Bourbaki

Exercise 6, § 4

Topological Vector Spaces, V, p. 76

Bourbaki marks this one as harder than the rest.

Let EE be a hilbertian space, M,NM, N two closed subspaces of EE and M∘,N∘M^{\circ}, N^{\circ} their respective orthogonal complements; let pM,pNp_M, p_N denote the orthogonal projections on MM and NN respectively, such that 1E−pM,1E−pN1_E - p_M, 1_E - p_N are the orthogonal projections on M∘M^{\circ} and N∘N^{\circ} respectively. Put uNM=(1E−pN)pMu_{NM} = (1_E - p_N) p_M and δ(M,N)=∥uNM∥\delta(M, N) = \|u_{NM}\|; we have δ(M,N)=δ(N∘,M∘)≤1\delta(M, N) = \delta(N^{\circ}, M^{\circ}) \leq 1; the relation δ(M,N)<1\delta(M, N) < 1 implies M∩N∘={0}M \cap N^{\circ} = \{0\}. a) Let M~\tilde{M} denote the orthogonal complement in MM of M∩N∘M \cap N^{\circ}, and let ε(M,N)=δ(M~,N)\varepsilon(M, N) = \delta(\tilde{M}, N). Show that (with the notations of exerc. 5) ℓ(uNM)=M∩N∘\ell(u_{NM}) = M \cap N^{\circ} and deduce that ε(M,N)=⟨uNM⟩≤δ(M,N)\varepsilon(M, N) = \langle u_{NM} \rangle \leq \delta(M, N); moreover, if M∩N∘={0}M \cap N^{\circ} = \{0\} (and in particular if δ(M,N)<1\delta(M, N) < 1), then ε(M,N)=δ(M,N)\varepsilon(M, N) = \delta(M, N). b) For a continuous linear mapping uu from EE into itself, we designate by conorm of uu the number c(u)=inf⁡∥u(x)∥/∥x∥c(u) = \inf \|u(x)\| / \|x\|, where xx ranges over the set of all vectors ≠0\neq 0 orthogonal to u−1(0)u^{-1}(0) (if u=0u = 0, put c(u)=1c(u) = 1). For u(E)u(E) to be closed in EE, it is necessary and sufficient that c(u)>0c(u) > 0 (I, p. 17, th. 1). We have c(u∗)=c(u)c(u^*) = c(u). c) Let vNM=pNpMv_{NM} = p_N p_M. Show that

ε(M,N)2+c(vNM)2=1 \varepsilon(M, N)^2 + c(v_{NM})^2 = 1
(observe that ∥uNM(x)∥2+∥vNM(x)∥2=∥pM⋅x∥2\|u_{NM}(x)\|^2 + \|v_{NM}(x)\|^2 = \|p_M \cdot x\|^2, and that the kernel of vNMv_{NM} is M∘+(M∩N∘)M^{\circ} + (M \cap N^{\circ}) and deduce that ⟨uNM⟩2≤1−c(vNM)2\langle u_{NM} \rangle^2 \leq 1 - c(v_{NM})^2). d) Deduce from b) and c) that ε(N,M)=ε(M,N)\varepsilon(N, M) = \varepsilon(M, N) and, using a), that ε(M∘,N∘)=ε(M,N)\varepsilon(M^{\circ}, N^{\circ}) = \varepsilon(M, N). e) Put g(M,N)=∥pM−pN∥g(M, N) = \|p_M - p_N\| (cf. V, p. 64, exerc. 17). Show that
g(M,N)=sup⁡(δ(M,N),δ(N,M)) g(M, N) = \sup(\delta(M, N), \delta(N, M))
(observe that pM−pN=(1E−pN)pM−pN(1E−pM)p_M - p_N = (1_E - p_N) p_M - p_N (1_E - p_M)); deduce that ε(M,N)≤g(M,N)\varepsilon(M, N) \leq g(M, N). If M∩N∘=N∩M∘={0}M \cap N^{\circ} = N \cap M^{\circ} = \{0\}, we have the relation ε(M,N)=δ(M,N)=δ(N,M)=g(M,N)\varepsilon(M, N) = \delta(M, N) = \delta(N, M) = g(M, N). If g(M,N)<1g(M, N) < 1, then M∩N∘=N∩M∘={0}M \cap N^{\circ} = N \cap M^{\circ} = \{0\}. f) Let QM,QNQ_M, Q_N be two continuous projections in EE, with respective images MM and NN; give another proof of the relation g(M,N)≤∥QM−QN∥g(M, N) \leq \|Q_M - Q_N\| (V, p. 64, exerc. 17). (Note that for all x∈Ex \in E, we have ∥(1E−QM)⋅x∥2+∥QM∗⋅x∥2=∥x∥2+∥(QM−QM∗)⋅x∥2\|(1_E - Q_M) \cdot x\|^2 + \|Q_M^* \cdot x\|^2 = \|x\|^2 + \|(Q_M - Q_M^*) \cdot x\|^2, and apply this relation to x=(pM−pN)⋅yx = (p_M - p_N) \cdot y, observing the relations (1E−QM)(pM−pN)=(QM−QN)pN(1_E - Q_M) (p_M - p_N) = (Q_M - Q_N) p_N and (pM−pN)QM=(1E−pN)(QM−QN)(p_M - p_N) Q_M = (1_E - p_N) (Q_M - Q_N).)

Solution

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Transcribed from the printed volume. Not translated.