L$ with $\|u_k\|=1$ as $k\in\N$ and observe that $$\label{eq1Lm1.5wk}
\langle u_k,x\rangle<\la u_k,x_k\ra\le\|u_k\|\cdot\|x_k\|=\varepsilon_k\;\mbox{\rm for all }\;x\in\Omega,$$ where $\varepsilon_k:=\|x_k\|\downarrow 0$ as $k\to\infty$. Since $\{u_k\}$ is bounded, we let $k\to\infty$ in (\[eq1Lm1.5wk\]) and suppose without loss of generality that $u_k\xrightarrow{w}a\in L$, where the symbol “$w$" indicates the weak convergence in the Hilbert space $X$. This implies that $$\label{eq2Lm1.5}
\sup\big\{\langle a,x\rangle\;\big|\;x\in\Omega\big\}\le 0.$$

To verify further the strict inequality $$\inf\big\{\langle a,x\rangle\;\big|\;x\in\Omega\big\}<0,$$ it suffices to show that there exists $\ox\in\Omega$ with $\langle a,\ox\rangle<0$. Arguing by contradiction, suppose that $\langle a,x\rangle\ge 0$ for all $x\in\Omega$ and deduce from that $\langle a,x\rangle=0$ whenever $x\in\Omega$. Since $a\in L=\overline{\aff(\Omega)}$, there exists a sequence $a_j\to a$ as $j\to\infty$ with $a_j\in\aff(\Omega)$. The latter inclusion can be equivalently rewritten as $$a_j=\sum_{i=1}^{m_j}\lambda^j_i\omega^j_i\;\mbox{ with }\;\sum_{i=1}^{m_j}\lambda^j_i=1\;\mbox{ and }\;\omega^j_i\in\Omega\;\mbox{ for }\;i=1,\ldots,m_j,$$ which readily implies the equalities $$\Vert a_j\Vert^2=\langle a_j,a_j\rangle=\sum_{i=1}^{m_j}\lambda^j_i\langle a_j,\omega^j_i\rangle=0,$$ and thus $a_j=0$. The passage to the limit as $j\to\infty$ gives us $a=0$.

Now we apply the result of [@z Theorem 3.1.2], which ensures by using the existence of $b_k\in\Omega$ and $u_k\in X$ such that $$\label{eq3Lm1.5}
u_k\in N(b_k;\Omega),\;\|b_k\|\le\sqrt{\varepsilon_k},\;\mbox{ and }\;\|u_k-w_k\|\le\sqrt{\varepsilon_k}.$$ Since $\|w_k\|=1$, it follows from that $\|u_k\|\to 1$. Furthermore, we get from $w_k\xrightarrow{w}0$, $\varepsilon_k\downarrow 0$, and that $u_k\xrightarrow{w}0$ as $k\to\infty$. Remembering that $\Omega$ enjoys the SNC property, 