Counterexample of non invertible operator












4














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







share|cite|improve this question






















  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago
















4














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







share|cite|improve this question






















  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago














4












4








4







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









sinbadh

6,312824




6,312824












  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago


















  • "Just" remove the wannabe inverse from a Banach space?
    – Hagen von Eitzen
    5 hours ago
















"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
5 hours ago




"Just" remove the wannabe inverse from a Banach space?
– Hagen von Eitzen
5 hours ago










1 Answer
1






active

oldest

votes


















5














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.






share|cite|improve this answer





















  • 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
    1 hour ago











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1 Answer
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active

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5














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.






share|cite|improve this answer





















  • 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
    1 hour ago
















5














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.






share|cite|improve this answer





















  • 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
    1 hour ago














5












5








5






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









mechanodroid

25.6k62245




25.6k62245












  • 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
    1 hour ago


















  • 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
    1 hour ago
















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
1 hour ago




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
1 hour ago


















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