By Vandebril R., Van Barel M., Golub G.

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8. CONJUGATE OR DUAL OF AN OPERATOR 55 and ε > 0 was arbitrary. 25), this establishes the result. The ∗-mapping T −→ T ∗ is thus a norm preserving map ∗ B(X, Y ) −→ B(Y ∗ , X ∗ ) . It has simple properties of its own, for example (λT + µS)∗ = λT ∗ + µS ∗ , ∗ ∗ ∀ S, T ∈ B(X, Y ) , ∗ ∀ S ∈ B(X, Y ) , T ∈ B(Y, Z) , (T S) = S T , ∗ (IX ) = IX ∗ , where IX is the identity mapping of X to itself. Examples. X = RN , T : X → X may be represented by an N × N matrix MT in the standard basis, say. Then T ∗ also has a matrix representation in the dual basis and MT ∗ = MTt the transpose of MT .

6. UNIFORM BOUNDEDNESS PRINCIPLE 49 Example. Closed does not imply bounded in general, even for linear operators. Take X = C(0, 1) with the max norm. Let T f = f for f ∈ D = C 1 (0, 1). Consider T as a mapping of D into X. T is not bounded . Let fn (x) = xn . Then fn = 1 for all n, but T fn = nxn−1 so T fn = n. X → f and fn → g. Then, by the Fundamental T is closed . Let {fn }∞ n=1 ⊂ D and suppose fn − Theorem of Calculus, t fn (τ ) dτ fn (t) = fn (0) + 0 for n = 1, 2, . . Taking the limit of this equation as n → ∞ yields t g(τ ) dτ , f (t) = f (0) + 0 so g = f , by another application of the Fundamental Theorem of Calculus.

Suppose X is a vector space. The algebraic dual of X is the set of all linear functionals on X, and is also a vector space. Suppose also that X is a NLS. Show that X has finite dimension if and only if the algebraic dual and the dual space X ∗ coincide. 26. Let X be a NLS and M a nonempty subset. The annihilator M a of M is defined to be the set of all bounded linear functionals f ∈ X ∗ such that f restricted to M is zero. Show that M a is a closed subspace of X ∗ . What are X a and {0}a ? 60 27.