— Let H(E) be the set of all continuous hermitian endomorphisms of the hilbertian space E. Let F be a non-empty, directed increasing and bounded subset of H(E).
(i) The set F has an upper bound u0 in H(E); we have
⟨x∣u0(x)⟩=u∈Fsup⟨x∣u(x)⟩for allx∈E.
(ii) The filter of sections of F converges to u0 in the space L(E) endowed with the topology of simple convergence.
Let Σ be the filter of sections of F; for every u∈H(E), let Φu be the continuous hermitian form on E defined by
Φu(x,y)=⟨x∣u(y)⟩. Let
Ψu(x)=Φu(x,x) for
u∈H(E) and
x∈E. By the polarization formulas (V, p. 2), we have
(19)4Φu(x,y)(20)4Φu(x,y)=Ψu(x+y)−Ψu(x−y)=Ψu(x+y)−Ψu(x−y)−iΨu(x+iy)+iΨu(x−iy)if K=Rif K=C.
For every x∈E, the mapping u↦Ψu(x) from into R is increasing and bounded, hence has a limit with respect to Σ. By the preceding formulas, the limit
u,ΣlimΦu(x,y)=Φ(x,y) exists for every pair
(x,y) of elements of
E. It is clear that
Φ is a hermitian form on
E. If
v1∈F and
v2 is a bound of
F, the hermitian forms
f1=Φ−Φv1 and
f2=Φv2−Φ are positive; there exists a real number
M≥0 such that
f1(x,x)+f2(x,x)=Φv2−v1(x,x)≤M∥x∥2, hence
f1(x,x)≤M∥x∥2,f2(x,x)≤M∥x∥2(x∈E); consequently the semi-norms
x↦fi(x,x)1/2 are continuous on
E. Since
f2−f1=Φv2+Φv1−2Φ, we conclude that
x↦Φ(x,x) is a continuous function on
E, and by formulas (19) and (20), that
Φ is continuous on
E×E. Therefore there exists (V, p. 16, cor. 2) an element
u0 of
H(E) such that
Φ=Φu0. Formula (18) is evidently satisfied, hence
u0 is the upper bound of
F in
H(E). This proves (i).
We have, by construction
u,Σlim⟨x∣(u0−u)(x)⟩=0for all x∈E. Let
v1∈F; given a
u∈F such that
u≥v1, let
v=u0−u. If we apply the Cauchy-Schwarz inequality to the positive hermitian form
Φv on
E, we get
∥v(x)∥4=∣Φv(v(x),x)∣2≤Φv(v(x),v(x))⋅Φv(x,x)=⟨v(x)∣v2(x)⟩⟨x∣v(x)⟩≤∥v∥3∥x∥2⟨x∣v(x)⟩≤∥u0−v1∥3∥x∥2⟨x∣v(x)⟩, since
∥v∥≤∥u0−v1∥ by V, p. 44, prop. 9. Then by (21) we get
limu,Σ∥(u0−u)(x)∥=0 for all
x∈E; which proves assertion (ii).
In particular, prop. 13 can be applied to the case of an increasing and bounded sequence (un)n∈N of elements of H(E). Then there exists an element v of H(E) characterized by
⟨x∣v(x)⟩=n→∞lim⟨x∣un(x)⟩=n∈Nsup⟨x∣un(x)⟩(x∈E), and we have
v(x)=limn→∞un(x) for all
x∈E. Moreover,
v is the upper bound of the set of the
un in
H(E).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
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