Is the operator norm invariant under a renorming of the Banach space?

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2
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Let $I colon X to Y$ a non-isometric isomorphism between Banach spaces. Let $T colon X to X$ be a bounded operator. Do we have the equality
$$
|T|_B(X)
=|ITI^-1|_B(Y) ?
$$
Here $B(X)$ is the space of bounded linear operators on $X$. Here "isomorphism" means bounded bijective linear map with bounded inverse.

functional-analysis banach-spaces normed-spaces

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edited Jul 18 at 19:43

asked Jul 18 at 19:22

Zouba

883517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Must $I$ or $ITI^-1$ be bounded?
  – mechanodroid
  Jul 18 at 19:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $I$ and $I^-1$ are bounded.
  – Zouba
  Jul 18 at 19:44
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Let $I colon X to Y$ a non-isometric isomorphism between Banach spaces. Let $T colon X to X$ be a bounded operator. Do we have the equality
$$
|T|_B(X)
=|ITI^-1|_B(Y) ?
$$
Here $B(X)$ is the space of bounded linear operators on $X$. Here "isomorphism" means bounded bijective linear map with bounded inverse.

functional-analysis banach-spaces normed-spaces

share|cite|improve this question

edited Jul 18 at 19:43

asked Jul 18 at 19:22

Zouba

883517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Must $I$ or $ITI^-1$ be bounded?
  – mechanodroid
  Jul 18 at 19:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $I$ and $I^-1$ are bounded.
  – Zouba
  Jul 18 at 19:44
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $I colon X to Y$ a non-isometric isomorphism between Banach spaces. Let $T colon X to X$ be a bounded operator. Do we have the equality
$$
|T|_B(X)
=|ITI^-1|_B(Y) ?
$$
Here $B(X)$ is the space of bounded linear operators on $X$. Here "isomorphism" means bounded bijective linear map with bounded inverse.

functional-analysis banach-spaces normed-spaces

share|cite|improve this question

edited Jul 18 at 19:43

asked Jul 18 at 19:22

Zouba

883517

Let $I colon X to Y$ a non-isometric isomorphism between Banach spaces. Let $T colon X to X$ be a bounded operator. Do we have the equality
$$
|T|_B(X)
=|ITI^-1|_B(Y) ?
$$
Here $B(X)$ is the space of bounded linear operators on $X$. Here "isomorphism" means bounded bijective linear map with bounded inverse.

functional-analysis banach-spaces normed-spaces

share|cite|improve this question

edited Jul 18 at 19:43

asked Jul 18 at 19:22

Zouba

883517

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 19:43

edited Jul 18 at 19:43

edited Jul 18 at 19:43

asked Jul 18 at 19:22

Zouba

883517

asked Jul 18 at 19:22

Zouba

883517

asked Jul 18 at 19:22

Zouba

883517

883517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Must $I$ or $ITI^-1$ be bounded?
  – mechanodroid
  Jul 18 at 19:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $I$ and $I^-1$ are bounded.
  – Zouba
  Jul 18 at 19:44
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Must $I$ or $ITI^-1$ be bounded?
  – mechanodroid
  Jul 18 at 19:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $I$ and $I^-1$ are bounded.
  – Zouba
  Jul 18 at 19:44
  

  
  
  
  

Must $I$ or $ITI^-1$ be bounded?
– mechanodroid
Jul 18 at 19:37

Must $I$ or $ITI^-1$ be bounded?
– mechanodroid
Jul 18 at 19:37

$I$ and $I^-1$ are bounded.
– Zouba
Jul 18 at 19:44

$I$ and $I^-1$ are bounded.
– Zouba
Jul 18 at 19:44

add a comment | 

2 Answers
2

active

oldest

votes

up vote
5
down vote



accepted

In general, we don't have $lVert TrVert_B(X) = lVert ITI^-1rVert_B(Y)$. Let $X = mathbbR^2$ endowed with the Euclidean norm, $Y = mathbbR^2$ endowed with an $ell^p$-norm for $p neq 2$, $I$ the identity and $T$ a rotation by an angle not an integer multiple of $pi/2$. On $X$, $T$ is an isometry, but on $Y$ its norm is greater than $1$.

share|cite|improve this answer

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

add a comment | 

up vote
2
down vote

The answer is no.

Consider the Banach space $X = (mathbbR^2, |cdot|_infty)$ and let $T : X to X$ be given as $T = beginbmatrix2 & 3 \ 0 & 1endbmatrix$.

Then

$$|T(x,y)|_infty = |(2x + 3y, y)|_infty = maxy le 2|x| + 3|y| le 5|(x,y)|_infty$$

and $T(1,1) = (5,1)$ so we conclude $|T| = 5$.

Now we shall diagonalize $T$. Define $I : X to X$ as $I = beginbmatrix1 & 1 \ 0 & 1endbmatrix$ which is an isomorphism, but it is not isometric because $I(1,1) = (2,1)$.

