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?











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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?











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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?











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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






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      edited 2 hours ago

























      asked 2 hours ago









      Taras Banakh

      14.6k12985




      14.6k12985




















          2 Answers
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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 $x_n$.






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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.






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            • 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
              1 hour ago











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            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 $x_n$.






            share|cite|improve this answer
























              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 $x_n$.






              share|cite|improve this answer






















                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 $x_n$.






                share|cite|improve this answer












                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 $x_n$.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered 1 hour ago









                Bill Johnson

                23.9k367116




                23.9k367116




















                    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.






                    share|cite|improve this answer




















                    • 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
                      1 hour ago















                    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.






                    share|cite|improve this answer




















                    • 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
                      1 hour ago













                    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.






                    share|cite|improve this answer












                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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
                      1 hour 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










                    • 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
                      1 hour 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
                    1 hour 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
                    1 hour ago


















                     

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