Dual norm of a functional defined in $c_0$
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I'd like a check in the following exercise
In $(c_0, Vert cdot Vert_{infty}) $ consider for every element $u=(u_1,u_2,u_3,ldots)$ in $c_0$ the functional $T(u)=sum_{n=1}^{infty}frac{1}{2^n}u_n$.
Check that $T$ is a continuous linear functional and compute $Vert T Vert$ (the dual norm). Can one find some $u in c_0$ s.t $Vert u Vert_{infty}=1$ and $T(u)=Vert T Vert$ ?
The first part is clear and I'v checked it also here. My problem is on how to formalize the last part: I strongly believe that there's no $u in c_0$ with that property, but I don't know how to move properly.
From the boundedness one can see that $Vert T Vert leq 1$, and choosing the "cut-off" sequence $u=(underbrace{1,...,1}_{N text{times}},0,0,ldots)$ I have that $sum_{n=1}^{infty} frac{u_{n}}{2^n}=1-2^{-N}$, and then the supremum over $N$ is 1, and than follows that $Vert T Vert =1$.
But how can I say that there is no one $u$ with that property? I would say that the only way for which $T(u)$ is equal to $1 (=Vert T Vert)$ is the cut-off sequence is made from all entries equal to $1$, which is a contradiction, since $u$ must belong to $c_0$. Is this enough?
sequences-and-series functional-analysis normed-spaces
asked Dec 10 '18 at 18:30
VoBVoB
734513
$endgroup$
add a comment |
$begingroup$
I'd like a check in the following exercise
In $(c_0, Vert cdot Vert_{infty}) $ consider for every element $u=(u_1,u_2,u_3,ldots)$ in $c_0$ the functional $T(u)=sum_{n=1}^{infty}frac{1}{2^n}u_n$.
Check that $T$ is a continuous linear functional and compute $Vert T Vert$ (the dual norm). Can one find some $u in c_0$ s.t $Vert u Vert_{infty}=1$ and $T(u)=Vert T Vert$ ?
The first part is clear and I'v checked it also here. My problem is on how to formalize the last part: I strongly believe that there's no $u in c_0$ with that property, but I don't know how to move properly.
From the boundedness one can see that $Vert T Vert leq 1$, and choosing the "cut-off" sequence $u=(underbrace{1,...,1}_{N text{times}},0,0,ldots)$ I have that $sum_{n=1}^{infty} frac{u_{n}}{2^n}=1-2^{-N}$, and then the supremum over $N$ is 1, and than follows that $Vert T Vert =1$.
But how can I say that there is no one $u$ with that property? I would say that the only way for which $T(u)$ is equal to $1 (=Vert T Vert)$ is the cut-off sequence is made from all entries equal to $1$, which is a contradiction, since $u$ must belong to $c_0$. Is this enough?
sequences-and-series functional-analysis normed-spaces
asked Dec 10 '18 at 18:30
VoBVoB
734513
$endgroup$
$begingroup$
What is $f(u)$?
$endgroup$
– user25959
Dec 10 '18 at 18:35
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Fixed, thanks !
$endgroup$
– VoB
Dec 10 '18 at 18:45
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Note that $T$ is defined and continuous with the same norm on $l_infty$. If $|x| =1$ and $|Tx| = 1$ then we must have $|x_k| = 1$ for all $k$. Hence if $x in c_0$ we must have $|Tx| < 1$.
$endgroup$
– copper.hat
Dec 10 '18 at 19:09
$begingroup$
@copper.hat I was looking for an explicit inequality, but thanks anyway !
$endgroup$
– VoB
Dec 10 '18 at 19:32
add a comment |
$begingroup$
I'd like a check in the following exercise
In $(c_0, Vert cdot Vert_{infty}) $ consider for every element $u=(u_1,u_2,u_3,ldots)$ in $c_0$ the functional $T(u)=sum_{n=1}^{infty}frac{1}{2^n}u_n$.
