Bourbaki

Proposition 11

Tag 036C, permanent. Cite it as [Bourbaki, Tag 036C].

Théories spectrales, chapter IV, § 5. Opérateurs partiels normaux et théorème spectral

Let T\mathscr{T} be a locally convex topology on L(E)\mathscr{L}(E) which is less fine than the Banach-space topology of L(E)\mathscr{L}(E).

a) Let f∈C0(R)f\in \mathscr{C}_0(\mathbf{R}). For every sequence (un)n∈N(u_n)_{n\in\mathbf{N}} in A(E)\mathscr{A}(E) which converges to uu for the T\mathscr{T} -resolvent topology, the sequence (f(un))n∈N(f(u_n))_{n\in\mathbf{N}} converges to f(u)f(u) in the space L(E)\mathscr{L}(E) endowed with the topology T\mathscr{T};

b) Suppose that every u∈A(E)u\in \mathscr{A}(E) admits a countable fundamental system of neighbourhoods for the T\mathscr{T} -resolvent topology. The mapping of C0(R)×A(E)\mathscr{C}_0(\mathbf{R})\times \mathscr{A}(E) into the space L(E)\mathscr{L}(E) endowed with the topology T\mathscr{T} defined by (f,u)↦f(u)(f, u)\mapsto f(u) is continuous.

Let us prove a). The topology T\mathscr{T} on L(E)\mathscr{L}(E) is the supremum of the topologies defined by the seminorms continuous for the topology T\mathscr{T} (EVT, II, p. 26, Cor. and following remark). It is therefore enough to prove that for every seminorm pp on L(E)\mathscr{L}(E) which is continuous for the topology T\mathscr{T}, the sequence p(f(un)−f(u))p(f(u_n)-f(u)) converges to 0 (TG, I, p. 51, Prop. 10).

Let pp be such a seminorm. Let L(E)p\mathscr{L}(E)_p denote the separated completed Banach space of the space L(E)\mathscr{L}(E) endowed with the seminorm pp, and let ϖ\varpi denote the canonical mapping of L(E)\mathscr{L}(E) into L(E)p\mathscr{L}(E)_p; it is continuous, and one has p(u)=∥ϖ(u)∥pp(u) =\|\varpi (u)\|_p for every u∈Lu\in \mathscr{L} (E), where the norm is that of the space L(E)p\mathscr{L}(E)_p.

Since T\mathscr{T} is less fine than the Banach-space topology of L\mathscr{L} (E), there exists c⩾0c\geqslant 0 such that p(u)⩽c∥u∥p(u)\leqslant c\|u\| for every u∈L(E)u\in \mathscr{L}(E).

Suppose first that f∈D(R)f\in \mathscr{D}(\mathbf{R}). Let f~\widetilde{f} be an almost analytic extension of ff. Let us denote by μ\mu the Lebesgue measure on C\mathbf{C}. By the Helffer–Sjöstrand formula (Theorem 2 of IV, p. 284), we have

ϖ(f(un))=−1π∫C∂f~∂z(λ)ϖ(R(un,λ))dμ(λ)(14) \varpi (f(u_n)) =-\frac{1}{\pi}\int_{\mathbf{C}}\frac{\partial\widetilde{f}}{\partial z}(\lambda )\varpi (R(u_n, \lambda ))d\mu(\lambda ) \tag{14}

for every n∈Nn\in \mathbf{N}.

Let λ∈C−R\lambda \in \mathbf{C}-\mathbf{R}. By assumption, the sequence of resolvents R(un,λ)R(u_n, \lambda ) converges to R(u,λ)R(u, \lambda ) in L(E)\mathscr{L}(E) endowed with the topology T\mathscr{T}, hence the sequence (ϖ(R(un,λ)))n(\varpi (R(u_n, \lambda )))_n converges to ϖ(R(u,λ))\varpi (R(u, \lambda )) in the Banach space L(E)p\mathscr{L}(E)_p.

