Complete dual of bornological space

If I understand properly, I doubt very much that this is true.

I the article On different types of non-distinguished Frechet spaes (Note di Mat. 10 (1990), 149-165), Bonet, Dierolf, and Fernandez write that for a Frechet space $X$ with dual $(X',\beta(X',X))$ the Mackey topology $\mu(X',X'')$ is bornological if and only if every linear form on $X'$ which is bounded on bounded sets is already continuous. Moreover, they show examples that this not always the case. It thus seems to me that $(E,\tau)=(X',\mu(X',X''))$ is a counterexample.

Jochen is quite right. I have another example, just using any irreflexive Banach space $A$. The space $E = (A^*,\mu(A^*,A))$ is Mackey, by definition. The bounded sets in $E$ are the same as the norm-bounded sets, because $A$ is Banach, and therefore barrelled, so all dual topologies on $A^*$ have the same bounded sets (because $\sigma(A^*,A)$-bounded $\Leftrightarrow$ equicontinuous). So the canonical embedding $i(A)$ of $A$ in the strong dual of $E$ is an isomorphism, and so $E^* \cong A$ is complete.

But $E$ is not bornological, because by the same characterization of the bounded sets of $E$, the bounded linear functionals on $E$ are exactly $A^{**}$, which contains $E^* = i(A)$ as a proper subspace by the assumption that $A$ be irreflexive.

I think this also shows that Gach's proof of (4) $\Rightarrow$ (1) is at fault when he says "apply (4.1.5)", because if we apply his proof to $E$, the $F$ obtained in the proof will be $A^*$ with its norm topology, and $A^*$ and $E$ don't have the same set of continuous linear functionals.