Closed linear span of the range of $\boldsymbol f$ and Pettis integrals of $\boldsymbol f$

Question

Let $X$ be a noncompact locally compact topological space, let $H$ by a complex Hilbert space and let $\boldsymbol f:X\to H$ be a continuous function vanishing at infinity whose support is equal to $X$.

It can be shown that closed convex hull of $\boldsymbol f(X)$ in $H$ is equal to the set of all the vectors $\int_X\boldsymbol f\,d\mu$ as $\mu$ ranges over the set of all Radon probability measures on $X$ (the proof can be adapted from Bourbaki INT III, § 3, prop. 5). So the linear subspace of $H$ spanned by the closed convex balanced hull of $\boldsymbol f(X)$ in $H$ is equal to the set of all the vectors $\int_X\boldsymbol f\,d\mu$ as $\mu$ ranges over the set of all bounded complex Radon measures on $X$.

For Baire reasons, if $\boldsymbol f(X)$ spans an infinite-dimensional linear subspace of $H$, then the closed linear subspace of $H$ it spans is strictly larger than the linear subspace of $H$ spanned by its closed convex balanced hull. Hence, I would like to know if the entirety of the closed linear span of $\boldsymbol f(X)$ in $H$ can be obtained by using unbounded complex Radon measures.

More precisely, my question is:

If $\xi$ is a vector in the closed linear space of $\boldsymbol f(X)$ in $H$, does there necessarily exist a complex Radon measure $\mu$ on $X$ such that $\boldsymbol f$ is Pettis-$\mu$-integrable and such that $\xi=\int_X\boldsymbol f\,d\mu$ ?

Answer 1

Let $X=\{1/n:n\in\mathbb{N}\}\cup\{0\}$ with the obvious topology. As $X$ is compact, any Radon measure is finite. Let $(e_n)_{n\in\mathbb N}$ be a ONB of a Hilbert space $H$. Define $f:X\to H$ by $f(n)=\frac1{n^2}e_n$, $f(0)=0$ and consider the vector $v=\sum_{n=1}^\infty \frac1ne_n$. If $v=\int_Xf(x)\ d\mu(x)$, then $\mu(n)=n$ which is not a Radon measure.