Bourbaki

Proposition 11

Tag 09B1, permanent. Cite it as [Bourbaki, Tag 09B1].

Integration, chapter VI, § 2. Vectorial measures . "

— Let F,G,HF, G, H be three Banach spaces, Φ\Phi a continuous bilinear mapping of F×GF \times G into HH. Let mm be a majorizable vectorial measure on TT, with values in GG. Then there exists one and only one continuous linear mapping IΦ,mI_{\Phi,m} of L‾F1(∣m∣)\overline{\mathscr{L}}^1_F(|m|) into HH such that, for every z∈Fz \in F and every numerical function hh integrable for ∣m∣|m|, one has IΦ,m(hz)=Φ(z,∫hdm)I_{\Phi,m}(hz) = \Phi(z, \int h dm). *Moreover,

∣IΦ,m(f)∣≤∥Φ∥∫∣f∣d∣m∣ |I_{\Phi,m}(f)| \leq \| \Phi \| \int |f| d|m|

for every function f∈L‾F1(∣m∣)f \in \overline{\mathscr{L}}^1_F(|m|).

If there exists such a mapping, its value for a step function ff over the ∣m∣|m|-integrable sets is uniquely determined: for, it is known that one can then write f=∑iaiφXif = \sum_i a_i \varphi_{X_i}, where the XiX_i are ∣m∣|m|-integrable and disjoint, and the ai∈Fa_i \in F (Ch. IV, §4, No. 9, Lemma). The value of IΦ,m(f)I_{\Phi,m}(f) must therefore be equal to ∑iΦ(ai,∫φXidm)\sum_i \Phi(a_i, \int \varphi_{X_i} dm). Now, we have (No. 3, Prop. 5)

∣∑iΦ(ai,∫φXidm)∣≤∥Φ∥⋅∑i∣ai∣⋅∣m∣(Xi)=∥Φ∥∫∣f∣d∣m∣, \left| \sum_i \Phi \left( a_i, \int \varphi_{X_i} dm \right) \right| \leq \| \Phi \| \cdot \sum_i |a_i| \cdot |m|(X_i) = \| \Phi \| \int |f| d|m|,

which shows first of all that the element ∑iΦ(ai,∫φXidm)\sum_i \Phi(a_i, \int \varphi_{X_i} dm) of HH does not depend on the particular expression of ff in the form ∑iaiφXi\sum_i a_i \varphi_{X_i}, hence that we may denote it by IΦ,m(f)I_{\Phi,m}(f). One verifies immediately that the mapping IΦ,mI_{\Phi,m} so defined is linear on the space EF\mathscr{E}_F of step functions over the ∣m∣|m|-integrable sets: for, it suffices to write two functions f,gf, g of EF\mathscr{E}_F in the form f=∑iaiφXif = \sum_i a_i \varphi_{X_i} and g=∑ibiφXig = \sum_i b_i \varphi_{X_i} with the same finite family of pairwise disjoint ∣m∣|m|-integrable sets XiX_i (which is possible by virtue of the Lemma of Ch. IV, §4, No. 9). The inequality (5) then shows that IΦ,mI_{\Phi,m} is continuous on EF\mathscr{E}_F, and since this subspace is dense in L‾F1\overline{\mathscr{L}}^1_F (Ch. IV, §4, No. 10, Cor. 1 of Prop. 19 and Ch. V, §1, No. 3), one deduces from this the existence and uniqueness of IΦ,mI_{\Phi,m}, as well as the inequality (4).

One says that IΦ,m(f)I_{\Phi,m}(f) is the integral of ff with respect to mm (relative to the bilinear mapping Φ\Phi); when the value of the bilinear mapping Φ\Phi at the point (x,y)(x,y) is denoted xyxy, we shall write ∫f dm\int f\,dm instead of IΦ,m(f)I_{\Phi,m}(f).

With the notations of No. 6, the integral ∫⟨f,g⟩ dμ\int \langle f, g \rangle\,d\mu is none other than IΦ,m(f)I_{\Phi,m}(f) with Φ(x,x′)=⟨x,x′⟩\Phi(x,x') = \langle x, x' \rangle and m=g⋅μm = g \cdot \mu.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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