Counterexample of non invertible operator
We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".
When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?
operator-theory operator-algebras
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
add a comment |
We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".
When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?
operator-theory operator-algebras
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
Dec 26 '18 at 19:42
add a comment |
We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".
When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?
operator-theory operator-algebras
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
We know the following classical statement: "Let $X$ a Banach space and $T:Xto X$ a bounded operator such that $|T|<1$. Then $I-T$ is invertible".
When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $|T|<1$ and $I-T$ is not invertible?
operator-theory operator-algebras
operator-theory operator-algebras
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
asked Dec 26 '18 at 19:40
sinbadhsinbadh
6,317824
6,317824
"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
Dec 26 '18 at 19:42
add a comment |
"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
Dec 26 '18 at 19:42
"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
Dec 26 '18 at 19:42
"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
Dec 26 '18 at 19:42
add a comment |
1 Answer
1
active
oldest
votes
Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$
However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$
Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.
For a more formal proof that $I-T$ is not invertible, we'll show that $e_1$ is not in the image of $I-T$.
Indeed, assume that $exists x = (x_n)_n in c_{00}$ such that $(I-T)x = e_1$. Then
$$(1,0,0, ldots) = e_1 = (I-T)x = left(x_1, x_2 - frac12 x_1, x_3-frac12x_2, ldotsright)$$
which implies $x_1 = 1$ and $x_{n+1} = frac12 x_n, forall n in mathbb{N}$. Induction gives $x_n = frac1{2^{n-1}},forall n in mathbb{N}$ or $$x = left(1, frac12, frac14, ldotsright)notin c_{00}$$
which is a contradiction.
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
1
This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
Dec 27 '18 at 0:00
-1: the example works, but the argument is sketchy. I have never seen a proof of "if $sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it.
– Martin Argerami
Dec 27 '18 at 2:51
1
@KaviRamaMurthy example is valid since this $T$ has a natural extension to all $ell^infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $sum T^n$. So $T$ is indeed invertible in all $ell^infty$, but it does not in $c_{00}$
– sinbadh
Dec 27 '18 at 4:59
1
@MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete.
– mechanodroid
Dec 27 '18 at 10:41
add a comment |
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Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$
However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$
Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.
For a more formal proof that $I-T$ is not invertible, we'll show that $e_1$ is not in the image of $I-T$.
Indeed, assume that $exists x = (x_n)_n in c_{00}$ such that $(I-T)x = e_1$. Then
$$(1,0,0, ldots) = e_1 = (I-T)x = left(x_1, x_2 - frac12 x_1, x_3-frac12x_2, ldotsright)$$
which implies $x_1 = 1$ and $x_{n+1} = frac12 x_n, forall n in mathbb{N}$. Induction gives $x_n = frac1{2^{n-1}},forall n in mathbb{N}$ or $$x = left(1, frac12, frac14, ldotsright)notin c_{00}$$
which is a contradiction.
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
1
This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
Dec 27 '18 at 0:00
-1: the example works, but the argument is sketchy. I have never seen a proof of "if $sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it.
– Martin Argerami
Dec 27 '18 at 2:51
1
@KaviRamaMurthy example is valid since this $T$ has a natural extension to all $ell^infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $sum T^n$. So $T$ is indeed invertible in all $ell^infty$, but it does not in $c_{00}$
– sinbadh
Dec 27 '18 at 4:59
1
@MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete.
– mechanodroid
Dec 27 '18 at 10:41
add a comment |
Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$
However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$
Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.
For a more formal proof that $I-T$ is not invertible, we'll show that $e_1$ is not in the image of $I-T$.
Indeed, assume that $exists x = (x_n)_n in c_{00}$ such that $(I-T)x = e_1$. Then
$$(1,0,0, ldots) = e_1 = (I-T)x = left(x_1, x_2 - frac12 x_1, x_3-frac12x_2, ldotsright)$$
which implies $x_1 = 1$ and $x_{n+1} = frac12 x_n, forall n in mathbb{N}$. Induction gives $x_n = frac1{2^{n-1}},forall n in mathbb{N}$ or $$x = left(1, frac12, frac14, ldotsright)notin c_{00}$$
which is a contradiction.
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
1
This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
Dec 27 '18 at 0:00
-1: the example works, but the argument is sketchy. I have never seen a proof of "if $sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it.
– Martin Argerami
Dec 27 '18 at 2:51
1
@KaviRamaMurthy example is valid since this $T$ has a natural extension to all $ell^infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $sum T^n$. So $T$ is indeed invertible in all $ell^infty$, but it does not in $c_{00}$
– sinbadh
Dec 27 '18 at 4:59
1
@MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete.
– mechanodroid
Dec 27 '18 at 10:41
add a comment |
Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$
However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$
Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.
