Bourbaki

Exercise 24, § 1

Tag 02Q6, permanent. Cite it as [Bourbaki, Tag 02Q6].

Spectral Theories, III, p. 112

Let E be a real or complex Banach space and u∈L(E)u\in \mathscr{L}(E). Suppose that the sequence (un)n∈N(u^n)_{n\in\mathbf{N}} is bounded and that uu is compact from E into E endowed with the weak topology σ(E,E′)\sigma (E,E'). For every integer N⩾1N\geqslant 1, put

vN=1N(1E+u+⋯+uN−1)∈L(E). v_N=\frac{1}{N}(1_E+u+\cdots+u^{N-1})\in\mathscr{L}(E).

a) Let F be the closure of the image of 1E−u1_E-u. For every x∈Fx\in F, one has vN(x)→0v_N(x)\rightarrow 0 in E when N→+∞N\rightarrow +\infty.

b) Let x∈Ex\in E. The sequence (vN(x))(v_N(x)) converges to an element y∈Ey\in E such that u(y)=yu(y) =y. (There exists a subsequence of (vN(x))(v_N(x)) which converges weakly to y∈Ey\in E; show that x−y∈Fx-y\in F using the Hahn–Banach theorem.)

c) The mapping w:E→Ew: E\rightarrow E such that w(x)w(x) is the element yy given by the preceding question is a continuous projector with image the proper space of uu relative to the eigenvalue 1.

d) Let z∈Cz\in \mathbf{C} such that ∣z∣=1|z|= 1. The sequence in L(E)\mathscr{L}(E) defined by

vz,N=1N(1E+z−1u+⋯+z−(N−1)uN−1) v_{z,N}=\frac{1}{N}(1_E+z^{-1}u+\cdots+z^{-(N-1)}u^{N-1})

converges in L(E)\mathscr{L}(E) endowed with the topology of simple convergence. Let wzw_z be its limit. Then wzw_z is a projector, one has wzw=wwz=zwzw_zw=ww_z=zw_z and wz1wz2=0w_{z_1}w_{z_2}= 0 if z1≠z2z_1\not =z_2.

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Machine translation of the English, by gpt-5-6. Not checked by a person.