# Characterization of operator ranges

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.)

$$\,$$

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 $$V$$.

$$\,$$

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.

## 1 Answer

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.

• 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.) – Jochen Glueck May 22 '19 at 20:48
• Unfortunately we were unable to continue collaboration, and part II never appeared. – Mikhail Ostrovskii May 23 '19 at 5:55