Bourbaki

Lemma 2

Tag 081N, permanent. Cite it as [Bourbaki, Tag 081N].

Topological Vector Spaces, chapter V, § 4. Some classes of operators in hilbertian spaces

Let (ei)i∈I(e_i)_{i \in I} be an orthonormal basis of E. Then

τ(u)=∑i∈I⟨ei∣u(ei)⟩ \tau(u) = \sum_{i \in I} \langle e_i|u(e_i) \rangle

for all u∈Lf(E)u \in \mathscr{L}_f(E).

It is enough to consider the case when u=ba∗u = ba^* with a,ba, b in E. Then

⟨ei∣u(ei)⟩=ei∗b⋅a∗ei=⟨ei∣a⟩‾⟨ei∣b⟩ \langle e_i|u(e_i) \rangle = e_i^* b \cdot a^* e_i = \overline{\langle e_i|a \rangle} \langle e_i|b \rangle

and lemma 2 follows from formula (22) and formula (3) of V, p. 22.

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.