Let $(X,\tau)$ be a locally convex space and $(X^{*},\tau_{s})$ be its topological dual space equipped with the strong topology. Denote by $S(X,X^{*})$ the collection of operators from $X$ to $X^{*}$ of the form $u^{*}Tu$ where $T$ is symmetric, positive, nuclear operator in a Hilbert space $\mathscr{H}$ and $u: X \to \mathscr{H}$ is continuous and linear.

In this paper, Theorem 2.1 (which is the so-called Bochner-Minlos Theorem for locally convex spaces) state the following:

Theorem: Let $C: X \to \mathbb{C}$ be a positive-definite functional with $C(0) = 1$. If $C$ is $\tau_{s}-(X,X^{*})$ continuous at $x=0$, then there exists a Radon measure $\mu$ on $(X^{*}, \mathbb{B}(X^{*}))$ such that $C(x)$, $x\in X$ is the characteristic functional of $\mu$.

Here $\mathbb{B}(X_{s}^{*})$ is the Borel $\sigma$-algebra on $(X^{*}, \tau_{s})$ and $\tau_{s}-(X,X^{*})$ is the weakest vector topology in $X$ with respect to which all quadractic forms $x \in X \mapsto \langle u^{*}Tu,x \rangle \in \mathbb{R}$, $u^{*}Tu \in S(X,X^{*})$ are continous.

The paper do not prove the result but refers to Theorem 4.1 (section VI.4.2) on this book. However, the cited theorem states the following:

Theorem: If a positive-definite functional $\chi: X^{*} \to \mathbb{C}$, $\chi(0) = 1$ is continuous in the topology $\tau_{s}-(X^{*},X)$ then it is the characteristic functional of a Radon probability measure on $X$.

The problem is that the spaces $X$ and $X^{*}$ are exchanged in both theorems. I found it particularly curious because I thought the theorem was only valid for measures on the dual space, as stated in the first theorem above. At first glance I thought that if we take $X$ to be $X^{*}$ in the first theorem we would have the second theorem and vice-versa. However, as far as I know (and I'm not an expert on locally convex spaces) it is not always true that $X^{**} = X$.

Question 1: So, what am I missing here? Are the hypothesis sufficient to garantee $X^{**} = X$? Does the theorem hold in both directions (i.e. when the roles of $X$ and $X^{*}$ are exchanged)?

Question 2: If we replace $\tau_{s}-(X,X^{*})$ by continuous (i.e. in the first theorem $C$ is continuous and in the second $\chi$ is continuous), do both theorems remain valid? When $X = \mathscr{S}(\mathbb{R}^{d})$ I know the first theorem holds in this case where $C$ is continuous, but I've never seen the second version of the theorem in this case. This is why I thought the second theorem would not hold in general, at least when $\chi$ is just continuous.

Let $(X,\tau)$ be a locally convex space and $(X^{*},\tau_{s})$ be its topological dual space equipped with the strong topology. Denote by $S(X,X^{*})$ the collection of operators from $X$ to $X^{*}$ of the form $u^{*}Tu$ where $T$ is symmetric, positive, nuclear operator in a Hilbert space $\mathscr{H}$ and $u: X \to \mathscr{H}$ is continuous and linear.

In Jianglun - Generalized Bochner theorem on locally convex spaces, Theorem 2.1 (which is the so called Bochner–Minlos Theorem for locally convex spaces) states the following:

Theorem: Let $C: X \to \mathbb{C}$ be a positive-definite functional with $C(0) = 1$. If $C$ is $\tau_{s}\text-(X,X^{*})$ continuous at $x=0$, then there exists a Radon measure $\mu$ on $(X^{*}, \mathbb{B}(X^{*}))$ such that $C(x)$, $x\in X$ is the characteristic functional of $\mu$.

Here $\mathbb{B}(X_{s}^{*})$ is the Borel $\sigma$-algebra on $(X^{*}, \tau_{s})$ and $\tau_{s}\text-(X,X^{*})$ is the weakest vector topology in $X$ with respect to which all quadractic forms $x \in X \mapsto \langle u^{*}Tu,x \rangle \in \mathbb{R}$, $u^{*}Tu \in S(X,X^{*})$ are continous.

The paper does not prove the result but refers to Theorem 4.1 (section VI.4.2) on Vakhania, Tarieladze, and Chobanyan - Probability Distributions on Banach Spaces. However, the cited theorem states the following:

Theorem: If a positive-definite functional $\chi: X^{*} \to \mathbb{C}$, $\chi(0) = 1$ is continuous in the topology $\tau_{s}\text-(X^{*},X)$ then it is the characteristic functional of a Radon probability measure on $X$.

The problem is that the spaces $X$ and $X^{*}$ are exchanged in both theorems. I found it particularly curious because I thought the theorem was only valid for measures on the dual space, as stated in the first theorem above. At first glance I thought that if we take $X$ to be $X^{*}$ in the first theorem we would have the second theorem and vice-versa. However, as far as I know (and I'm not an expert on locally convex spaces) it is not always true that $X^{**} = X$.

Question 1: So, what am I missing here? Are the hypothesis sufficient to guarantee $X^{**} = X$? Does the theorem hold in both directions (i.e. when the roles of $X$ and $X^{*}$ are exchanged)?

Question 2: If we replace "$\tau_{s}\text-(X,X^{*})$ continuous" by "continuous" (i.e. in the first theorem $C$ is continuous and in the second $\chi$ is continuous), do both theorems remain valid? When $X = \mathscr{S}(\mathbb{R}^{d})$ I know the first theorem holds in this case where $C$ is continuous, but I've never seen the second version of the theorem in this case. This is why I thought the second theorem would not hold in general, at least when $\chi$ is just continuous.