Assume $(X,\Vert\cdot\Vert)$ is a normed space with the dual space $X'$. I want to show that the polar set of the open disk $U_{r}^{X'}(0)$ is equal to the closed disk $K_{\frac{1}{r}}^{X}(0)$, e.i.: ${}^{\circ}(U_{r}^{X'}(0)) = K_{\frac{1}{r}}^{X}(0)$
I have already proven that ${}^{\circ}(K_{r}^{X'}(0)) = K_{\frac{1}{r}}^{X}(0)$ and from $N \subseteq M$ follows ${}^{\circ}M \subseteq {}^{\circ}N$. Hence $K_{\frac{1}{r}}^{X}(0) = {}^{\circ}(K_{r}^{X'}(0)) \subseteq {}^{\circ}(U_{r}^{X'}(0))$.
The polar set is defined as ${}^{\circ}M = \{x\in X \vert \forall y \in X': Re(y(x))\leq 1\}$ and for balanced sets this is equivalent to ${}^{\circ}M = \{x\in X \vert \forall y \in X': \vert y(x) \vert \leq 1\}$
Can someone help me with the second inclusion?
The converse follows from the fact that $\|x\|=\sup \{|y(x)|: \|y\| \leq 1\}$ $\,\,\,$(1). Let $x$ belong to the annihilator of $U_r^{X'}(0)$. Then $|y(x)| \leq 1$ whenever $\|y\|<r$. If $\|y\|\leq r$ then $\|(1-t)y\| <r$ for any $t \in (0,1)$ so we get $|(1-t)y(x)| \leq 1$. Letting $t$ decrease to $0$ we get $|y(x)| \leq 1$. Replacing $y$ by $ry$ we see that $|y(x)| \leq 1/r$ whenver $\|y\|\leq 1$. Hence $\|x\| \leq 1/r$ by (1).