Is every operator range a Baire space in the relative topology?

Question

Let $X$ be a Banach space and let $U\subseteq X$ be a (not necessarily closed) linear subspace. One says that $U$ is an operator range if there is another Banach space $E$, and a bounded linear map $T:E\to X$, such that $T(E)=U$.

Even though operator ranges are clearly not necessarily closed, they have many special properties, e.g. those highligted in Reference request: Baire's theorem for operator ranges.

In fact I was led to this post because I am after an answer to the following:

Question. Is every operator range a Baire space (in the sense that the intersection of countably many open dense sets is dense) in the relative topology?

The title of the post mentioned above is tantalizingly close but unfortunately it doesn't offer any answer to my question, neither was I able to find it anywhere else. Any ideas?

In case this question turns out to be senstitive to the kind of Banach spaces considered, I am really interested in separable-Hilbert-operator-ranges, that is, subspaces of separable Hilbert spaces which coincide with the range of a bounded linear map defined on a separable Hilbert space.

Answer 1

It follows from the open mapping theorem in the general version of theorem 2.11 in Rudin's Functional Analysis that

If the range $T(X)$ of continuous linear operator $T:X\to Y$ from an F-space to a topological vector space is Baire then $T$ is open onto its range which is closed in $Y$.

For the proof you just have to observe that every Baire spaces is of second category in itself so that you can apply the OMT to the corestriction $T:X\to T(X)$ where $T(X)$ has the relative topology from $Y$.