Proposition 19.1.5.label Let $E$ be a vector space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then for any $h \in E$,
\[\lim_{t \downto 0}\frac{f(x + th) - f(x)}{t}= \inf_{t > 0}\frac{f(x + th) - f(x)}{t}\]
exists in $[-\infty, \infty]$.
Proof. Let $0 < s \le t$, then since $f$ is convex,
\begin{align*}f(x + sh)&\le \paren{1 - \frac{s}{t}}f(x) + \frac{s}{t}f(x + th) \\ f(x + sh) - f(x)&\le \frac{s}{t}[f(x + th) - f(x)] \\ \frac{f(x + sh) - f(x)}{s}&\le \frac{f(x + th) - f(x)}{t}\end{align*}
$\square$
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