Let G be a Hilbert space and v∈Lc(F;G). Let n∈IE∩IF.
a) Let r∈R+∗. We have
i=0∑nαi(v∘u)r⩽i=0∑nαi(v)rαi(u)r
b) Let p,q,r∈R+∗ be such that p1+q1=r1. Then
(i=0∑nαi(v∘u)r)1/r⩽(i=0∑nαi(v)p)1/p(i=0∑nαi(u)q)1/q
c) Suppose that F = E. For every integer m⩾2, we have
i=0∑nαi(um)⩽(i=0∑nαi(u)2)m/2
Let r∈R+∗ and let g be the function defined on R+ by g(x)=xr. The function g∘ exp is convex (Lemma 6), and therefore assertion a) follows from Prop. 11 applied to the function g.
Assertion b) follows from a) and Hölder's inequality (INT, I, Prop. 4).
Let m⩾2 be an integer. Apply b) with r=1,p=q=2 and v=um−1. We find
i=0∑nαi(um)⩽(i=0∑nαi(um−1)2)1/2(i=0∑nαi(u)2)1/2
Let us prove c) by induction on m⩾2. The preceding inequality establishes the assertion when m=2. Suppose that m⩾3 and that the assertion is valid for m−1; since one has
i=0∑nαi(um−1)2⩽(i=0∑mαi(um−1))2
the above inequality implies
i=0∑nαi(um)⩽(i=0∑nαi(um−1))1/2(i=0∑nαi(u)2)1/2⩽(i=0∑nαi(u)2)m/2
applying the induction hypothesis.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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