Proof. It is sufficient to show that both operators map weakly convergent sequences to strongly convergent sequences.

Let $\seq{x_n^*}\subset E^{*}$ such that $x_{n}^{*} \to x \in E^{*}$ weakly, then since $\wh \wien: E^{*} \to \complex$ is continuous with respect to sequential weak* convergence,

as $n \to \infty$. Thus $\iota^{*} x_{n}^{*} \to \iota^{*}x^{*}$ strongly in $H$.

To see the compactness of $\iota$, let $\seq{h_n}\subset H$ such that $h_{n} \to 0$ weakly. By the Uniform Boundedness Principle, there exists $R \ge 0$ such that $\sup_{n \in \nat}\norm{h_n}_{H} \le R$. Let $\seq{x_n^*}\subset E^{*}$ be a weak*-dense subset of $E^{*}$ with $\norm{x_n^*}_{E^*}= 1$ for all $n \in \nat$. Let $\eps > 0$, then by compactness of $\seq{\iota^*x^*_n}\subset H$, there exists $N \in \nat$ such that $\seq{\iota^*x^*_n}\subset \bigcup_{n = 1}^{N} B_{H}(\iota^{*}x^{*}_{n}, \eps)$. In which case,

Given that $h_{n} \to 0$ weakly, $\max_{1 \le k \le N}\angles{h_n, x_k^*}_{E} \to 0$ as $n \to \infty$. As $\eps > 0$ is arbitrary, $\norm{h_n}_{E} \to 0$ as $n \to \infty$.$\square$