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If it exists, the inverse of a compact linear operator in infinite dimensional space cannot be bouded
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I have been reading some posts on here that I think are related such as this and this. I am still having a tough time coming up with a nice proof for my question.
Question: If a compact linear operator $T:X rightarrow X$ on an infinite dimensional normed space $X$ has an inverse which is defined on all of $X$, show that the inverse cannot be bounded.
I found this in $8.3.8$ of Erwin Kreyszig functional analysis.
functional-analysis inverse compact-operators
edited Jul 24 at 12:12
mechanodroid
22.2k52041
asked Jul 24 at 2:06
MathIsHard
1,122415
add a comment |Â
up vote
4
down vote
favorite
I have been reading some posts on here that I think are related such as this and this. I am still having a tough time coming up with a nice proof for my question.
Question: If a compact linear operator $T:X rightarrow X$ on an infinite dimensional normed space $X$ has an inverse which is defined on all of $X$, show that the inverse cannot be bounded.
I found this in $8.3.8$ of Erwin Kreyszig functional analysis.
functional-analysis inverse compact-operators
edited Jul 24 at 12:12
mechanodroid
22.2k52041
asked Jul 24 at 2:06
MathIsHard
1,122415
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have been reading some posts on here that I think are related such as this and this. I am still having a tough time coming up with a nice proof for my question.
Question: If a compact linear operator $T:X rightarrow X$ on an infinite dimensional normed space $X$ has an inverse which is defined on all of $X$, show that the inverse cannot be bounded.
I found this in $8.3.8$ of Erwin Kreyszig functional analysis.
functional-analysis inverse compact-operators
edited Jul 24 at 12:12
mechanodroid
22.2k52041
asked Jul 24 at 2:06
MathIsHard
1,122415
I have been reading some posts on here that I think are related such as this and this. I am still having a tough time coming up with a nice proof for my question.
Question: If a compact linear operator $T:X rightarrow X$ on an infinite dimensional normed space $X$ has an inverse which is defined on all of $X$, show that the inverse cannot be bounded.
I found this in $8.3.8$ of Erwin Kreyszig functional analysis.
functional-analysis inverse compact-operators
edited Jul 24 at 12:12
mechanodroid
22.2k52041
asked Jul 24 at 2:06
MathIsHard
1,122415
edited Jul 24 at 12:12
mechanodroid
22.2k52041
edited Jul 24 at 12:12
mechanodroid
22.2k52041
edited Jul 24 at 12:12
mechanodroid
22.2k52041
22.2k52041
asked Jul 24 at 2:06
MathIsHard
1,122415
asked Jul 24 at 2:06
MathIsHard
1,122415
asked Jul 24 at 2:06
MathIsHard
1,122415
1,122415
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
Suppose $T^-1$ existed and was bounded. Then $T^-1(B_X) subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields
$$B_X = T(T^-1(B_X)) subseteq T(KB_X) = KT(B_X).$$
Since $overlineT(B_X)$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.
answered Jul 24 at 2:19
Theo Bendit
12k1843
@ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks.
– MathIsHard
Jul 24 at 2:23
Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you.
– MathIsHard
Jul 24 at 2:25
3
Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact.
– Theo Bendit
Jul 24 at 2:30
1
@MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact.
– Theo Bendit
Jul 24 at 2:32
1
@ Theo Bendit thank you very much. I really appreciate the help.
– MathIsHard
Jul 24 at 2:33
add a comment |Â
up vote
0
down vote
It is known that the compact operators $mathbbK(X)$ form a two-sided ideal in the algebra of bounded operators $mathbbB(X)$.
So if $T in mathbbK(X)$ has an inverse $T^-1 in mathbbB(X)$, we have
$$TT^-1 = I$$
so it would follow that $I in mathbbK(X)$, and the identity map $I$ is not compact when $X$ is infinite-dimensional.
Therefore $T^-1$ cannot be bounded.
Actually, a compact operator $T in mathbbK(X)$ when $X$ is a Banach space cannot even be surjective, let alone bijective.
Indeed, a surjective bounded map $T in mathbbB(X)$ is open by the Open mapping theorem, so $T(B(0,1))$ is an open set. Therefore $T(overlineB(0,1))$ contains a closed ball which isn't compact in an infinite-dimensional space $X$. Hence $T(overlineB(0,1))$ cannot be precompact so $T$ is not compact.
answered Jul 24 at 12:09
mechanodroid
22.2k52041
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Suppose $T^-1$ existed and was bounded. Then $T^-1(B_X) subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields
$$B_X = T(T^-1(B_X)) subseteq T(KB_X) = KT(B_X).$$
Since $overlineT(B_X)$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.
answered Jul 24 at 2:19
Theo Bendit
12k1843
@ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks.
