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14. If f ∈ Lp0 for some 0 < p0 ≤ ∞, and its support E := {x ∈ X : f (x) = 0} has finite measure, show that f ∈ Lp for all 0 < p < p0 , and that f pLp → µ(E) as p → 0. 2. Linear functionals on Lp . Given an exponent 1 ≤ p ≤ ∞, define the dual exponent 1 ≤ p ≤ ∞ by the formula p1 + p1 = 1 (thus p = p/(p − 1) for 1 < p < ∞, while 1 and ∞ are duals of each other). 26) λg (f ) := f g dµ X is well-defined on Lp ; the functional is also clearly linear. Furthermore, H¨ older’s inequality also tells us that this functional is continuous.

Let 1 ≤ p < ∞, and assume µ is σ-finite. Let λ : Lp → C be a continuous linear functional. Then there exists a unique g ∈ Lp such that λ = λg . 5). Both theorems start with an abstract function µ : X → R or λ : Lp → C, and create a function out of it. Indeed, we shall see shortly that the two theorems are essentially equivalent to each other. 5, once we introduce the notion of a dual space. 17 (Continuity is equivalent to boundedness for linear operators). Let T : X → Y be a linear transformation from one normed vector space (X, X ) to another (Y, Y ).

These operations respect almost everywhere equivalence, and so Lp becomes a (complex) vector space. Next, we set up the norm structure. 18) fr Lp = f r Lpr for all 0 < p, r < ∞. 3. Let 0 < p < ∞ and f, g ∈ Lp . (i) (Non-degeneracy) f Lp = 0 if and only if f = 0. 34 1. Real analysis (ii) (Homogeneity) cf c. Lp = |c| f Lp for all complex numbers (iii) ((Quasi-)triangle inequality) We have f +g Lp ≤ C( f Lp + g Lp ) for some constant C depending on p. If p ≥ 1, then we can take C = 1 (this fact is also known as Minkowski’s inequality).

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An epsilon of room: pages from year three of a mathematical blog by Tao T.

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