Exercise 2.3 Let ($2. E, u) be a positive finite measure space. Show that L, ($2, u()* can be identified with Lo($2, #). (This completes the proof of Theorem 2.13.) The following proves the theorem, but need to be completed, see highlighting please. Function spaces from Functional Analysis. Theorem 2.13 Let ($2. E. u) be a positive finite measure space. If p and q are conjugate exponents, where p e [1, 00), then L,(u)* = L,(u). Proof Start by assuming the scalars are real. We begin with the case p e (1, 00). Let g E La(u), where - + - = 1, and define a scalar-valued function d on Lp(u) by D.(f) = | f8du, fe Lp(u). (2.6) We will show first that d is a linear functional on Lp(u) such that IIdyll = Igllq. and then we will show that all linear functionals on Lo() can be achieved in this way. We note that o, is linear (by the linearity of the integral) and |0, (f)| S IIfllpllglla (by Holder's Inequality). Thus, 4 is a bounded linear functional and IIll s Iglq. In order to show equality of the norms, it suffices to find a function f in Lp(u) such that IIf Ilp = 1 and such that do(f) = Iglly. First, define a scalar-valued function h on ? by letting h(x) = [g(x)|4-1 (sign g(x)) for each x e ?. Thenwhere (q - 1)p = q follows from the assumption that _ + _ = 1. It follows that h is in Lp(u). Next, observe that be(h) = Igl;. Now, let f = TAG 4. Then IIflip = 1 and = liglla- Therefore, is a bounded linear functional on Lp() and Idyll = Ilglly. We now wish to show that any bounded linear functional in Lp(u)* can be written as in (2.6) for some g in Lo(u). To that end, let y be a bounded linear functional in the dual space Lp(u)*. Define a measure v on ($2, E) by v(A) = V(XA) for all A E E. It is routine to show that v is finitely additive (by the linearity of v ), and it is countably additive by the continuity of v. We also claim that v