By M. Matsuda

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C) Deﬁne f (x) = x−1/2 if 0 < x ≤ 1 and f (0) = 0. Does 1 0 f (x) dx exist? 9. More exercises involving integration: (a) Put simple lower and upper bounds on the family of integrals 1 I(α, β) = 0 dx (xβ + 1)α where α, β ≥ 0. (b) Show that π/2 ln(1/ sin t) dt < ∞. 0 (c) A function f (t) is of exponential order on [0, ∞) if there exist positive numbers b and C such that whenever t ≥ 0, |f (t)| ≤ Cebt . Show that the Laplace transform of f (t) given by ∞ F (s) = f (t)e−st dt 0 exists if f (t) is of exponential order.

Choosing bi = ai for all i, we see that the square of the arithmetic mean never exceeds the mean of the squares. With functions f (x) and g(x), analogous operations yield b f (x)g(x) dx ≥ a 1 b−a b b f (x) dx a g(x) dx a if f (x) and g(x) are either both increasing or both decreasing on [a, b]. If one function is increasing and the other is decreasing, the inequality sign is reversed. 20) holds for all x1 , x2 ∈ (a, b) and every p ∈ (0, 1). In the case of strict inequality for x1 = x2 , f is strictly convex on (a, b).

Suppose for each j, 1 ≤ j ≤ n, that aj1 , . . , ajm are nonzero numbers. Suppose δ1 , . . , δn are positive numbers such that δ1 + · · · + δn = 1. For each j denote m |aji |. Sj = i=1 Then m |a1i |δ1 · · · |ani |δn ≤ S1δ1 · · · Snδn . 5 H¨ older’s Inequality 41 Proof. 4) to each summand. With n = 2 write δ1 = 1/p, δ2 = 1/q, and let a1i = |ai |p and a2i = |bi |q for i = 1, . . , m. 8) becomes m 1/p m |ai bi | ≤ i=1 1/q m |ai |p |bi |q i=1 . 10) i=1 This special case is also commonly referred to as H¨older’s inequality, and we can give another proof based on Young’s inequality.