Dual of "Dual of Fréchet Space with Weak*-Topology" Equals Dual of "Dual of Fréchet Space with Topology of Compact Convergence" Let $X$ be a Fréchet space. I know that:


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*Closed convex, balanced, hulls of compact subsets of $X$ are compact. 


Let ${X^*}$ denote the (topological) dual space. I know that:


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*${X^*}$ is also the dual of "$X$ with the weak topology (induced by ${X^*}$)".

*Weakly bounded sets in $X$ are bounded sets in $X$.
Now, ${X^*}$ can be given the weak*-topology (induced by $X$). I know that:


*

*The dual of "${X^*}$ with the weak* topology" is $X$ (as vector spaces).


${X^*}$ can also be given the topology of uniform convergence on compact subsets of $X$. 
Is it possible to prove directly - without delving into the theory of locally convex topologies on dual pairs and the Mackey-Arens theorem - that in this special case, starting with a Fréchet space $X$, the dual of "${X^*}$ with the topology of uniform convergence on compact subsets of $X$" is $X$ (as vector spaces)?
 A: The answer is yes - there is a direct proof. Here is a sketch of a proof - steps (iv) and (v) below are adapted from Joseph L. Taylor's "Several Complex Variables with Connections to Algebraic Geometry and Lie Groups", Section 11.11. All the parenthetical references, like "(3.7)", are to Rudin's Functional Analysis book (Second Edition); Rudin's book includes neither the theory of locally convex topologies on dual pairs nor the Mackey-Arens theorem. 
(i) If $X$ is a Fréchet space then the topology of compact convergence on ${X^*}$ is the same as the locally convex topology induced by the seminorms ${\sup _{x \in K}}\left| {f(x)} \right|$, $K$ relatively compact. This follows from (1.34)-(1.38). Let $X_c^*$ denote $X^*$ with these coincident topologies.
(ii) If $X$ is a Fréchet space then the closed convex balanced hull of a compact set is compact. This is (3.20), tweaked to included balanced sets.
(iii) For any locally convex space $Y$, (3.11) gives that the $Y$-topology, or weak*-topology, denoted $Y_w^*$, makes ${Y^*}$ into a locally convex space whose dual space is $Y$: $Y \cong {\left( {Y_w^*} \right)^*}$ as vector spaces; in this case we identify $Y$ with its image in ${\left( {Y_w^*} \right)^*}$, consisting of the evaluation functionals. The $Y$-topology has local subbase elements of the form $\left\{ {f \in {Y^*}:\left| {f(y)} \right| < r} \right\}$. 
(iv) If $Y$ is a Fréchet space then a local base element for the topology of compact convergence is $\left\{ {f \in {Y^*}:{{\sup }_{y \in K}}\left| {f(y)} \right| < r} \right\}$, $K$ relatively compact in $Y$. Evidently, the $Y$-topology is weaker than the topology of compact convergence from (iii), so a continuous linear functional on $Y_w^*$ must be continuous on $Y_c^*$. If $f(x) = f(y)$ for all $f \in {Y^*}$, then $x = y$ by (3.4). Putting these facts together with (3.14) gives that there is an embedding $Y \cong {\left( {Y_w^*} \right)^*} \subset {\left( {Y_c^*} \right)^*}$ via $y \mapsto {\Lambda_y}$ where ${\Lambda_y}(f) = f(y)$, or, if we identify $y$ with ${\Lambda_y}$, we simply write $y(f) = f(y)$. If ${\left( {Y_c^*} \right)^*}$ is given the weak*-topology, then this map is continuous: a local subbase element of $\left( {Y_c^*} \right)_w^*$ has the form $\left\{ {\Lambda  \in {{\left( {Y_c^*} \right)}^*}:\left| {\Lambda (f)} \right| < r} \right\}$ for some $f \in {Y^*}$; the inverse image of this set under the embedding is $\left\{ {y \in Y:\left| {f(y)} \right| < r} \right\}$, which is open in $Y$. In particular, a compact subset $K$ in $Y$ maps to a compact subset of $\left( {Y_c^*} \right)_w^*$.
(v) For a Fréchet space $X$, we prove that $X \cong {\left( {X_w^*} \right)^*}={\left( {X_c^*} \right)^*}$: let $\lambda  \in {(X_c^*)^*}$, then there is a $0$-neighborhood $V$ in $X_c^*$ such that $\left| {\lambda (f)} \right| < 1$ for all $f \in V$. We may assume that the neighborhood $V$ has the form $\left\{ {f \in X_c^*:\mathop {\sup }\limits_{x \in K} \left| {f(x)} \right| < 1} \right\}$ for some relatively compact $K$. Without loss of generality, by (ii) we may assume that $K$ is compact, convex, and balanced; by (iv), $K$ is compact, convex, and balanced in $\left( {X_c^*} \right)_w^*$. 
We know that every continuous linear functional on $\left( {X_c^*} \right)_w^*$ is given by an element of $X_c^*$, because ${\left( {\left( {X_c^*} \right)_w^*} \right)^*}$ can be identified with $X_c^*$ by (iii). If $\lambda $ is not an element of $K$, then it follows by the convex separation theorem (3.7) that there is an $f \in {\left( {\left( {X_c^*} \right)_w^*} \right)^*} \equiv X_c^*$ (identification) such that $\left| {f(\lambda )} \right| > 1$ and $\left| {f(x)} \right| \le 1$ for all $x \in K$; multiplying by an appropriate scalar, we can assume without loss of generality that $\left| {f(\lambda )} \right| > 1$ and $\left| {f(x)} \right| < r < 1$ for all $x \in K$. But $f \in X_c^*$, so $f \in V$. This is a contradiction, since $\left| {f(\lambda )} \right| = \left| {\lambda (f)} \right| < 1$ if $f \in V$. We conclude that $\lambda $ does belong to $K$, and, in particular, $\lambda $ belongs to $X$.