For every n∈Nn\in \mathbf{N}, we have

∥∂f~∂z(λ)R(un,λ)∥⩽∣1I(λ)∂f~∂z(λ)∣ \|\frac{\partial\widetilde{f}}{\partial z}(\lambda )R(u_n, \lambda )\|\leqslant |\frac{1}{\mathscr{I} (\lambda)}\frac{\partial\widetilde{f}}{\partial z}(\lambda )|

(Proposition 17 of IV, p. 248), hence

∥∂f~∂z(λ)ϖ(R(un,λ))∥p⩽c∣1I(λ)∂f~∂z(λ)∣ \|\frac{\partial\widetilde{f}}{\partial z}(\lambda )\varpi (R(u_n, \lambda ))\|_p\leqslant c|\frac{1}{\mathscr{I} (\lambda)}\frac{\partial\widetilde{f}}{\partial z}(\lambda )|

Estimate (13) of IV, p. 281 shows that the right-hand side of this inequality is a bounded function for λ∈C−R\lambda \in \mathbf{C}-\mathbf{R}; it is integrable on C\mathbf{C} since f~\widetilde{f} has compact support. It follows from Lebesgue's theorem (INT, IV, p. 137, § 3, no7^o7, Theorem 6) and from the Helffer–Sjöstrand formula applied to ff that ϖ(f(un))\varpi (f(u_n)) converges to ϖ(f(u))\varpi (f(u)), hence p(f(un)−f(u))p(f(u_n)-f(u)) tends to 0.

Suppose now that f∈C0(R)f\in \mathscr{C}_0(\mathbf{R}). Let ε>0\varepsilon >0. There exists a function fεf_{\varepsilon} in D(R)\mathscr{D}(\mathbf{R}) such that ∥fε−f∥∞⩽ε(\|f_{\varepsilon}-f\|_{\infty}\leqslant \varepsilon (cf. Proposition 4, a) of IV, p. 202 and INT, III, p. 45, § 1, no2^o2, Proposition 3). By Proposition 5, c) of IV, p. 275, we then have ∥f(u)−fε(u)∥⩽ε\|f(u)-f_{\varepsilon}(u)\|\leqslant \varepsilon and ∥f(un)−fε(un)∥⩽ε\|f(u_n)-f_{\varepsilon}(u_n)\|\leqslant \varepsilon for every n∈Nn\in \mathbf{N}. Consequently, we obtain

p(f(un)−f(u))⩽2cε+p(fε(un)−fε(u)) p(f(u_n)-f(u))\leqslant 2c\varepsilon +p(f_{\varepsilon}(u_n)-f_{\varepsilon}(u))

for every n∈Nn\in \mathbf{N}. Since fε∈D(R)f_{\varepsilon}\in \mathscr{D}(\mathbf{R}), the sequence (p(fε(un)−fε(u)))n∈N(p(f_{\varepsilon}(u_n)-f_{\varepsilon}(u)))_{n\in\mathbf{N}} converges to 0 by the preceding case, and therefore p(f(un)−f(u))p(f(u_n)-f(u)) converges to 0. This completes the proof of a).

Let us prove assertion b). Under the assumption concerning T\mathscr{T}, it follows from TG, IX, p. 17, Proposition 10 and the following remark, and from assertion a), that the mapping u↦f(u)u\mapsto f(u) from A(E)\mathscr{A}(E) into L(E)\mathscr{L}(E) is continuous when f∈C0(R)f\in \mathscr{C}_0(\mathbf{R}). Since the mappings f↦f(u)f\mapsto f(u) from C0(R)\mathscr{C}_0(\mathbf{R}) into L(E)\mathscr{L}(E) are continuous of norm ⩽1\leqslant 1 for every u∈A(E)u\in \mathscr{A}(E) (Proposition 5, c) of IV, p. 275), assertion b) follows from TG, X, p. 13, Corollary 3.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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