Check that $T$ is a continuous linear functional and compute $Vert T Vert$ (the dual norm). Can one find some $u in c_0$ s.t $Vert u Vert_{infty}=1$ and $T(u)=Vert T Vert$ ?
The first part is clear and I'v checked it also here. My problem is on how to formalize the last part: I strongly believe that there's no $u in c_0$ with that property, but I don't know how to move properly.
From the boundedness one can see that $Vert T Vert leq 1$, and choosing the "cut-off" sequence $u=(underbrace{1,...,1}_{N text{times}},0,0,ldots)$ I have that $sum_{n=1}^{infty} frac{u_{n}}{2^n}=1-2^{-N}$, and then the supremum over $N$ is 1, and than follows that $Vert T Vert =1$.
But how can I say that there is no one $u$ with that property? I would say that the only way for which $T(u)$ is equal to $1 (=Vert T Vert)$ is the cut-off sequence is made from all entries equal to $1$, which is a contradiction, since $u$ must belong to $c_0$. Is this enough?
sequences-and-series functional-analysis normed-spaces
asked Dec 10 '18 at 18:30
VoBVoB
734513
$endgroup$
I'd like a check in the following exercise
In $(c_0, Vert cdot Vert_{infty}) $ consider for every element $u=(u_1,u_2,u_3,ldots)$ in $c_0$ the functional $T(u)=sum_{n=1}^{infty}frac{1}{2^n}u_n$.
Check that $T$ is a continuous linear functional and compute $Vert T Vert$ (the dual norm). Can one find some $u in c_0$ s.t $Vert u Vert_{infty}=1$ and $T(u)=Vert T Vert$ ?
The first part is clear and I'v checked it also here. My problem is on how to formalize the last part: I strongly believe that there's no $u in c_0$ with that property, but I don't know how to move properly.
From the boundedness one can see that $Vert T Vert leq 1$, and choosing the "cut-off" sequence $u=(underbrace{1,...,1}_{N text{times}},0,0,ldots)$ I have that $sum_{n=1}^{infty} frac{u_{n}}{2^n}=1-2^{-N}$, and then the supremum over $N$ is 1, and than follows that $Vert T Vert =1$.
But how can I say that there is no one $u$ with that property? I would say that the only way for which $T(u)$ is equal to $1 (=Vert T Vert)$ is the cut-off sequence is made from all entries equal to $1$, which is a contradiction, since $u$ must belong to $c_0$. Is this enough?
sequences-and-series functional-analysis normed-spaces
sequences-and-series functional-analysis normed-spaces
asked Dec 10 '18 at 18:30
VoBVoB
734513
asked Dec 10 '18 at 18:30
VoBVoB
734513
edited Dec 10 '18 at 18:46
VoB
asked Dec 10 '18 at 18:30
VoBVoB
734513
asked Dec 10 '18 at 18:30
VoBVoB
734513
asked Dec 10 '18 at 18:30
VoBVoB
734513
734513
$begingroup$
What is $f(u)$?
$endgroup$
– user25959
Dec 10 '18 at 18:35
$begingroup$
Fixed, thanks !
$endgroup$
– VoB
Dec 10 '18 at 18:45
$begingroup$
Note that $T$ is defined and continuous with the same norm on $l_infty$. If $|x| =1$ and $|Tx| = 1$ then we must have $|x_k| = 1$ for all $k$. Hence if $x in c_0$ we must have $|Tx| < 1$.
$endgroup$
– copper.hat
Dec 10 '18 at 19:09
$begingroup$
@copper.hat I was looking for an explicit inequality, but thanks anyway !
$endgroup$
– VoB
Dec 10 '18 at 19:32
add a comment |
$begingroup$
What is $f(u)$?
$endgroup$
– user25959
Dec 10 '18 at 18:35
$begingroup$
Fixed, thanks !
$endgroup$
– VoB
Dec 10 '18 at 18:45
$begingroup$
Note that $T$ is defined and continuous with the same norm on $l_infty$. If $|x| =1$ and $|Tx| = 1$ then we must have $|x_k| = 1$ for all $k$. Hence if $x in c_0$ we must have $|Tx| < 1$.
$endgroup$
– copper.hat
Dec 10 '18 at 19:09
$begingroup$
@copper.hat I was looking for an explicit inequality, but thanks anyway !
$endgroup$
– VoB
Dec 10 '18 at 19:32
$begingroup$
What is $f(u)$?
$endgroup$
– user25959
Dec 10 '18 at 18:35
$begingroup$
What is $f(u)$?
$endgroup$
– user25959
Dec 10 '18 at 18:35
$begingroup$
Fixed, thanks !
$endgroup$
– VoB
Dec 10 '18 at 18:45
$begingroup$
Fixed, thanks !
$endgroup$
– VoB
Dec 10 '18 at 18:45
$begingroup$
Note that $T$ is defined and continuous with the same norm on $l_infty$. If $|x| =1$ and $|Tx| = 1$ then we must have $|x_k| = 1$ for all $k$. Hence if $x in c_0$ we must have $|Tx| < 1$.
$endgroup$
– copper.hat
Dec 10 '18 at 19:09
$begingroup$
Note that $T$ is defined and continuous with the same norm on $l_infty$. If $|x| =1$ and $|Tx| = 1$ then we must have $|x_k| = 1$ for all $k$. Hence if $x in c_0$ we must have $|Tx| < 1$.
$endgroup$
– copper.hat
Dec 10 '18 at 19:09
$begingroup$
@copper.hat I was looking for an explicit inequality, but thanks anyway !
$endgroup$
– VoB
Dec 10 '18 at 19:32
$begingroup$
@copper.hat I was looking for an explicit inequality, but thanks anyway !
$endgroup$
– VoB
Dec 10 '18 at 19:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
While you don't yet have a "proof" the idea is all there:
Suppose for contradiction that $T(u)=1$. Take any $epsilon<1$. By definition of $c_0$ there is some $N$ such that $|u_n| < epsilon$ for all $ngeq N$.
Then $|T(u)| leq left(sumlimits_{n=1}^{N} + sumlimits_{n=N+1}^{infty}right)| frac{1}{2^n}u_n| leq (1-2^{-N}) + epsilon2^{-N}<1$, a contradiction.
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
$endgroup$
$begingroup$
Thanks, I was searching for a contradiction, and this makes the point.
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Your proof uses, in the second inequality, also the fact that $Vert u Vert_{infty}=1$, right? (and of course that $|u_n|<epsilon $ for $n geq N$)
$endgroup$
– VoB
Dec 10 '18 at 19:48
$begingroup$
Yes, first inequality is triangle inequality, second inequality is as you said.
$endgroup$
– user25959
Dec 10 '18 at 19:53
add a comment |
$begingroup$
Since
$ Vert uVert_infty = 1,$ we must have $|u_n|leq 1$ for all $n$.
Evidently, we cannot have $u_n = 1$ for all $n$, since in that case $u notin c_0$. Hence $|u_{n_0}| < 1$ for some $n_0$. Thus
$$ left| sum_{n=1}^infty frac{u_n}{2^n} right| = sum_{n=1}^infty frac{|u_n|}{2^n} = sum_{substack{ngeq 1\nnot=n_0}} frac{|u_n|}{2^n}+frac{|u_{n_0}|}{2^{n_0}} < sum_{substack{ngeq 1\nnot=n_0}} frac{1}{2^n} + frac{1}{2^{n_0}} = sum_{n=1}^infty frac{1}{2^n}=1, $$
for any $u in c_0$, so in particular equality cannot hold since the inequality is strict.
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
$endgroup$
$begingroup$
It's quite clear, but I can't understand the second-last inequality: why does $frac{1}{2^{n0}}$ disappear?
