Bourbaki

Exercise 28, § 6

Tag 02W5, permanent. Cite it as [Bourbaki, Tag 02W5].

Spectral Theories, III, p. 141

It is assumed in this exercise that conditions (i) to (iv) are satisfied. (a) We fix x0∈C˚x_0\in \mathring{C}. For λ∈Uf\lambda \in U_f, let A(λ)=A(\lambda ) = inf{M∈R+∣x(λ)≼CMx0}\{M\in \mathbf{R}_+|x(\lambda )\preccurlyeq_CMx_0\}. Let C0∈R+C_0\in \mathbf{R}_+ be such that f≼CC0u(x0)f\preccurlyeq_CC_0u(x_0). For every λ\lambda in UfU_f, the interval [λ,λ+C0/A(λ)[[\lambda , \lambda + C_0/A(\lambda )[ is contained in Uf(U_f(cf. exercise 26, question d)).

(b) Deduce that the function λ↦A(λ)\lambda \mapsto A(\lambda ) tends to +∞+\infty when λ∈Uf\lambda \in U_f tends to λm\lambda_m.

For every λ\lambda in UfU_f, we denote by x^(λ)=x(λ)/A(λ)\widehat{x}(\lambda ) =x(\lambda )/A(\lambda ).

We now suppose that condition (vi) is satisfied.

(c) Prove that the set of values of the mapping x^\widehat{x} possesses a value of adherence xmx_m such that

0≼Cxm≼Cx00\preccurlyeq_Cx_m\preccurlyeq_Cx_0 and xm=λmu(xm)x_m=\lambda_mu(x_m).

(d) Prove that xm≠0x_m\not = 0. (If we had xm=0x_m= 0, let U be a neighbourhood of 0 in E such that x0+U⊂Cx_0+ U\subset C; for every μ⩾0\mu\geqslant 0, there would exist a sequence (λn)(\lambda_n) in UfU_f converging to λm\lambda_m such that x0−μx^(λn)≽C0x_0-\mu\widehat{x}(\lambda_n)\succcurlyeq_C0, in contradiction with the definition of A(λ).A(\lambda ).)

(e) Let y∈C−{0}y\in C-\{0\} such that y=λmu(y)y=\lambda_mu(y). Put θ=\theta = sup{γ∣γy≼Cxm}\{\gamma |\gamma y\preccurlyeq_Cx_m\}. We have θ∈R+∗\theta \in \mathbf{R}^*_+ and xm=θyx_m=\theta y. (We have θy≼Cxm\theta y\preccurlyeq_Cx_m and θy=λmu(θy)\theta y=\lambda_mu(\theta y); if xm≠θyx_m\not =\theta y, then we would have xm−θy≻C0x_m-\theta y\succ_C0 and there would exist ε>0\varepsilon >0 such that xm−θy≽Cεyx_m-\theta y\succcurlyeq_C\varepsilon y, in contradiction with the definition of θ.\theta .)

(f) Let y∈C−{0}y\in C-\{0\} and μ∈R\mu\in \mathbf{R} such that y=μu(y)y=\mu u(y). Then we have μ=λm\mu=\lambda_m. (Note that 0<μ0< \mu and μ⩾λm\mu\geqslant \lambda_m; if we had μ>λm\mu > \lambda_m, let θ=\theta = sup{γ∣γxm≼Cy}\{\gamma |\gamma x_m\preccurlyeq_Cy\}; we would then have θ∈R+∗\theta \in \mathbf{R}_+^*, then θxm≼Cy\theta x_m\preccurlyeq_Cy, whence θxm=λmu(θxm)≺Cμu(θxm)≼C\theta x_m=\lambda_mu(\theta x_m)\prec_C\mu u(\theta x_m)\preccurlyeq_C μu(y)=y\mu u(y) =y, in contradiction with the definition of θ.\theta .)

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.