– MathIsHard
Jul 24 at 2:23
Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you.
– MathIsHard
Jul 24 at 2:25
3
Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact.
– Theo Bendit
Jul 24 at 2:30
1
@MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact.
– Theo Bendit
Jul 24 at 2:32
1
@ Theo Bendit thank you very much. I really appreciate the help.
– MathIsHard
Jul 24 at 2:33
add a comment |Â
up vote
8
down vote
accepted
Suppose $T^-1$ existed and was bounded. Then $T^-1(B_X) subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields
$$B_X = T(T^-1(B_X)) subseteq T(KB_X) = KT(B_X).$$
Since $overlineT(B_X)$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.
answered Jul 24 at 2:19
Theo Bendit
12k1843
@ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks.
– MathIsHard
Jul 24 at 2:23
Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you.
– MathIsHard
Jul 24 at 2:25
3
Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact.
– Theo Bendit
Jul 24 at 2:30
1
@MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact.
– Theo Bendit
Jul 24 at 2:32
1
@ Theo Bendit thank you very much. I really appreciate the help.
– MathIsHard
Jul 24 at 2:33
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Suppose $T^-1$ existed and was bounded. Then $T^-1(B_X) subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields
$$B_X = T(T^-1(B_X)) subseteq T(KB_X) = KT(B_X).$$
Since $overlineT(B_X)$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.
answered Jul 24 at 2:19
Theo Bendit
12k1843
Suppose $T^-1$ existed and was bounded. Then $T^-1(B_X) subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields
$$B_X = T(T^-1(B_X)) subseteq T(KB_X) = KT(B_X).$$
Since $overlineT(B_X)$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.
answered Jul 24 at 2:19
Theo Bendit
12k1843
edited Jul 24 at 3:08
answered Jul 24 at 2:19
Theo Bendit
12k1843
answered Jul 24 at 2:19
Theo Bendit
12k1843
answered Jul 24 at 2:19
Theo Bendit
12k1843
12k1843
@ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks.
– MathIsHard
Jul 24 at 2:23
Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you.
– MathIsHard
Jul 24 at 2:25
3
Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact.
– Theo Bendit
Jul 24 at 2:30
1
@MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact.
– Theo Bendit
Jul 24 at 2:32
1
@ Theo Bendit thank you very much. I really appreciate the help.
– MathIsHard
Jul 24 at 2:33
add a comment |Â
@ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks.
– MathIsHard
Jul 24 at 2:23
Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you.
– MathIsHard
Jul 24 at 2:25
3
Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact.
– Theo Bendit
Jul 24 at 2:30
1
@MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact.
– Theo Bendit
Jul 24 at 2:32
1
@ Theo Bendit thank you very much. I really appreciate the help.
– MathIsHard
Jul 24 at 2:33
@ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks.
– MathIsHard
Jul 24 at 2:23
@ Theo thank you for your answer. I like your proof, I am not familiar with the notation of $B_X$. Is that a bounded sequence in X? Thanks.
– MathIsHard
Jul 24 at 2:23
Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you.
– MathIsHard
Jul 24 at 2:25
Do you mind expanding why the closed subset of a compact set is compact implies finite-dimensional? This is a new topic for me... thank you.
– MathIsHard
Jul 24 at 2:25
3
3
Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact.
– Theo Bendit
Jul 24 at 2:30
Sorry! $B_X$ refers to the closed unit ball (i.e. the set of points with norm less than or equal to $1$). Then $KB_X$ is the set of points with norm less than or equal to $K$. Being bounded means that $B_X$ maps to a bounded a set, i.e. a subset of $KB_X$ for some $K$. Compactness of a map means that $B_X$ maps to a set whose closure is compact.
– Theo Bendit
Jul 24 at 2:30
1
1
@MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact.
– Theo Bendit
Jul 24 at 2:32
@MathIsHard A standard result in topology states that if we have a closed subset of a compact set/space, then the subset is compact (this is true for sequentially compactness as well and is not hard to prove). Less easy to prove, a normed linear space is $X$ is finite-dimensional if and only if $B_X$ is compact.
