Theorem 18.3.3. Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $f \in C([a, b]; E)$, $g \in C([a, b]; \real)$, $N \subset [a, b]$ be at most countable, and $B \subset E$ be a closed convex set such that:
$f, g$ are right-differentiable on $[a, b] \setminus N$.
For each $x \in [a, b] \setminus N$, $D^{+}f(x) \in D^{+}g(x)B$.
$g$ is non-decreasing.
then
\[f(b) - f(a) \in [g(b) - g(a)]B\]
Proof. Let $\phi \in E^{*}$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,
\begin{align*}\lim_{t \downto 0}\frac{\phi(f(x + t) - f(x))}{t}&\le \lim_{t \downto 0}\sup_{z \in B}\frac{\phi(g(x + t) - g(x)z)}{t}\\ D^{+}(\phi \circ f)(x)&\le \sup_{z \in B}\phi(z) \cdot \lim_{t \downto 0}\frac{\phi(g(x + t) - g(x))}{t}\\ D^{+}(\phi \circ f)(x)&\le D^{+}g(x) \cdot \sup_{z \in B}\phi(z)\end{align*}
By Lemma 18.3.2,
\[\phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)\]
Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the Hahn-Banach theorem, there exists $\phi \in E^{*}$ such that
\[\phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)\]
which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.$\square$