Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace - Part II

Question

This is a follow-up to this previous question, but under stronger assumptions.

Let $(X, \mu)$ be a (say, $\sigma$-finite) measure space, let $g \in L^2$ (say, over the real scalar field). Let $\tilde g: X \to \mathbb{R}$ be a measurable function and assume that there exists a norm dense vector subspace $D$ of $L^2$ with the following two properties:

1. The subspace $D$ is an operator range in $L^2$, i.e., there exists a complete norm on $D$ under which the injection $D \hookrightarrow L^2$ is continuous.

2. For every $f \in D$ the function $\tilde g f$ is integrable, and we have $\int \tilde g f \, d\mu = \int g f \, d\mu$.

Question. Does it follow that $\tilde g = g$ almost everywhere?

Remarks.

• Without the assumption that $D$ be an operator range, the answer is no. Two counterexamples were given by Piotr Hajlasz and Gro-Tsen in the answers to question linked at the beginning of the post.

• The complete norm on $D$ is not required to render $D$ a Hilbert space (although this might be an interesting variant of the question).

• As explained in the previous question, it suffices to show that $\tilde g \in L^2$.

• As also explained in the previous question, the answer is yes if $D$ is a lattice ideal in $L^2$ (even if $D$ is not assumed to be an operator range).

Answer 1

Gro-Tsen's answer to your previous question provides a counterexample if you define $D$ to be all vectors in $\ell_2$ that are of the form $\sum_n a_n f_n$, where $f_n = e_n + e_{n+1}$, $(e_n)$ is the unit vector basis for $\ell_2$, and $\sum |a_n| < \infty.$ $D$ is the range of a bounded linear operator from $\ell_1 $ into $\ell_2$.