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# Introduction to Fourier Analysis on Euclidean Spaces by Elias M. Stein, Guido Weiss By Elias M. Stein, Guido Weiss

The authors current a unified therapy of easy themes that come up in Fourier research. Their goal is to demonstrate the position performed via the constitution of Euclidean areas, fairly the motion of translations, dilatations, and rotations, and to encourage the examine of harmonic research on extra basic areas having a similar constitution, e.g., symmetric areas.

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Extra info for Introduction to Fourier Analysis on Euclidean Spaces

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.