Then $$ITI^-1 = beginbmatrix 2 & 0 \ 0 & 1endbmatrix$$

so $|ITI^-1| = 2$.

share|cite|improve this answer

answered Jul 18 at 21:46

mechanodroid

22.2k52041

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

In general, we don't have $lVert TrVert_B(X) = lVert ITI^-1rVert_B(Y)$. Let $X = mathbbR^2$ endowed with the Euclidean norm, $Y = mathbbR^2$ endowed with an $ell^p$-norm for $p neq 2$, $I$ the identity and $T$ a rotation by an angle not an integer multiple of $pi/2$. On $X$, $T$ is an isometry, but on $Y$ its norm is greater than $1$.

share|cite|improve this answer

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

add a comment | 

up vote
5
down vote



accepted

In general, we don't have $lVert TrVert_B(X) = lVert ITI^-1rVert_B(Y)$. Let $X = mathbbR^2$ endowed with the Euclidean norm, $Y = mathbbR^2$ endowed with an $ell^p$-norm for $p neq 2$, $I$ the identity and $T$ a rotation by an angle not an integer multiple of $pi/2$. On $X$, $T$ is an isometry, but on $Y$ its norm is greater than $1$.

share|cite|improve this answer

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

In general, we don't have $lVert TrVert_B(X) = lVert ITI^-1rVert_B(Y)$. Let $X = mathbbR^2$ endowed with the Euclidean norm, $Y = mathbbR^2$ endowed with an $ell^p$-norm for $p neq 2$, $I$ the identity and $T$ a rotation by an angle not an integer multiple of $pi/2$. On $X$, $T$ is an isometry, but on $Y$ its norm is greater than $1$.

share|cite|improve this answer

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

In general, we don't have $lVert TrVert_B(X) = lVert ITI^-1rVert_B(Y)$. Let $X = mathbbR^2$ endowed with the Euclidean norm, $Y = mathbbR^2$ endowed with an $ell^p$-norm for $p neq 2$, $I$ the identity and $T$ a rotation by an angle not an integer multiple of $pi/2$. On $X$, $T$ is an isometry, but on $Y$ its norm is greater than $1$.

share|cite|improve this answer

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

answered Jul 18 at 19:54

Daniel Fischer♦

171k16154274

171k16154274

add a comment | 

add a comment | 

up vote
2
down vote

The answer is no.

Consider the Banach space $X = (mathbbR^2, |cdot|_infty)$ and let $T : X to X$ be given as $T = beginbmatrix2 & 3 \ 0 & 1endbmatrix$.

Then

$$|T(x,y)|_infty = |(2x + 3y, y)|_infty = maxy le 2|x| + 3|y| le 5|(x,y)|_infty$$

and $T(1,1) = (5,1)$ so we conclude $|T| = 5$.

Now we shall diagonalize $T$. Define $I : X to X$ as $I = beginbmatrix1 & 1 \ 0 & 1endbmatrix$ which is an isomorphism, but it is not isometric because $I(1,1) = (2,1)$.

Then $$ITI^-1 = beginbmatrix 2 & 0 \ 0 & 1endbmatrix$$

so $|ITI^-1| = 2$.

share|cite|improve this answer

answered Jul 18 at 21:46

mechanodroid

22.2k52041

add a comment | 

up vote
2
down vote

The answer is no.

Consider the Banach space $X = (mathbbR^2, |cdot|_infty)$ and let $T : X to X$ be given as $T = beginbmatrix2 & 3 \ 0 & 1endbmatrix$.

Then

$$|T(x,y)|_infty = |(2x + 3y, y)|_infty = maxy le 2|x| + 3|y| le 5|(x,y)|_infty$$

and $T(1,1) = (5,1)$ so we conclude $|T| = 5$.

Now we shall diagonalize $T$. Define $I : X to X$ as $I = beginbmatrix1 & 1 \ 0 & 1endbmatrix$ which is an isomorphism, but it is not isometric because $I(1,1) = (2,1)$.

Then $$ITI^-1 = beginbmatrix 2 & 0 \ 0 & 1endbmatrix$$

so $|ITI^-1| = 2$.

share|cite|improve this answer

answered Jul 18 at 21:46

mechanodroid

22.2k52041

add a comment | 

up vote
2
down vote

up vote
2
down vote

The answer is no.

Consider the Banach space $X = (mathbbR^2, |cdot|_infty)$ and let $T : X to X$ be given as $T = beginbmatrix2 & 3 \ 0 & 1endbmatrix$.

Then

$$|T(x,y)|_infty = |(2x + 3y, y)|_infty = maxy le 2|x| + 3|y| le 5|(x,y)|_infty$$

and $T(1,1) = (5,1)$ so we conclude $|T| = 5$.

Now we shall diagonalize $T$. Define $I : X to X$ as $I = beginbmatrix1 & 1 \ 0 & 1endbmatrix$ which is an isomorphism, but it is not isometric because $I(1,1) = (2,1)$.

Then $$ITI^-1 = beginbmatrix 2 & 0 \ 0 & 1endbmatrix$$

so $|ITI^-1| = 2$.

share|cite|improve this answer

answered Jul 18 at 21:46

mechanodroid

22.2k52041

The answer is no.

Consider the Banach space $X = (mathbbR^2, |cdot|_infty)$ and let $T : X to X$ be given as $T = beginbmatrix2 & 3 \ 0 & 1endbmatrix$.

Then

$$|T(x,y)|_infty = |(2x + 3y, y)|_infty = maxy le 2|x| + 3|y| le 5|(x,y)|_infty$$

and $T(1,1) = (5,1)$ so we conclude $|T| = 5$.

Now we shall diagonalize $T$. Define $I : X to X$ as $I = beginbmatrix1 & 1 \ 0 & 1endbmatrix$ which is an isomorphism, but it is not isometric because $I(1,1) = (2,1)$.

Then $$ITI^-1 = beginbmatrix 2 & 0 \ 0 & 1endbmatrix$$

so $|ITI^-1| = 2$.

share|cite|improve this answer

answered Jul 18 at 21:46

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 21:46

mechanodroid

22.2k52041

answered Jul 18 at 21:46

mechanodroid

22.2k52041

answered Jul 18 at 21:46

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment | 

 

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