By Constantin Niculescu

Thorough advent to a huge zone of arithmetic includes contemporary effects contains many workouts

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**Example text**

K∈ X Remark. The following particular case 1 n n xk k=1 1 n n k=1 1 xk (M + m)2 (1 + (−1)n+1 )(M − m)2 − 4M m 8M mn2 ≤ represents an improvement on Schweitzer’s inequality for odd n. 2. Let ak , bk , ck , mk , Mk , mk , Mk be positive numbers with mk < Mk and mk < Mk for k ∈ {1, . . , n} and let p > 1. Prove that the maximum of n n ak xpk n p bk ykp k=1 k=1 ck xk yk k=1 for xk ∈ [mk , Mk ] and yk ∈ [mk , Mk ] (k ∈ {1, . . , n}) is attained at a 2n-tuple whose components are endpoints. 3. Assume that f : I → R is strictly convex and continuous and g : I → R is continuous.

2 (The Rogers–H¨ older inequality for p > 1) Let p, q ∈ (1, ∞) with 1/p + 1/q = 1, and let f ∈ Lp (µ) and g ∈ Lq (µ). 5) f g dµ ≤ f Lp g Lq . 6) X and thus X The above result extends in a straightforward manner to the pairs p = 1, q = ∞ and p = ∞, q = 1. 6) should be reversed. See Exercises 3 and 4. 6) is called the Cauchy–Buniakovski– Schwarz inequality. Proof. The ﬁrst inequality is trivial. If f or g is zero µ-almost everywhere, then the second inequality is trivial. Otherwise, using Young’s inequality, we have |f (x)| |g(x)| 1 |f (x)|p 1 |g(x)|q · ≤ · + · p f Lp g Lq p f Lp q g qLq for all x in X, such that f g ∈ L1 (µ).

16) follows easily from the monotonicity of the subdifferential, while maximality can be established by reductio ad absurdum. 2, x f (x) − f (c) = ϕ(t) dt = f ∗∗ (x) − f ∗∗ (c) c for all x, c ∈ I. It remains to ﬁnd a c ∈ I for which f (c) = f ∗∗ (c). Choose z ∈ I and y ∈ I ∗ such that y ∈ ∂f (z). By (iii), this means z ∈ (∂f )−1 (y) = ∂f ∗ (y). According to (i), applied for f and f ∗ , we have zy = f (z) + f ∗ (y) = f ∗ (y) + f ∗∗ (z), that is, f (z) = f ∗∗ (z). 3, if f is diﬀerentiable, then its conjugate can be determined by eliminating x from the equations f (x) + f ∗ (y) = xy and f (x) = y.