Proposition 26.9.2.label Let $E$ be a separated locally convex space over $K \in \RC$, $U \subset K$ be open, $Y$ be a Hausdorff space, and $f: U \times Y \to E$. If $f$ is differentiable in the first variable and $\frac{df}{dx}\in C(U \times Y; E)$, then
\[\frac{f(x + h, y) - f(x, y)}{h}\to \frac{df}{dx}(x, y)\]
as $h \to 0$, uniformly on compact sets.
Proof. Let $A \subset U$ and $B \subset Y$ be compact, then by the Mean Value Theorem, for any $(x, y) \in A \times B$ and $h \in \real$ with $x + h$,
\begin{align*}&\frac{f(x + h, y) - f(x, y)}{h}- \frac{df}{dx}(x, y) \\&\in \overline{\text{Conv}}\bracs{\frac{df}{dx}(x + k, y) - \frac{df}{dx}(x, y) \bigg | k \in B_K(0, |h|)}\end{align*}
Let $\eps > 0$ such that $A + B_{K}(0, |\eps|) \subset U$, then since $\frac{df}{dx}\in C(U \times Y; E)$, $\frac{df}{dx}|_{(A + B_K(0, |\eps|)) \times B}$ is uniformly continuous[1]. Since $E$ is locally convex,
\[\frac{f(x + h, y) - f(x, y)}{h}- \frac{df}{dx}(x, y) \to 0\]
uniformly on $A \times B$.$\square$
- $K$ is a compact Hausdorff space, which comes with a unique uniform structure. keyboard_return
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