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If $X$ is a seaparable Banach space and $Y$ is a subspace of $X$, not necessarily closed, can one always find an bounded operator with range $Y$? It is easy when $Y$ is closed and complemented, what if it's not?

Edit 1: I modified the question to add that $X$ is separable. It seems that the general case is open.

https://mathoverflow.net/questions/101253/surjectivity-of-operators-on-linfty

Edit 2: If $Y$ is not closed, Jonas Meyer answer below shows that the answer is 'No'. What about when $Y$ is closed?

Edit 3: A followup question has been posted about the case when $Y$ is closed: Bounded operators with prescribed range - part II

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  • Could you clarify this for me -- must the range of the operator include all of Y or could there be a y in Y such that y is not in L(X)? Or perhaps the answer is trivial of all of Y is not included? – Betty Mock Aug 24 '13 at 02:13
  • another thought - a bounded operator must be continuous, and if Y is open wouldn't any continuous operator that covers Y also cover the closure of Y? – Betty Mock Aug 24 '13 at 02:19
  • +1 Good question. In the simplest case where $Y$ is one-dimensional, the answer, Hahn-Banach, is already non-trivial. – Michael Aug 24 '13 at 06:15
  • @BettyMock I want Range(T)=$Y$. When not all of $Y$ is included, you can simply take a finite dimensional subspace of $Y$. I don't know whether it is easier when you require an infinite dimensional subspace of $Y$. – Markus Aug 24 '13 at 15:52
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    @BettyMock Proper subspaces of a topological vector space (TVS) always have empty interior, so $Y$ cannot be open. Even if you mean "not-close" instead of "open", what you say is still not true. Bounded operators can have non-closed ranges. – Markus Aug 24 '13 at 16:13
  • @Michael When $Y$ is closed and complemented, in particular when $Y$ is one-dimensional, then one can simply defined $T$ as the identity on $Y$ and $0$ on the complement. I think you are right, to show that every finite dimensional subspace is complemented, one needs HB. – Markus Aug 24 '13 at 16:20
  • @BettyMock: the simplest example is on a separable Hilbert space with orthonormal basis ${e_n}$, the operator $e_n\mapsto \frac1n,e_n$ is compact with dense (but non-closed) range. – Martin Argerami Aug 24 '13 at 17:37

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In part 3 of t.b.'s answer here it is shown that if a bounded operator between Banach spaces has finite codimensional range, then it has closed range. If $Y$ is the kernel of an unbounded linear functional $X\to\mathbb C$, then $Y$ has finite codimension but is not closed, hence $Y$ is not the range of a bounded operator.

This is generalized in "Algebraic complements and ranges of linear operators" by Enflo and Smith, 2011, where the main result is stated as follows:

Let $B$ be an infinite-dimensional Banach space and let $C$ be any non-empty, closed subspace of $B$ whose algebraic complement, $N$, is non-closed. Then, $N$ is not the range of a continuous linear operator on a Banach space.

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