Theorem 26.2.8 (Interchange of Limits and Derivatives).label Let $E$ be a TVS over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$. Let $\fF \subset 2^{\tilde D_\sigma^n(U; F)}$ be a filter such that:
- (a)
There exists $f: U \to F$ such that $\fF \to f$ pointwise.
- (b)
For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_{\sigma}(E; F)$ such that for all $x \in U$, and $A \in \sigma$ with $x + [0, 1]A \subset U$, $D_{\sigma}^{k}(\fF) \to f^{(k)}$ uniformly on $x + [0, 1]A$.
then $f \in \tilde D_{\sigma}^{n}(U; F)$ and $D^{k}_{\sigma} f = f^{(k)}$ for all $1 \le k \le n$. In particular, if $\sigma$ is saturated, then $(b)$ may be replaced by
- (b)
For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_{\sigma}(E; F)$ such that $D_{\sigma}^{k}(\fF) \to f^{(k)}$ uniformly on every $A \in \sigma$.
Proof. Assume without loss of generality that $n = 1$. For any $\varphi \in \tilde D^{1}_{\sigma}(U; F)$, $x \in U$, and $h \in E$ such that $x + h \in U$,
Since $\fF \to f$ pointwise, for any $S \in \fF$,
By the Mean Value Theorem, for any $g \in \tilde D^{1}_{\sigma}(U; F)$,
Hence
so for any $t \in (0, 1)$ and $A \in \sigma$,
and
In addition, since $D_{\sigma}(\fF) \to f^{(1)}$ pointwise,
as well.
Now, let $V \in \cn_{F}(0)$ be convex and circled. Using assumption (b), let $S \in \fF$ such that for any $\varphi \in S$,
Fix $\varphi \in S$, then as $\varphi$ is differentiable at $x$, there exists $\delta \in (0, 1)$ such that
for all $t \in (0, \delta)$.
So
for all $t \in (0, \delta)$. Therefore $f$ is $\tilde \sigma$-differentiable at $x$ with $D_{\sigma} f(x) = f^{(1)}(x)$.$\square$