Functional Analysis-Math 920 (Spring 2003) by Abbas C.

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By Abbas C.

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Extra resources for Functional Analysis-Math 920 (Spring 2003)

Sample text

2 Let X be a Banach space, and let Y be a normed vector space. Assume that T ⊂ L(X, Y ) with sup T x < ∞ ∀ x ∈ X. T ∈T Then sup T < ∞. T ∈T Proof: We define X := {f ∈ C 0 (X, R) | f (x) = T x , T ∈ T }. By the above proposition there are a constant C < ∞ and a ball Bε (x0 ) so that T x ≤ C ∀ T ∈ T , |x − x0 | ≤ ε. e. T ≤ 2C/ε. 1 If X, Y are metric spaces then a map f : X → Y is called open if U ⊂ X open =⇒ f (U ) ⊂ Y open.

Xn )|dξ, R which we write shortly as |u(x)| ≤ |∂i u|dξi . R This actually proves our exercise above. For i = 1, . . , n these are n inequalities which we all multiply with each other. Hence 1 n−1 n n |u(x)| n−1 ≤ |∂i u|dξi . R i=1 We integrate over the variable x1 and obtain |u| n n−1 1 n−1 dξ1 ≤ |∂1 u|dξ1 R 1 n−1 n |∂i u|dξi · R R i=2 dξ1 . R We use H¨ older’s inequality in the form n |f2 · · · fn | ≤ n fi Lpi with i=2 1/pi = 1 i=2 where 1 n−1 |∂i u|dξi fi := and pi = n − 1. R This implies 1 n−1 n |∂i u|dξi R i=2 1 n−1 n dξ1 ≤ R |∂i u|dξ1 dξi i=2 R2 so that 1 n−1 n |u| n−1 dξ1 ≤ R 1 n−1 n |∂1 u|dξ1 |∂i u|dξ1 dξi R i=2 44 R2 .

7) αp + β p ≤ (α2 + β 2 )p/2 ∀ α, β ≥ 0. 8) a+b 2 + ≤ for all a, b ∈ R and 2 ≤ p. 8) is trivially true for β = 0. 8) for the case β = 1. Then there is nothing more to do because the function F : [0, ∞) −→ R F (x) := (x2 + 1)p/2 − xp − 1 is an increasing function with F (0) = 0. 8). Take now α= a−b a+b , β= 2 2 and recall that the function x → |x|p/2 is convex for p ≥ 2. Then a+b 2 p a−b + 2 p 2 ≤ a+b 2 = a2 + b2 2 ≤ a−b + 2 p/2 1 (|a|p + |b|p ). 7 Let Ω ⊂ Rn be a domain and 2 ≤ p < ∞. Then the space Lp (Ω) is uniformly convex.