Bourbaki

Exercise 14, § 4

Topological Vector Spaces, V, p. 79

Let E be a complex hilbertian space, and T a Hilbert-Schmidt endomorphism of E. a) Let R and L be positive Hilbert-Schmidt endomorphisms such that R2=T∗TR^2 = T*T and L2=TT∗L^2 = TT^* (V, p. 57, cor. 3); let R=abs⁡(T)R = \operatorname{abs}(T), and call this the « absolute value » of T (cf. V, p. 75, exerc. 4); we have L=abs⁡(T∗)L = \operatorname{abs}(T^*). Show that Ker⁡(T)=Ker⁡(R)\operatorname{Ker}(T) = \operatorname{Ker}(R) and L(E)‾=T(E)‾\overline{L(E)} = \overline{T(E)}. There exists one, and only one isometry V from R(E)R(E) onto T(E)T(E) such that T=VRT = VR; if we extend V by continuity to R(E)‾\overline{R(E)}, then to an operator U∈L(E)U \in \mathscr{L}(E) by taking U.x=0U.x = 0 on the orthogonal complement of R(E)R(E), we also have T=URT = UR (polar decomposition of T). Then R=U∗T=U∗UR=RU∗UR = U*T = U^*UR = RU^*U and L=URU∗,T=LU∗L = URU^*, T = LU^*. If T belongs to L1(E)\mathscr{L}^1(E), then so does R=abs⁡(T)R = \operatorname{abs}(T), and T is the product of two Hilbert-Schmidt endomorphisms.

b) If TT belongs to L1(E)\mathscr{L}^1(E), show that Tr⁡(abs⁡(T))=sup⁡(∑i∣⟨ai∣T.bi⟩∣)\operatorname{Tr}(\operatorname{abs}(T)) = \sup (\sum_i | \langle a_i|T.b_i \rangle |) where, on the right hand side, (ai)(a_i) and (bi)(b_i) range over the set of orthonormal bases of E (use the polar decomposition of T). Show that if we put ∥T∥1=Tr⁡(abs⁡(T))\| T \|_1 = \operatorname{Tr}(\operatorname{abs}(T)), then ∥T∥1\| T \|_1 is a norm on the space L1(E)\mathscr{L}^1(E), such that ∥T∥2≤∥T∥1\| T \|_2 \leq \| T \|_1. c) Conversely, if T∈L(E)T \in \mathscr{L}(E) is such that, for every pair ((ai),(bi))((a_i), (b_i)) of orthonormal bases of E, the sum ∑i∣⟨ai∣T.bi⟩∣\sum_i | \langle a_i|T.b_i \rangle | is finite, then T∈L1(E)T \in \mathscr{L}^1(E) (first observe that T is a Hilbert-Schmidt endomorphism, then use the polar decomposition of T). d) Let (Tv)(T_v) be a sequence of Hilbert-Schmidt endomorphisms (resp. of elements of L1(E)\mathscr{L}^1(E)) such that for every pair of points x, y of E, the sequence (⟨x∣Tv.y⟩)(\langle x|T_v.y \rangle) converges to ⟨x∣T.y⟩\langle x|T.y \rangle, where T is a linear mapping from E into itself; in addition, assume that the sequence of norms ∥Tv∥2\| T_v \|_2 (resp. ∥Tv∥1\| T_v \|_1) is bounded. Show that T is a Hilbert-Schmidt endomorphism (resp. an element of L1(E)\mathscr{L}^1(E)) (use b) and c)). e) Deduce from d) that the space L1(E)\mathscr{L}^1(E) is a Banach space for the norm ∥T∥1\| T \|_1. f) For an endomorphism T∈L(E)T \in \mathscr{L}(E) to belong to L1(E)\mathscr{L}^1(E) it is necessary and sufficient that, for at least one orthonormal basis (ei)(e_i) of E, the sum ∑i∥T.ei∥\sum_i \| T.e_i \| is finite (with the notations of a), note that ∣⟨ei∣R.ei⟩∣≤∥T.ei∥|\langle e_i|R.e_i \rangle| \leq \| T.e_i \|. g) In the space ℓc2\ell_c^2, let (en)(e_n) be the canonical orthonormal basis, and let a=∑n=0∞1n+1ena = \sum_{n=0}^\infty \frac{1}{n+1} e_n; if F is the 1-dimensional subspace C.a\mathbf{C}.a, the orthoprojector pFp_F has finite trace, but the series ∑n∥pF.en∥\sum_n \| p_F.e_n \| is not convergent.

Solution

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