diff --git a/src/fa/convex/def.tex b/src/fa/convex/def.tex
index 4a037e1..94a83ab 100644
--- a/src/fa/convex/def.tex
+++ b/src/fa/convex/def.tex
@@ -41,6 +41,31 @@
     
 \end{proof}
 
+\begin{lemma}
+\label{lemma:convex-domain}
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then $f$ is convex if and only if $\bracs{f < \infty}$ is convex and $f|_{\bracs{f < \infty}}$ is a convex function. 
+\end{lemma}
+\begin{proof}
+    If $f$ is convex, then for any $x, y \in \bracs{f < \infty}$ and $t \in [0, 1]$, 
+    \[
+    f((1 - t)x + ty) \le (1 - t)f(x) + tf(y) < \infty
+    \]
+
+    so $\bracs{f < \infty}$ is convex, and the restriction $f|_{\bracs{f < \infty}}$ is a convex function. 
+
+    On the other hand, for any $x, y \in E$ and $t \in [0, 1]$, if $f(x) = \infty$ or $f(y) = \infty$, then
+    \[
+    f((1 - t)x + ty) \le = \infty (1 - t)f(x) + tf(y)
+    \]
+
+    Otherwise, $x, y \in \bracs{f < \infty}$, and
+    \[
+    f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)
+    \]
+
+    by convexity of $f|_{\bracs{f < \infty}}$. 
+\end{proof}
+
 \begin{lemma}
 \label{lemma:convex-reverse}
     Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$,