Mixing
Image unit ball under Isometric Isomorphism
$begingroup$
Let $X,Y$ be a normed spaces, not necessarily finitely dimensional and let $Tin B(X,Y)$ be an isometric isomorphism. I want to show the same holds for the dual of $T$, $T'in B(Y',X')$. The isomorphism part is easy, since one can easily show that $T'$ is invertible whenever $T$ is. Now, I want to show that $T'$ is an isometry. Is the following correct?
$
|T'(y')|=sup{|y'(Tx)|,big|,|x|leq1}=sup{|y'(Tx)|,big| |Tx|leq 1}=|y'|
$
I am not sure about the part where I substituted $|Tx|$ for $|x|$, but I think it is allowed since $T$ is an isometry.
linear-algebra functional-analysis operator-theory duality-theorems dual-spaces
edited Jan 30 at 18:54
Aweygan
14.7k21442
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be a normed spaces, not necessarily finitely dimensional and let $Tin B(X,Y)$ be an isometric isomorphism. I want to show the same holds for the dual of $T$, $T'in B(Y',X')$. The isomorphism part is easy, since one can easily show that $T'$ is invertible whenever $T$ is. Now, I want to show that $T'$ is an isometry. Is the following correct?
$
|T'(y')|=sup{|y'(Tx)|,big|,|x|leq1}=sup{|y'(Tx)|,big| |Tx|leq 1}=|y'|
$
I am not sure about the part where I substituted $|Tx|$ for $|x|$, but I think it is allowed since $T$ is an isometry.
linear-algebra functional-analysis operator-theory duality-theorems dual-spaces
edited Jan 30 at 18:54
Aweygan
14.7k21442
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be a normed spaces, not necessarily finitely dimensional and let $Tin B(X,Y)$ be an isometric isomorphism. I want to show the same holds for the dual of $T$, $T'in B(Y',X')$. The isomorphism part is easy, since one can easily show that $T'$ is invertible whenever $T$ is. Now, I want to show that $T'$ is an isometry. Is the following correct?
$
|T'(y')|=sup{|y'(Tx)|,big|,|x|leq1}=sup{|y'(Tx)|,big| |Tx|leq 1}=|y'|
$
I am not sure about the part where I substituted $|Tx|$ for $|x|$, but I think it is allowed since $T$ is an isometry.
linear-algebra functional-analysis operator-theory duality-theorems dual-spaces
edited Jan 30 at 18:54
Aweygan
14.7k21442
$endgroup$
Let $X,Y$ be a normed spaces, not necessarily finitely dimensional and let $Tin B(X,Y)$ be an isometric isomorphism. I want to show the same holds for the dual of $T$, $T'in B(Y',X')$. The isomorphism part is easy, since one can easily show that $T'$ is invertible whenever $T$ is. Now, I want to show that $T'$ is an isometry. Is the following correct?
$
|T'(y')|=sup{|y'(Tx)|,big|,|x|leq1}=sup{|y'(Tx)|,big| |Tx|leq 1}=|y'|
$
I am not sure about the part where I substituted $|Tx|$ for $|x|$, but I think it is allowed since $T$ is an isometry.
linear-algebra functional-analysis operator-theory duality-theorems dual-spaces
linear-algebra functional-analysis operator-theory duality-theorems dual-spaces
edited Jan 30 at 18:54
Aweygan
14.7k21442
edited Jan 30 at 18:54
Aweygan
14.7k21442
edited Jan 30 at 18:54
Aweygan
14.7k21442
edited Jan 30 at 18:54
Aweygan
14.7k21442
edited Jan 30 at 18:54
Aweygan
14.7k21442
14.7k21442
asked Jan 30 at 18:48
user408856
add a comment |
add a comment |
1 Answer
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Yes, it is correct. As $T$ is an isometry, $|Tx|=|x|$ for all $xin X$. Hence we have an equality of sets
$${|y'(Tx)|:|x|leq1}={|y'(Tx)|:|Tx|leq1},$$
and taking suprema yields the desired equality.
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes, it is correct. As $T$ is an isometry, $|Tx|=|x|$ for all $xin X$. Hence we have an equality of sets
$${|y'(Tx)|:|x|leq1}={|y'(Tx)|:|Tx|leq1},$$
and taking suprema yields the desired equality.
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
$endgroup$
add a comment |
$begingroup$
Yes, it is correct. As $T$ is an isometry, $|Tx|=|x|$ for all $xin X$. Hence we have an equality of sets
$${|y'(Tx)|:|x|leq1}={|y'(Tx)|:|Tx|leq1},$$
and taking suprema yields the desired equality.
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
$endgroup$
add a comment |
$begingroup$
Yes, it is correct. As $T$ is an isometry, $|Tx|=|x|$ for all $xin X$. Hence we have an equality of sets
$${|y'(Tx)|:|x|leq1}={|y'(Tx)|:|Tx|leq1},$$
and taking suprema yields the desired equality.
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
$endgroup$
Yes, it is correct. As $T$ is an isometry, $|Tx|=|x|$ for all $xin X$. Hence we have an equality of sets
$${|y'(Tx)|:|x|leq1}={|y'(Tx)|:|Tx|leq1},$$
and taking suprema yields the desired equality.
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
answered Jan 30 at 18:53
AweyganAweygan
14.7k21442
14.7k21442
add a comment |
add a comment |
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