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Compact images of nowhere dense closed convex sets in a Hilbert space
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Let $B=-B$ be a nowhere dense bounded closed convex set in the Hilbert space $ell_2$ such that the linear hull of $B$ is dense in $ell_2$.
Question. Is there a non-compact linear bounded operator $T:ell_2to ell_2$ such that $T(B)$ is compact?
fa.functional-analysis banach-spaces operator-theory hilbert-spaces compactness
asked 2 hours ago
Taras Banakh
14.6k12985
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Let $B=-B$ be a nowhere dense bounded closed convex set in the Hilbert space $ell_2$ such that the linear hull of $B$ is dense in $ell_2$.
Question. Is there a non-compact linear bounded operator $T:ell_2to ell_2$ such that $T(B)$ is compact?
fa.functional-analysis banach-spaces operator-theory hilbert-spaces compactness
asked 2 hours ago
Taras Banakh
14.6k12985
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $B=-B$ be a nowhere dense bounded closed convex set in the Hilbert space $ell_2$ such that the linear hull of $B$ is dense in $ell_2$.
Question. Is there a non-compact linear bounded operator $T:ell_2to ell_2$ such that $T(B)$ is compact?
fa.functional-analysis banach-spaces operator-theory hilbert-spaces compactness
asked 2 hours ago
Taras Banakh
14.6k12985
Let $B=-B$ be a nowhere dense bounded closed convex set in the Hilbert space $ell_2$ such that the linear hull of $B$ is dense in $ell_2$.
Question. Is there a non-compact linear bounded operator $T:ell_2to ell_2$ such that $T(B)$ is compact?
fa.functional-analysis banach-spaces operator-theory hilbert-spaces compactness
fa.functional-analysis banach-spaces operator-theory hilbert-spaces compactness
asked 2 hours ago
Taras Banakh
14.6k12985
asked 2 hours ago
Taras Banakh
14.6k12985
edited 2 hours ago
asked 2 hours ago
Taras Banakh
14.6k12985
asked 2 hours ago
Taras Banakh
14.6k12985
asked 2 hours ago
Taras Banakh
14.6k12985
14.6k12985
add a comment |Â
add a comment |Â
2 Answers
2
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3
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Your revised assumption is that the norm (rather than semi-norm after the revision) on $ell_2$ given by sup-ing against vectors in $B$ is not equivalent to the usual norm on any finite codimensional subspace. So there is an ON sequence $x_n$ in $ell_2$ s.t. $sup langle x_n,brangle : bin B to 0$ as $nto infty$. Take for $T$ the orthogonal projection onto the closed span of the $x_n$.
answered 1 hour ago
Bill Johnson
23.9k367116
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No, take $B$ to be the set of vectors in the unit ball of $l^2$ whose first coordinate is zero. If $T: l^2 to l^2$ is a bounded operator and $T(B)$ is compact, then letting $C$ be the set of vectors in the unit ball which are nonzero only in the first coordinate (plus the zero vector), we have that $C$ is compact and therefore $T(C)$ is also compact. Thus $T(B+C) = T(B) + T(C)$ is compact, and $B + C$ contains the unit ball, so $T$ is a compact operator.
answered 2 hours ago
Nik Weaver
19.1k145119
Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case.
– Taras Banakh
2 hours ago
... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$.
– Nik Weaver
2 hours ago
I had in mind infinite codimension in the sense that the linear hull is a dense subspace of the first Baire category and hence of infinite codimension in the algebraic sense. This was the true setting. For my purposes it suffices to assume that $B$ is the image of the unit ball of the Hilbert space under a non-open bounded operator $ell_2toell_2$.
– Taras Banakh
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your revised assumption is that the norm (rather than semi-norm after the revision) on $ell_2$ given by sup-ing against vectors in $B$ is not equivalent to the usual norm on any finite codimensional subspace. So there is an ON sequence $x_n$ in $ell_2$ s.t. $sup langle x_n,brangle : bin B to 0$ as $nto infty$. Take for $T$ the orthogonal projection onto the closed span of the $x_n$.
answered 1 hour ago
Bill Johnson
23.9k367116
add a comment |Â
up vote
3
down vote
accepted
Your revised assumption is that the norm (rather than semi-norm after the revision) on $ell_2$ given by sup-ing against vectors in $B$ is not equivalent to the usual norm on any finite codimensional subspace. So there is an ON sequence $x_n$ in $ell_2$ s.t. $sup langle x_n,brangle : bin B to 0$ as $nto infty$. Take for $T$ the orthogonal projection onto the closed span of the $x_n$.
answered 1 hour ago
Bill Johnson
23.9k367116
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your revised assumption is that the norm (rather than semi-norm after the revision) on $ell_2$ given by sup-ing against vectors in $B$ is not equivalent to the usual norm on any finite codimensional subspace. So there is an ON sequence $x_n$ in $ell_2$ s.t. $sup langle x_n,brangle : bin B to 0$ as $nto infty$. Take for $T$ the orthogonal projection onto the closed span of the $x_n$.
answered 1 hour ago
Bill Johnson
23.9k367116
Your revised assumption is that the norm (rather than semi-norm after the revision) on $ell_2$ given by sup-ing against vectors in $B$ is not equivalent to the usual norm on any finite codimensional subspace. So there is an ON sequence $x_n$ in $ell_2$ s.t. $sup langle x_n,brangle : bin B to 0$ as $nto infty$. Take for $T$ the orthogonal projection onto the closed span of the $x_n$.
