diff --git a/src/fa/lc/convex.tex b/src/fa/lc/convex.tex
index e3bb4ad..ab01dcc 100644
--- a/src/fa/lc/convex.tex
+++ b/src/fa/lc/convex.tex
@@ -7,6 +7,63 @@
     Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.
 \end{definition}
 
+\begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}]
+\label{lemma:convex-interior}
+    Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then
+    \[
+    \bracs{tx + (1 - t)y|t \in (0, 1)} \subset A^o
+    \]
+\end{lemma}
+\begin{proof}
+    Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^o \in \cn^o(y)$. Since $y \in \ol{A}$, $\alpha A^o \cap A \ne \emptyset$, so there exists $z \in A^o$ such that $\alpha z \in A$.
+
+    Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
+    \[
+    \mu z + (1 - \mu)\alpha z = \frac{\alpha z}{\alpha - 1} + \frac{(\alpha - 1 - \alpha)\alpha z}{\alpha - 1} = 0
+    \]
+    By (TVS1) and (TVS2),
+    \[
+    U = \underbrace{\bracs{\mu w + (1 - \mu)\alpha z|w \in A^o}}_{\subset A} \in \cn^o(0)
+    \]
+    so $0 \in A^o$.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:convex-gymnastics}
+    Let $E$ be a TVS over $K \in \RC$, $A, B \subset E$ be convex, then the following sets are convex:
+    \begin{enumerate}
+        \item $A^o$.
+        \item $\ol{A}$.
+        \item $A + B$.
+        \item For any $\lambda \in K$, $\lambda A$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By \ref{lemma:convex-interior}.
+
+    (2): Let $x, y \in \ol{A}$. By \ref{definition:closure}, there exists filters $\fF, \mathfrak{G} \subset 2^A$ such that $\fF$ converges to $x$ and $\mathfrak{G}$ converges to $y$. In which case,
+    \[
+    \fU = \bracs{tE + (1 - t)F|t \in [0, 1], E \in \fF, F \in \mathfrak{G}} \subset 2^A
+    \]
+    converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
+\end{proof}
+
+\begin{proposition}[{{\cite[II.1.3]{SchaeferWolff}}}]
+\label{proposition:convex-interior-closure}
+    Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$.
+\end{proposition}
+\begin{proof}
+    Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$,
+    \[
+    y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}
+    \]
+    by \ref{lemma:convex-interior}.
+\end{proof}
+
+
+
+
+
 \begin{definition}[Sublinear Functional]
 \label{definition:sublinear-functional}
     Let $E$ be a vector space over $K \in \RC$, then a \textbf{sublinear functional} is a mapping $\rho: E \to \real$ such that: