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 $T:\ell_2\to \ell_2$ such that $T(B)$ is compact?
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asked Oct 31, 2018 at 19:45
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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. $\sup \{\langle x_n,b\rangle : b\in B \} \to 0$ as $n\to \infty$. Take for $T$ the orthogonal projection onto the closed span of the $x_n$.
answered Oct 31, 2018 at 20:21
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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 Oct 31, 2018 at 20:04
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$\begingroup$ Thank you for your answer. But this is a trivial case (finite codimension). Let me add one more condition to exclude this case. $\endgroup$ Commented Oct 31, 2018 at 20:09
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$\begingroup$ ... but if $B$ has infinite codimension then there will be a bounded noncompact operator which vanishes on $B$. $\endgroup$ Commented Oct 31, 2018 at 20:11
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$\begingroup$ 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 $\ell_2\to\ell_2$. $\endgroup$ Commented Oct 31, 2018 at 20:14