By Constantin Niculescu
Thorough creation to a big region of arithmetic comprises fresh effects contains many workouts
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Example text
Prove the discrete Berwald inequality: 1 n+1 n ak ≥ k=0 3(n − 1) 4(n + 1) 1/2 1 n+1 n a2k 1/2 k=0 for every concave sequence a0 , a1 , . . , an of nonnegative numbers. 4 above), if the Berwald inequality works for two concave sequences, then it also works for all linear combinations of them, with positive coefficients. Then apply the assertion (ii) of the preceding exercise. 7 Conjugate Convex Functions The aim of this section is to develop a concept of duality between convex functions which makes possible an easy handling of some problems.
4. (A symmetric form of Rogers–H¨older inequality) Let p, q, r be nonzero real numbers such that 1/p + 1/q = 1/r. (i) Prove that the inequality n λk |ak bk |r 1/r n ≤ k=1 λk |ak |p 1/p k=1 n λk |bk |q 1/q k=1 holds in each of the following three cases: p > 0, q > 0, r > 0; (ii) p < 0, q > 0, r < 0; p > 0, q < 0, r < 0. Prove that the opposite inequality holds in each of the following cases: p > 0, q < 0, r > 0; p < 0, q > 0, r > 0; n p < 0, q < 0, r < 0. Here λ1 , . . , λn > 0, k=1 λk = 1, and a1 , .
Proof. 7) follows immediately by integrating the inequality |f + g| ≤ |f | + |g|. For p ∈ (1, ∞) we have |f + g|p ≤ (|f | + |g|)p ≤ (2 sup{|f |, |g|})p ≤ 2p (|f |p + |g|p ), which shows that f + g ∈ Lp (µ). 2, f +g p Lp |f + g|p dµ ≤ = X |f + g|p−1 |f | dµ + X X 1/q 1/p ≤ |f |p dµ |f + g|(p−1)q dµ X X 1/p |g|p dµ + X =( f Lp + g |f + g|p−1 |g| dµ Lp ) f +g 1/q |f + g|(p−1)q dµ X p/q Lp , where 1/p + 1/q = 1, and it remains to observe that p − p/q = 1. 7) if and only if there is a positive measurable function ϕ such that f (x)ϕ(x) = g(x) almost everywhere on the set {x | f (x)g(x) = 0}.