– Theo Bendit
Jul 24 at 2:32
1
1
@ Theo Bendit thank you very much. I really appreciate the help.
– MathIsHard
Jul 24 at 2:33
@ Theo Bendit thank you very much. I really appreciate the help.
– MathIsHard
Jul 24 at 2:33
add a comment |Â
up vote
0
down vote
It is known that the compact operators $mathbbK(X)$ form a two-sided ideal in the algebra of bounded operators $mathbbB(X)$.
So if $T in mathbbK(X)$ has an inverse $T^-1 in mathbbB(X)$, we have
$$TT^-1 = I$$
so it would follow that $I in mathbbK(X)$, and the identity map $I$ is not compact when $X$ is infinite-dimensional.
Therefore $T^-1$ cannot be bounded.
Actually, a compact operator $T in mathbbK(X)$ when $X$ is a Banach space cannot even be surjective, let alone bijective.
Indeed, a surjective bounded map $T in mathbbB(X)$ is open by the Open mapping theorem, so $T(B(0,1))$ is an open set. Therefore $T(overlineB(0,1))$ contains a closed ball which isn't compact in an infinite-dimensional space $X$. Hence $T(overlineB(0,1))$ cannot be precompact so $T$ is not compact.
answered Jul 24 at 12:09
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
It is known that the compact operators $mathbbK(X)$ form a two-sided ideal in the algebra of bounded operators $mathbbB(X)$.
So if $T in mathbbK(X)$ has an inverse $T^-1 in mathbbB(X)$, we have
$$TT^-1 = I$$
so it would follow that $I in mathbbK(X)$, and the identity map $I$ is not compact when $X$ is infinite-dimensional.
Therefore $T^-1$ cannot be bounded.
Actually, a compact operator $T in mathbbK(X)$ when $X$ is a Banach space cannot even be surjective, let alone bijective.
Indeed, a surjective bounded map $T in mathbbB(X)$ is open by the Open mapping theorem, so $T(B(0,1))$ is an open set. Therefore $T(overlineB(0,1))$ contains a closed ball which isn't compact in an infinite-dimensional space $X$. Hence $T(overlineB(0,1))$ cannot be precompact so $T$ is not compact.
answered Jul 24 at 12:09
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is known that the compact operators $mathbbK(X)$ form a two-sided ideal in the algebra of bounded operators $mathbbB(X)$.
So if $T in mathbbK(X)$ has an inverse $T^-1 in mathbbB(X)$, we have
$$TT^-1 = I$$
so it would follow that $I in mathbbK(X)$, and the identity map $I$ is not compact when $X$ is infinite-dimensional.
Therefore $T^-1$ cannot be bounded.
Actually, a compact operator $T in mathbbK(X)$ when $X$ is a Banach space cannot even be surjective, let alone bijective.
Indeed, a surjective bounded map $T in mathbbB(X)$ is open by the Open mapping theorem, so $T(B(0,1))$ is an open set. Therefore $T(overlineB(0,1))$ contains a closed ball which isn't compact in an infinite-dimensional space $X$. Hence $T(overlineB(0,1))$ cannot be precompact so $T$ is not compact.
answered Jul 24 at 12:09
mechanodroid
22.2k52041
It is known that the compact operators $mathbbK(X)$ form a two-sided ideal in the algebra of bounded operators $mathbbB(X)$.
So if $T in mathbbK(X)$ has an inverse $T^-1 in mathbbB(X)$, we have
$$TT^-1 = I$$
so it would follow that $I in mathbbK(X)$, and the identity map $I$ is not compact when $X$ is infinite-dimensional.
Therefore $T^-1$ cannot be bounded.
Actually, a compact operator $T in mathbbK(X)$ when $X$ is a Banach space cannot even be surjective, let alone bijective.
Indeed, a surjective bounded map $T in mathbbB(X)$ is open by the Open mapping theorem, so $T(B(0,1))$ is an open set. Therefore $T(overlineB(0,1))$ contains a closed ball which isn't compact in an infinite-dimensional space $X$. Hence $T(overlineB(0,1))$ cannot be precompact so $T$ is not compact.
answered Jul 24 at 12:09
mechanodroid
22.2k52041
answered Jul 24 at 12:09
mechanodroid
22.2k52041
answered Jul 24 at 12:09
mechanodroid
22.2k52041
answered Jul 24 at 12:09
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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