Let u be a continuous and positive endomorphism of the hilbertian space E. Let
Tr(u)=p∈Fsupτ(pup) (upper bound in R+), where F is the set of all finite rank orthoprojectors on E. We say that Tr(u) is the trace of u.
Let p be the orthoprojector from E onto a finite dimensional vector subspace of E, and let (x1,...,xm) be an orthonormal basis of F. We have established the relation
τ(pup)=i=1∑m⟨xi∣u(xi)⟩. Consequently, we can define the trace by the formula
Tr(u)=x1,...,xmsupi=1∑m⟨xi∣u(xi)⟩, where
(x1,...,xm) ranges over the set of all finite orthonormal sequences of vectors of
E.
By lemma 3 (V, p. 48), we have
Tr(u)=i∈I∑⟨ei∣u(ei)⟩ for every orthonormal basis
(ei)i∈I of
E. From this, we deduce
Tr(u+v)=Tr(u)+Tr(v) Tr(λu)=λ⋅Tr(u) for all continuous and positive endomorphisms
u and
v of
E and for every real number
λ≥0 (we make the convention
0.(+∞)=0 in (27)). Let
ϕ be an isomorphism from
E onto a hilbertian space
F; since
ϕ transforms every orthonormal basis of
E into an orthonormal basis of
F, we get from (25) that
Tr(ϕuϕ−1)=Tr(u).
Let (uα)α∈A be a non-empty directed increasing and bounded family of continuous and positive endomorphisms of E; let u=supαuα, then ⟨x∣u(x)⟩=supα⟨x∣uα(x)⟩ for all x∈E (V, p. 46, prop. 13). We have Tr(u)=supJ⊂I∑i∈J⟨ei∣u(ei)⟩, where J ranges over all finite subsets of I, hence
Tr(u)=αsupTr(uα)foru=αsupuα.
Let pF be the orthoprojector from E onto the hilbertian subspace F; there exists an orthonormal basis (ei)i∈I of E and a subset J of I, such that (ei)i∈J is an orthonormal basis of F. We have Tr(pFupF)=∑i∈J⟨ei∣u(ei)⟩. This formula has two consequences : firstly, we have Tr(pFupF)≤Tr(u); secondly, taking u=1E, we get
Tr(pF)={dimF+∞if F is finite dimensionalif not.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
Nothing.