$endgroup$
– VoB
Dec 10 '18 at 19:27
$begingroup$
Ok, it's clear, you summed up for all $n in mathbb{N}$, now I've seen ! Thanks
$endgroup$
– VoB
Dec 10 '18 at 19:31
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
While you don't yet have a "proof" the idea is all there:
Suppose for contradiction that $T(u)=1$. Take any $epsilon<1$. By definition of $c_0$ there is some $N$ such that $|u_n| < epsilon$ for all $ngeq N$.
Then $|T(u)| leq left(sumlimits_{n=1}^{N} + sumlimits_{n=N+1}^{infty}right)| frac{1}{2^n}u_n| leq (1-2^{-N}) + epsilon2^{-N}<1$, a contradiction.
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
$endgroup$
$begingroup$
Thanks, I was searching for a contradiction, and this makes the point.
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Your proof uses, in the second inequality, also the fact that $Vert u Vert_{infty}=1$, right? (and of course that $|u_n|<epsilon $ for $n geq N$)
$endgroup$
– VoB
Dec 10 '18 at 19:48
$begingroup$
Yes, first inequality is triangle inequality, second inequality is as you said.
$endgroup$
– user25959
Dec 10 '18 at 19:53
add a comment |
$begingroup$
While you don't yet have a "proof" the idea is all there:
Suppose for contradiction that $T(u)=1$. Take any $epsilon<1$. By definition of $c_0$ there is some $N$ such that $|u_n| < epsilon$ for all $ngeq N$.
Then $|T(u)| leq left(sumlimits_{n=1}^{N} + sumlimits_{n=N+1}^{infty}right)| frac{1}{2^n}u_n| leq (1-2^{-N}) + epsilon2^{-N}<1$, a contradiction.
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
$endgroup$
$begingroup$
Thanks, I was searching for a contradiction, and this makes the point.
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Your proof uses, in the second inequality, also the fact that $Vert u Vert_{infty}=1$, right? (and of course that $|u_n|<epsilon $ for $n geq N$)
$endgroup$
– VoB
Dec 10 '18 at 19:48
$begingroup$
Yes, first inequality is triangle inequality, second inequality is as you said.
$endgroup$
– user25959
Dec 10 '18 at 19:53
add a comment |
$begingroup$
While you don't yet have a "proof" the idea is all there:
Suppose for contradiction that $T(u)=1$. Take any $epsilon<1$. By definition of $c_0$ there is some $N$ such that $|u_n| < epsilon$ for all $ngeq N$.
Then $|T(u)| leq left(sumlimits_{n=1}^{N} + sumlimits_{n=N+1}^{infty}right)| frac{1}{2^n}u_n| leq (1-2^{-N}) + epsilon2^{-N}<1$, a contradiction.
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
$endgroup$
While you don't yet have a "proof" the idea is all there:
Suppose for contradiction that $T(u)=1$. Take any $epsilon<1$. By definition of $c_0$ there is some $N$ such that $|u_n| < epsilon$ for all $ngeq N$.
Then $|T(u)| leq left(sumlimits_{n=1}^{N} + sumlimits_{n=N+1}^{infty}right)| frac{1}{2^n}u_n| leq (1-2^{-N}) + epsilon2^{-N}<1$, a contradiction.
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
answered Dec 10 '18 at 18:45
user25959user25959
1,573816
1,573816
$begingroup$
Thanks, I was searching for a contradiction, and this makes the point.
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Your proof uses, in the second inequality, also the fact that $Vert u Vert_{infty}=1$, right? (and of course that $|u_n|<epsilon $ for $n geq N$)
$endgroup$
– VoB
Dec 10 '18 at 19:48
$begingroup$
Yes, first inequality is triangle inequality, second inequality is as you said.
$endgroup$
– user25959
Dec 10 '18 at 19:53
add a comment |
$begingroup$
Thanks, I was searching for a contradiction, and this makes the point.