answered 1 hour ago
Bill Johnson
23.9k367116
answered 1 hour ago
Bill Johnson
23.9k367116
answered 1 hour ago
Bill Johnson
23.9k367116
answered 1 hour ago
Bill Johnson
23.9k367116
23.9k367116
add a comment |Â
add a comment |Â
up vote
3
down vote
No, take $B$ to be the set of vectors in the unit ball of $l^2$ whose first coordinate is zero. If $T: l^2 to l^2$ is a bounded operator and $T(B)$ is compact, then letting $C$ be the set of vectors in the unit ball which are nonzero only in the first coordinate (plus the zero vector), we have that $C$ is compact and therefore $T(C)$ is also compact. Thus $T(B+C) = T(B) + T(C)$ is compact, and $B + C$ contains the unit ball, so $T$ is a compact operator.
answered 2 hours ago
Nik Weaver
19.1k145119
Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case.
– Taras Banakh
2 hours ago
... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$.
– Nik Weaver
2 hours ago
I had in mind infinite codimension in the sense that the linear hull is a dense subspace of the first Baire category and hence of infinite codimension in the algebraic sense. This was the true setting. For my purposes it suffices to assume that $B$ is the image of the unit ball of the Hilbert space under a non-open bounded operator $ell_2toell_2$.
– Taras Banakh
2 hours ago
add a comment |Â
up vote
3
down vote
No, take $B$ to be the set of vectors in the unit ball of $l^2$ whose first coordinate is zero. If $T: l^2 to l^2$ is a bounded operator and $T(B)$ is compact, then letting $C$ be the set of vectors in the unit ball which are nonzero only in the first coordinate (plus the zero vector), we have that $C$ is compact and therefore $T(C)$ is also compact. Thus $T(B+C) = T(B) + T(C)$ is compact, and $B + C$ contains the unit ball, so $T$ is a compact operator.
answered 2 hours ago
Nik Weaver
19.1k145119
Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case.
– Taras Banakh
2 hours ago
... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$.
– Nik Weaver
2 hours ago
I had in mind infinite codimension in the sense that the linear hull is a dense subspace of the first Baire category and hence of infinite codimension in the algebraic sense. This was the true setting. For my purposes it suffices to assume that $B$ is the image of the unit ball of the Hilbert space under a non-open bounded operator $ell_2toell_2$.
– Taras Banakh
2 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
No, take $B$ to be the set of vectors in the unit ball of $l^2$ whose first coordinate is zero. If $T: l^2 to l^2$ is a bounded operator and $T(B)$ is compact, then letting $C$ be the set of vectors in the unit ball which are nonzero only in the first coordinate (plus the zero vector), we have that $C$ is compact and therefore $T(C)$ is also compact. Thus $T(B+C) = T(B) + T(C)$ is compact, and $B + C$ contains the unit ball, so $T$ is a compact operator.
answered 2 hours ago
Nik Weaver
19.1k145119
No, take $B$ to be the set of vectors in the unit ball of $l^2$ whose first coordinate is zero. If $T: l^2 to l^2$ is a bounded operator and $T(B)$ is compact, then letting $C$ be the set of vectors in the unit ball which are nonzero only in the first coordinate (plus the zero vector), we have that $C$ is compact and therefore $T(C)$ is also compact. Thus $T(B+C) = T(B) + T(C)$ is compact, and $B + C$ contains the unit ball, so $T$ is a compact operator.
answered 2 hours ago
Nik Weaver
19.1k145119
answered 2 hours ago
Nik Weaver
19.1k145119
answered 2 hours ago
Nik Weaver
19.1k145119
answered 2 hours ago
Nik Weaver
19.1k145119
19.1k145119
Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case.
– Taras Banakh
2 hours ago
... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$.
– Nik Weaver
2 hours ago
I had in mind infinite codimension in the sense that the linear hull is a dense subspace of the first Baire category and hence of infinite codimension in the algebraic sense. This was the true setting. For my purposes it suffices to assume that $B$ is the image of the unit ball of the Hilbert space under a non-open bounded operator $ell_2toell_2$.
– Taras Banakh
2 hours ago
add a comment |Â
Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case.
– Taras Banakh
2 hours ago
... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$.
– Nik Weaver
2 hours ago
I had in mind infinite codimension in the sense that the linear hull is a dense subspace of the first Baire category and hence of infinite codimension in the algebraic sense. This was the true setting. For my purposes it suffices to assume that $B$ is the image of the unit ball of the Hilbert space under a non-open bounded operator $ell_2toell_2$.
– Taras Banakh
2 hours ago
Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case.
– Taras Banakh
2 hours ago
Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case.
– Taras Banakh
2 hours ago
... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$.
– Nik Weaver
2 hours ago
... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$.
– Nik Weaver
2 hours ago
I had in mind infinite codimension in the sense that the linear hull is a dense subspace of the first Baire category and hence of infinite codimension in the algebraic sense. This was the true setting. For my purposes it suffices to assume that $B$ is the image of the unit ball of the Hilbert space under a non-open bounded operator $ell_2toell_2$.
– Taras Banakh
2 hours ago
I had in mind infinite codimension in the sense that the linear hull is a dense subspace of the first Baire category and hence of infinite codimension in the algebraic sense. This was the true setting. For my purposes it suffices to assume that $B$ is the image of the unit ball of the Hilbert space under a non-open bounded operator $ell_2toell_2$.
– Taras Banakh
2 hours ago
add a comment |Â
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