For a more formal proof that $I-T$ is not invertible, we'll show that $e_1$ is not in the image of $I-T$.
Indeed, assume that $exists x = (x_n)_n in c_{00}$ such that $(I-T)x = e_1$. Then
$$(1,0,0, ldots) = e_1 = (I-T)x = left(x_1, x_2 - frac12 x_1, x_3-frac12x_2, ldotsright)$$
which implies $x_1 = 1$ and $x_{n+1} = frac12 x_n, forall n in mathbb{N}$. Induction gives $x_n = frac1{2^{n-1}},forall n in mathbb{N}$ or $$x = left(1, frac12, frac14, ldotsright)notin c_{00}$$
which is a contradiction.
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} to c_{00}$ be the unilateral shift. Consider $T = frac12 S$. Then $|T| = frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by
$$(I-T)^{-1} = sum_{n=0}^infty T^n = sum_{n=0}^infty frac1{2^n}S^n$$
However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = sum_{n=0}^infty frac1{2^n}S^ne_1 = sum_{n=0}^infty frac1{2^n}e_{n+1} = left(1, frac12, frac14, ldotsright)notin c_{00}$$
Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.
For a more formal proof that $I-T$ is not invertible, we'll show that $e_1$ is not in the image of $I-T$.
Indeed, assume that $exists x = (x_n)_n in c_{00}$ such that $(I-T)x = e_1$. Then
$$(1,0,0, ldots) = e_1 = (I-T)x = left(x_1, x_2 - frac12 x_1, x_3-frac12x_2, ldotsright)$$
which implies $x_1 = 1$ and $x_{n+1} = frac12 x_n, forall n in mathbb{N}$. Induction gives $x_n = frac1{2^{n-1}},forall n in mathbb{N}$ or $$x = left(1, frac12, frac14, ldotsright)notin c_{00}$$
which is a contradiction.
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
edited Dec 27 '18 at 10:40
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
answered Dec 26 '18 at 19:51
mechanodroidmechanodroid
27k62446
27k62446
1
This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
Dec 27 '18 at 0:00
-1: the example works, but the argument is sketchy. I have never seen a proof of "if $sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it.
– Martin Argerami
Dec 27 '18 at 2:51
1
@KaviRamaMurthy example is valid since this $T$ has a natural extension to all $ell^infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $sum T^n$. So $T$ is indeed invertible in all $ell^infty$, but it does not in $c_{00}$
– sinbadh
Dec 27 '18 at 4:59
1
@MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete.
– mechanodroid
Dec 27 '18 at 10:41
add a comment |
1
This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
Dec 27 '18 at 0:00
-1: the example works, but the argument is sketchy. I have never seen a proof of "if $sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it.
– Martin Argerami
Dec 27 '18 at 2:51
1
@KaviRamaMurthy example is valid since this $T$ has a natural extension to all $ell^infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $sum T^n$. So $T$ is indeed invertible in all $ell^infty$, but it does not in $c_{00}$
– sinbadh
Dec 27 '18 at 4:59
1
@MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete.
– mechanodroid
Dec 27 '18 at 10:41
1
1
This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
Dec 27 '18 at 0:00
This doesn't look like a valid proof to me. You are assuming that if $(I+T)^{-1}$ exists it has to be given by the series $sum T^{n}$. May be true, but certainly not obvious to me.
– Kavi Rama Murthy
Dec 27 '18 at 0:00
-1: the example works, but the argument is sketchy. I have never seen a proof of "if $sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it.
– Martin Argerami
Dec 27 '18 at 2:51
-1: the example works, but the argument is sketchy. I have never seen a proof of "if $sum_nT^n$ doesn't converge, then $I-T$ is not invertible". If such a proof exists, I would like to see it.
– Martin Argerami
Dec 27 '18 at 2:51
1
1
@KaviRamaMurthy example is valid since this $T$ has a natural extension to all $ell^infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $sum T^n$. So $T$ is indeed invertible in all $ell^infty$, but it does not in $c_{00}$
– sinbadh
Dec 27 '18 at 4:59
@KaviRamaMurthy example is valid since this $T$ has a natural extension to all $ell^infty$, which is complete. Therefore, by the original statement about Banach spaces, this $T$ is invertible and its inverse is $sum T^n$. So $T$ is indeed invertible in all $ell^infty$, but it does not in $c_{00}$
– sinbadh
Dec 27 '18 at 4:59
1
1
@MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete.
– mechanodroid
Dec 27 '18 at 10:41
@MartinArgerami You are right, I have added an explicit proof that $I-T$ is not surjective. I wanted to demonstrate where the original argument fails if the space is not complete.
– mechanodroid
Dec 27 '18 at 10:41
add a comment |
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"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
Dec 26 '18 at 19:42