By Sneddon I.N.

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By Zorn’s lemma we conclude that there exists (h, P ) ∈ S that is maximal for ; if we can show that P = X then we are done. Suppose otherwise; then P is a proper subspace of X and there exists a proper extension of h, by the first part of this proof. This contradicts the maximality of (h, P ). 9. (Bohnenblust-Sobczyk) Let p be a seminorm on the vector space X and suppose that M is a subspace of X. , |Φ(x)| p(x) for all x ∈ X). 34 Dual Spaces Proof Suppose first that X is a real vector space. Note that a seminorm is a sublinear functional, so we may apply the Hahn-Banach theorem to obtain Φ ∈ X ′ such that Φ|M = φ and Φ(x) p(x) for all x ∈ X, but also −Φ(x) p(−x) = p(x) ∀ x ∈ X by the homogeneity of the seminorm p.

Let M be a finite-dimensional subspace of the normed space X and let N be a closed subspace of X such that X = M ⊕ N. Prove that if φ0 is a linear functional on M then φ : M ⊕ N → F; m + n → φ0 (m) ∀ m ∈ M, n ∈ N is an element of the dual space X ∗ . 3. Prove that a normed vector space X is separable if its dual X ∗ is. ] Find a separable Banach space E such that E ∗ is not separable. ] Prove that a reflexive Banach space E is separable if and only if E ∗ is. 4. reflexive. 5. Prove that any infinite-dimensional normed space has a discontinuous linear functional defined on it.

12. Let L1 (R) denote the space of (equivalence classes of) complex-valued, Lebesgue-integrable functions on the real line, with norm · 1: L1 (R) → R+ ; f → R |f |. This is a commutative Banach algebra when equipped with the convolution product: f ⋆ g : R → R; t → R f (t − s)g(s) ds. ) This algebra lacks a unit; it is easy u to see that L1 (R) is isomorphic to the algebra given by adjoining the Dirac measure δ0 : by definition (f ⋆ δ0 )(t) = “ R f (t − s)δ0 (s) ds ” = f (t) ∀t ∈ R and δ0 ⋆ δ0 = δ0 .