Theorem 11.3.4 (Integration by Parts). Let $[a, b] \subset \real$, $E_{1}, E_{2}, H$ be TVSs over $F \in \RC$, and $E_{1} \times E_{2} \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
Let $f: [a, b] \to E_{1}$ and $G: [a, b] \to E_{2}$, then $f \in RS([a, b], G)$ if and only if $G \in RS([a, b], f)$. In which case,
Proof. Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_{F}(0)$, then there exits $P_{0} = \seqfz{x_j}\in \scp([a, b])$ such that $S(P, c, f, G) - \int_{a}^{b} fdG \in U$ for all $(P, c) \in \scp_{t}([a, b])$ with $P \ge P_{0}$. Let
then for any $(Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_{t}([a, b])$ with $Q \ge Q_{0}$,
by Lemma 11.3.3, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_{0}$, and $\int_{a}^{b} fdG - S(Q', d', G, f) \in U$.$\square$