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My question is motivated by the following little proposition:

Proposition. For a vector subspace $V$ of a Banach space $(X, \|\cdot\|_X)$ the following assertions are equivalent:

(i) There exists a Banach space $Z$ and a bounded linear operator $T: Z \to X$ with range $V$.

(ii) There exists a complete norm $\|\cdot\|_V$ on $V$ such that the canonical embedding of $(V, \|\cdot\|_V)$ into $(X,\|\cdot\|_X)$ is continuous.

(See below for a proof.)

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Question. Are (i) and (ii) also equivalent to the following assertion (iii)?

(iii) There exists a bounded linear operator $S: X \to X$ with range $V$.

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Proof of the Proposition. Obviously (ii) implies (i), so assume that (i) holds. Let $\tilde T: Z / \ker T \to X$ denote the injective operator induced by $T$; then $\tilde T$ also has range $V$. The inverse ${\tilde T}^{-1}$ is a closed linear operator $X \supseteq V \to Z / \ker T$, so $V$ becomes a Banach space with respect to the graph norm given by $\|x\|_V := \|x\|_X + \|{\tilde T}^{-1}x\|_{Z / \ker T}$ for all $x \in V$.

Remark. For Hilbert spaces results of this type can, for instance, be found in the paper "Fillmore and Williams: On Operator Ranges (1971)". In fact, the above proof is an adaptation of an argument that appears in the proof of Theorem 1.1 of this paper.

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The answer to your question is "No". It can be seen in the following way: If there exists an operator $S$ mentioned in the Question, then, using the standard techniques, one can show that $V$ has to be isomorphic to a quotient space of $X$. So it remains to show that there exists $X$ and an operator range in $X$ for which this condition fails. This can be done by using injective nuclear operators with non-closed range from any separable Banach space $V$ into $X$ and by picking $V$ and $X$ in such a way that $V$ is not a quotient of $X$. For example, let $X$ be a separable Hilbert space and $V$ be a separable Banach space which is not isomorphic to a Hilbert space.

Example 4.12 in Cross, R. W.; Ostrovskiĭ, M. I.; Shevchik, V. V. [Operator ranges in Banach spaces. I. Math. Nachr. 173 (1995), 91–114] is a much stronger example. In the same paper you can find more details on the proof sketched above.

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  • $\begingroup$ Thanks a lot for your answer and for the reference to your paper! By the way, is there a Part II of the paper? (I wasn't able to find it.) $\endgroup$ – Jochen Glueck May 22 at 20:48
  • $\begingroup$ Unfortunately we were unable to continue collaboration, and part II never appeared. $\endgroup$ – Mikhail Ostrovskii May 23 at 5:55

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