Bourbaki

Exercise 5, § 4

Topological Vector Spaces, V, p. 76

Let E,FE, F be two Hilbert spaces, and uu a continuous linear mapping from EE into FF. Let ℓ(u)\ell(u) denote the set of all x∈Ex \in E such that ∥u(x)∥=∥u∥⋅∥x∥\|u(x)\| = \|u\| \cdot \|x\|. a) Show that ℓ(u)\ell(u) is the closed vector subspace of EE which is the kernel of u∗u−∥u∥21Eu^*u - \|u\|^2 1_E, and is orthogonal to the kernel of uu. b) Show that the restriction of uu to ℓ(u)\ell(u) is a bijection from ℓ(u)\ell(u) onto ℓ(u∗)\ell(u^*), whose inverse bijection is the restriction of ∥u∥−2u∗\|u\|^{-2} u^* to ℓ(u∗)\ell(u^*); moreover the image of the orthogonal complement (ℓ(u))∘(\ell(u))^{\circ} under uu is contained in (ℓ(u∗))∘(\ell(u^*))^{\circ}. If u1u_1 is the restriction of uu to (ℓ(u))∘(\ell(u))^{\circ}, considered as a mapping from (ℓ(u))∘(\ell(u))^{\circ} into (ℓ(u∗))∘(\ell(u^*))^{\circ}, the adjoint u1∗u_1^* is the restriction of u∗u^* to (ℓ(u∗))∘(\ell(u^*))^{\circ}; if ℓ(u)≠E\ell(u) \neq E, let ⟨u⟩\langle u \rangle, called the sub-norm of uu, denote the norm ∥u1∥\|u_1\|; if ℓ(u)=E\ell(u) = E, we put ⟨u⟩=0\langle u \rangle = 0. Then ⟨u∗⟩=⟨u⟩\langle u^* \rangle = \langle u \rangle.

Solution

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Transcribed from the printed volume. Not translated.