The dual of $E$, equipped with weak topology (induced by $E'$) is identical to $E'$

Question

Let $E$ be a locally convex Hausdorff topological vector space.

We have proofed, that $E \cong (E', \sigma(E', E))'$, the latter is a dual of a weak dual of $E$. The corollary says, that the dual of $E$, equipped with $\sigma(E, E')$, is identical to $E'$, but I don't understand, how the first theorem implies this statement.

Answer 1

If we apply your statement to $(E', \sigma(E',E))$, we end up with \begin{align}(E', \sigma(E',E)) &\cong ((E',\sigma(E',E))',\sigma((E',\sigma(E',E))',(E',\sigma(E',E)))'\\ &\cong (E,\sigma(E,E'))'. \end{align}