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Your proof uses, in the second inequality, also the fact that $Vert u Vert_{infty}=1$, right? (and of course that $|u_n|<epsilon $ for $n geq N$)
$endgroup$
– VoB
Dec 10 '18 at 19:48
$begingroup$
Yes, first inequality is triangle inequality, second inequality is as you said.
$endgroup$
– user25959
Dec 10 '18 at 19:53
$begingroup$
Thanks, I was searching for a contradiction, and this makes the point.
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Thanks, I was searching for a contradiction, and this makes the point.
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Your proof uses, in the second inequality, also the fact that $Vert u Vert_{infty}=1$, right? (and of course that $|u_n|<epsilon $ for $n geq N$)
$endgroup$
– VoB
Dec 10 '18 at 19:48
$begingroup$
Your proof uses, in the second inequality, also the fact that $Vert u Vert_{infty}=1$, right? (and of course that $|u_n|<epsilon $ for $n geq N$)
$endgroup$
– VoB
Dec 10 '18 at 19:48
$begingroup$
Yes, first inequality is triangle inequality, second inequality is as you said.
$endgroup$
– user25959
Dec 10 '18 at 19:53
$begingroup$
Yes, first inequality is triangle inequality, second inequality is as you said.
$endgroup$
– user25959
Dec 10 '18 at 19:53
add a comment |
$begingroup$
Since
$ Vert uVert_infty = 1,$ we must have $|u_n|leq 1$ for all $n$.
Evidently, we cannot have $u_n = 1$ for all $n$, since in that case $u notin c_0$. Hence $|u_{n_0}| < 1$ for some $n_0$. Thus
$$ left| sum_{n=1}^infty frac{u_n}{2^n} right| = sum_{n=1}^infty frac{|u_n|}{2^n} = sum_{substack{ngeq 1\nnot=n_0}} frac{|u_n|}{2^n}+frac{|u_{n_0}|}{2^{n_0}} < sum_{substack{ngeq 1\nnot=n_0}} frac{1}{2^n} + frac{1}{2^{n_0}} = sum_{n=1}^infty frac{1}{2^n}=1, $$
for any $u in c_0$, so in particular equality cannot hold since the inequality is strict.
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
$endgroup$
$begingroup$
It's quite clear, but I can't understand the second-last inequality: why does $frac{1}{2^{n0}}$ disappear?
$endgroup$
– VoB
Dec 10 '18 at 19:27
$begingroup$
Ok, it's clear, you summed up for all $n in mathbb{N}$, now I've seen ! Thanks
$endgroup$
– VoB
Dec 10 '18 at 19:31
add a comment |
$begingroup$
Since
$ Vert uVert_infty = 1,$ we must have $|u_n|leq 1$ for all $n$.
Evidently, we cannot have $u_n = 1$ for all $n$, since in that case $u notin c_0$. Hence $|u_{n_0}| < 1$ for some $n_0$. Thus
$$ left| sum_{n=1}^infty frac{u_n}{2^n} right| = sum_{n=1}^infty frac{|u_n|}{2^n} = sum_{substack{ngeq 1\nnot=n_0}} frac{|u_n|}{2^n}+frac{|u_{n_0}|}{2^{n_0}} < sum_{substack{ngeq 1\nnot=n_0}} frac{1}{2^n} + frac{1}{2^{n_0}} = sum_{n=1}^infty frac{1}{2^n}=1, $$
for any $u in c_0$, so in particular equality cannot hold since the inequality is strict.
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
$endgroup$
$begingroup$
It's quite clear, but I can't understand the second-last inequality: why does $frac{1}{2^{n0}}$ disappear?
$endgroup$
– VoB
Dec 10 '18 at 19:27
$begingroup$
Ok, it's clear, you summed up for all $n in mathbb{N}$, now I've seen ! Thanks
$endgroup$
– VoB
Dec 10 '18 at 19:31
add a comment |
$begingroup$
Since
$ Vert uVert_infty = 1,$ we must have $|u_n|leq 1$ for all $n$.
