Let $V$ be a topological vector space over some locally compact field $\Bbb K$. Let $V'$ denote all continuous functionals on $V$, the weak topology is defined to be the coarsest topology on $V$ making the elements of $V'$ continuous.
Definition (weakly-complete-1)
$V$ is called weakly complete if it is complete in the uniform structure given by the weak topology.
I have seen another definition:
Definition (weakly-complete-2)
A topological vector-space $V$ is called weakly complete if it is isomorphic to $\Bbb K^J$ for some set $J$ (here take the product topology).
I think the mumbo-jumbo you do for $\Bbb K \in\{\Bbb R, \Bbb C\}$ also works for the other cases to show that $(\Bbb K^J, \bigoplus_J \Bbb K)$ are a dual pair which you can use to show that if $V$ is weakly-complete-1 if it is weakly-complete-2. Is there a way to turn this around?
Question
If $V$ is a Banach space that is weakly-complete-1, does it follow that $V$ equipped with its weak topology is weakly-complete-2? Meaning $(V,\tau_{\text{weak}})\cong \Bbb K^J$ for some $J$?
I think a reflexive Banach-space is weakly-complete-1, so every reflexive Banach-space must be (weakly) isomorphic to $\Bbb K^J$ for some $J$ if this were true. For separable Banach-spaces you may only have $J=\Bbb N$ for cardinality reasons, so this would imply that all separable reflexive Banach-spaces are weakly isomorphic, that seems unlikely!
If the answer is false, is there some other connection between the two notions?
If $E$ and $F$ are Banach spaces and $\phi:E\to F$ is weakly continuous, then it is norm-continuous by the uniform boundedness principle.
Therefore the reasoning given by the OP implies that all separable reflexive Banach spaces are in fact norm-isomorphic, which is patently false. The conclusion is that weakly-complete-2 is not equivalent to weakly-complete-1.