Proposition 13.2.4.label Let $K\suf\R^{n}$ be a convex body then every boundary point admits a supporting hyperplane.
Proof. Let $x_{0}\in\pd K=\pd (K^{c})$. Let $\curl{y_i}_{i\in\N}\suf K^{c}$ such that $y_{i}\to x_{0}$. BY Theorem 13.1.25 each $y_{i}\nin K$ can be strongly separated from $K$. So for each $i\in\N,\exists a_{i}\in \mathbb{S}^{n-1},\al_{i}$ such that
\begin{align*}\inner{x}{a_i}<\inner{y_i}{a_i}&&\fall x\in K\end{align*}
By compactness of $\mathbb{S}^{n-1}$ there exists a converging subsequence $a_{i_j}\to a\mathbb{S}^{n-1}$. For each $x\in K$ passing to the limit gives
\begin{align*}\inner{x}{a}\leq \inner{x_0}{a}\end{align*}
Since $x_{0}\in K$ we have $H\define \curl{x\in\R^n:\inner{x}{a}=\inner{x_0}{a}}$ is a supporting hyperplane of $K$ through $x_{0}$.$\square$
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