Lemma 18.1.3.label 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.
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 $f|_{\bracs{f < \infty}}$.$\square$
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