{align}$$
whereas we know that
$$
||x-y|| = sup {|x_1-y_1|,|x_2-y_2|,dots}.
$$
All you need to know to solve the problem is that norm for $c_0$ is the supremum norm.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 8:20









BigbearZzzBigbearZzz

8,88821652




8,88821652












  • $begingroup$
    Do we need to have $sup?$, seeing that $c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:23










  • $begingroup$
    @Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:24










  • $begingroup$
    The norm on $c_0$ is defined to be $||x|| = sup_{iinBbb N} |x_i|$.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:25










  • $begingroup$
    Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:26










  • $begingroup$
    Thanks about the norm info. That is an addition. Thanks!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:27


















  • $begingroup$
    Do we need to have $sup?$, seeing that $c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:23










  • $begingroup$
    @Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:24










  • $begingroup$
    The norm on $c_0$ is defined to be $||x|| = sup_{iinBbb N} |x_i|$.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:25










  • $begingroup$
    Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct?
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:26










  • $begingroup$
    Thanks about the norm info. That is an addition. Thanks!
    $endgroup$
    – Omojola Micheal
    Dec 14 '18 at 8:27
















$begingroup$
Do we need to have $sup?$, seeing that $c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:23




$begingroup$
Do we need to have $sup?$, seeing that $c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:23












$begingroup$
@Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:24




$begingroup$
@Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:24












$begingroup$
The norm on $c_0$ is defined to be $||x|| = sup_{iinBbb N} |x_i|$.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:25




$begingroup$
The norm on $c_0$ is defined to be $||x|| = sup_{iinBbb N} |x_i|$.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:25












$begingroup$
Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:26




$begingroup$
Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:26












$begingroup$
Thanks about the norm info. That is an addition. Thanks!
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:27




$begingroup$
Thanks about the norm info. That is an addition. Thanks!
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:27


















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