Evidently, we cannot have $u_n = 1$ for all $n$, since in that case $u notin c_0$. Hence $|u_{n_0}| < 1$ for some $n_0$. Thus
$$ left| sum_{n=1}^infty frac{u_n}{2^n} right| = sum_{n=1}^infty frac{|u_n|}{2^n} = sum_{substack{ngeq 1\nnot=n_0}} frac{|u_n|}{2^n}+frac{|u_{n_0}|}{2^{n_0}} < sum_{substack{ngeq 1\nnot=n_0}} frac{1}{2^n} + frac{1}{2^{n_0}} = sum_{n=1}^infty frac{1}{2^n}=1, $$
for any $u in c_0$, so in particular equality cannot hold since the inequality is strict.
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
$endgroup$
Since
$ Vert uVert_infty = 1,$ we must have $|u_n|leq 1$ for all $n$.
Evidently, we cannot have $u_n = 1$ for all $n$, since in that case $u notin c_0$. Hence $|u_{n_0}| < 1$ for some $n_0$. Thus
$$ left| sum_{n=1}^infty frac{u_n}{2^n} right| = sum_{n=1}^infty frac{|u_n|}{2^n} = sum_{substack{ngeq 1\nnot=n_0}} frac{|u_n|}{2^n}+frac{|u_{n_0}|}{2^{n_0}} < sum_{substack{ngeq 1\nnot=n_0}} frac{1}{2^n} + frac{1}{2^{n_0}} = sum_{n=1}^infty frac{1}{2^n}=1, $$
for any $u in c_0$, so in particular equality cannot hold since the inequality is strict.
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
answered Dec 10 '18 at 18:44
MisterRiemannMisterRiemann
5,8171625
5,8171625
$begingroup$
It's quite clear, but I can't understand the second-last inequality: why does $frac{1}{2^{n0}}$ disappear?
$endgroup$
– VoB
Dec 10 '18 at 19:27
$begingroup$
Ok, it's clear, you summed up for all $n in mathbb{N}$, now I've seen ! Thanks
$endgroup$
– VoB
Dec 10 '18 at 19:31
add a comment |
$begingroup$
It's quite clear, but I can't understand the second-last inequality: why does $frac{1}{2^{n0}}$ disappear?
$endgroup$
– VoB
Dec 10 '18 at 19:27
$begingroup$
Ok, it's clear, you summed up for all $n in mathbb{N}$, now I've seen ! Thanks
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
It's quite clear, but I can't understand the second-last inequality: why does $frac{1}{2^{n0}}$ disappear?
$endgroup$
– VoB
Dec 10 '18 at 19:27
$begingroup$
It's quite clear, but I can't understand the second-last inequality: why does $frac{1}{2^{n0}}$ disappear?
$endgroup$
– VoB
Dec 10 '18 at 19:27
$begingroup$
Ok, it's clear, you summed up for all $n in mathbb{N}$, now I've seen ! Thanks
$endgroup$
– VoB
Dec 10 '18 at 19:31
$begingroup$
Ok, it's clear, you summed up for all $n in mathbb{N}$, now I've seen ! Thanks
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– VoB
Dec 10 '18 at 19:31
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$begingroup$
What is $f(u)$?
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– user25959
Dec 10 '18 at 18:35
$begingroup$
Fixed, thanks !
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– VoB
Dec 10 '18 at 18:45
$begingroup$
Note that $T$ is defined and continuous with the same norm on $l_infty$. If $|x| =1$ and $|Tx| = 1$ then we must have $|x_k| = 1$ for all $k$. Hence if $x in c_0$ we must have $|Tx| < 1$.
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– copper.hat
Dec 10 '18 at 19:09
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@copper.hat I was looking for an explicit inequality, but thanks anyway !
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– VoB
Dec 10 '18 at 19:32