Bourbaki

Example 5

Tag 07XL, permanent. Cite it as [Bourbaki, Tag 07XL].

Topological Vector Spaces, chapter V, § 1. Prehilbertian spaces and hilbertian spaces

\* Let D be the open disc with centre 0 and radius 1 in C\mathbf{C}. The Hardy space H2(D)H^2(D) consists of all holomorphic functions f:D→Cf : D \to \mathbf{C} for which

sup⁡0<R<1∫01∣f(R⋅e(θ))∣2dθ<+∞. \sup_{0 < R < 1} \int_0^1 |f(R \cdot e(\theta))|^2 d\theta < + \infty .
If f1f_1 and f2f_2 belong to H2(D)H^2(D), the limit
⟨f1∣f2⟩=lim⁡R→1∫01f1(R⋅e(θ))‾⋅f2(R⋅e(θ)) dθ \langle f_1 | f_2 \rangle = \lim_{R \to 1} \int_0^1 \overline{f_1(R \cdot e(\theta))} \cdot f_2(R \cdot e(\theta)) \, d\theta
exists; the mapping (f1,f2)↦⟨f1∣f2⟩(f_1, f_2) \mapsto \langle f_1 | f_2 \rangle assigns a complex hilbertian space structure to the vector space H2(D)H^2(D).

For a function f:D→Cf : D \to \mathbf{C} to belong to H2(D)H^2(D), it is necessary and sufficient that there exists a sequence (an)n∈N(a_n)_{n \in \mathbf{N}} of complex numbers such that ∑n=0∞∣an∣2<+∞\sum_{n=0}^\infty |a_n|^2 < + \infty and that

f(z)=∑n=0∞anzn f(z) = \sum_{n=0}^\infty a_n z^n
for all z∈Dz \in D. Then we have ∥f∥2=∑n=0∞∣an∣2\|f\|^2 = \sum_{n=0}^\infty |a_n|^2 which gives an isomorphism from H2(D)H^2(D) onto the hilbertian space ℓ2\ell^2 (V, p. 4).

Every Hausdorff prehilbertian space is isomorphic to an everywhere dense subspace of a hilbertian space determined up to an isomorphism. Precisely :

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed volume. Not translated.