Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ has no fixed points
$begingroup$
As a follow-up to my previous question Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ is non-expansive. 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 $T$ has no fixed points.
Remark: We have that
$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$
MY TRIAL
Suppose for contradiction that $T$ has fixed points in $X$, then there exists $uin c_0$ s.t. $T(u)=u.$ That is,
begin{align} T(u_1,u_2,cdots)=(u_1,u_2,cdots), end{align}
Please, how do I draw out a contradiction from this?
functional-analysis analysis normed-spaces fixed-point-theorems
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
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add a comment |
$begingroup$
As a follow-up to my previous question Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ is non-expansive. 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 $T$ has no fixed points.
Remark: We have that
$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$
MY TRIAL
Suppose for contradiction that $T$ has fixed points in $X$, then there exists $uin c_0$ s.t. $T(u)=u.$ That is,
begin{align} T(u_1,u_2,cdots)=(u_1,u_2,cdots), end{align}
Please, how do I draw out a contradiction from this?
functional-analysis analysis normed-spaces fixed-point-theorems
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
$endgroup$
add a comment |
$begingroup$
As a follow-up to my previous question Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ is non-expansive. 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 $T$ has no fixed points.
Remark: We have that
$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$
MY TRIAL
Suppose for contradiction that $T$ has fixed points in $X$, then there exists $uin c_0$ s.t. $T(u)=u.$ That is,
begin{align} T(u_1,u_2,cdots)=(u_1,u_2,cdots), end{align}
Please, how do I draw out a contradiction from this?
functional-analysis analysis normed-spaces fixed-point-theorems
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
$endgroup$
As a follow-up to my previous question Show that $T:,c_0to c_0;;$, $xmapsto T(x)=(1,x_1,x_2,cdots),$ is non-expansive. 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 $T$ has no fixed points.
Remark: We have that
$$c_0={bar{x}=(x_1,x_2,cdots) :x_nto 0;text{as};nto infty}.$$
MY TRIAL
Suppose for contradiction that $T$ has fixed points in $X$, then there exists $uin c_0$ s.t. $T(u)=u.$ That is,
begin{align} T(u_1,u_2,cdots)=(u_1,u_2,cdots), end{align}
Please, how do I draw out a contradiction from this?
functional-analysis analysis normed-spaces fixed-point-theorems
functional-analysis analysis normed-spaces fixed-point-theorems
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
asked Dec 14 '18 at 8:55
Omojola MichealOmojola Micheal
1,875324
1,875324
add a comment |
add a comment |
1 Answer
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$(1,u_1,u_2,...)=(u_1,u_2,..)$ implies $1=u_1,u_1=u_2$ etc so $u_n=1$ for all $n$. But then $(u_n) notin c_0$.
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
$begingroup$
Sorry Sir, how's $u_nnotin c_0$?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:58
$begingroup$
Let me guess, it's because $u_n$ does not go to zero and $ntoinfty.$ Therefore, $u_nnotin c_o. $ I'm I right?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 9:01
$begingroup$
Yes, $u_n to 1$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:02
add a comment |
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1 Answer
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$begingroup$
$(1,u_1,u_2,...)=(u_1,u_2,..)$ implies $1=u_1,u_1=u_2$ etc so $u_n=1$ for all $n$. But then $(u_n) notin c_0$.
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
$begingroup$
Sorry Sir, how's $u_nnotin c_0$?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:58
$begingroup$
Let me guess, it's because $u_n$ does not go to zero and $ntoinfty.$ Therefore, $u_nnotin c_o. $ I'm I right?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 9:01
$begingroup$
Yes, $u_n to 1$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:02
add a comment |
$begingroup$
$(1,u_1,u_2,...)=(u_1,u_2,..)$ implies $1=u_1,u_1=u_2$ etc so $u_n=1$ for all $n$. But then $(u_n) notin c_0$.
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
$begingroup$
Sorry Sir, how's $u_nnotin c_0$?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:58
$begingroup$
Let me guess, it's because $u_n$ does not go to zero and $ntoinfty.$ Therefore, $u_nnotin c_o. $ I'm I right?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 9:01
$begingroup$
Yes, $u_n to 1$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:02
add a comment |
$begingroup$
$(1,u_1,u_2,...)=(u_1,u_2,..)$ implies $1=u_1,u_1=u_2$ etc so $u_n=1$ for all $n$. But then $(u_n) notin c_0$.
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
$(1,u_1,u_2,...)=(u_1,u_2,..)$ implies $1=u_1,u_1=u_2$ etc so $u_n=1$ for all $n$. But then $(u_n) notin c_0$.
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 8:57
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
$begingroup$
Sorry Sir, how's $u_nnotin c_0$?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:58
$begingroup$
Let me guess, it's because $u_n$ does not go to zero and $ntoinfty.$ Therefore, $u_nnotin c_o. $ I'm I right?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 9:01
$begingroup$
Yes, $u_n to 1$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:02
add a comment |
$begingroup$
Sorry Sir, how's $u_nnotin c_0$?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:58
$begingroup$
Let me guess, it's because $u_n$ does not go to zero and $ntoinfty.$ Therefore, $u_nnotin c_o. $ I'm I right?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 9:01
$begingroup$
Yes, $u_n to 1$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:02
$begingroup$
Sorry Sir, how's $u_nnotin c_0$?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:58
$begingroup$
Sorry Sir, how's $u_nnotin c_0$?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 8:58
$begingroup$
Let me guess, it's because $u_n$ does not go to zero and $ntoinfty.$ Therefore, $u_nnotin c_o. $ I'm I right?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 9:01
$begingroup$
Let me guess, it's because $u_n$ does not go to zero and $ntoinfty.$ Therefore, $u_nnotin c_o. $ I'm I right?
$endgroup$
– Omojola Micheal
Dec 14 '18 at 9:01
$begingroup$
Yes, $u_n to 1$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:02
$begingroup$
Yes, $u_n to 1$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:02
add a comment |
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