Victor Guillemin's Analysis II PDF

By Victor Guillemin

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Int Qi , i = 1, 2, 3 . . is a cover of U , 2. Each Qi ⊂ Iα for some α, 3. For every point p ∈ U , there exists a neighborhood Up of p such that Up ∩ Qi = φ for all i > Np . Proof. Take an exhaustion A1 , A2 , A3 , . . of U . By definition, the exhaustion satisfies ⎧ ⎨ Ai ⊆ Int Ai+1 Ai is compact ⎩ ∪Ai = U. We previously showed that you can always find an exhaustion. Let Bi = Ai − Int Ai−1 . For each x ∈ Bi , let Qx be a rectangle with x ∈ Int Qx such that Qx ⊆ Uα , for some alpha, and Qx ⊂ Int Ai+1 − Ai−2 .

You should check that the standard axioms for vector spaces are satisfied. There is a natural projection from V to V /W : π : V → V /W, v → v + W. 5) The map π is a linear map, it is surjective, and ker π = W . Also, Im π = V /W , so dim V /W = dim Im π = dim V − dim ker π = dim V − dim W. 6) Dual Space 13. The dual space construction: Let V be an n­dimensional vector space. Define V ∗ to be the set of all linear functions � : V → R. Note that if �1 , �2 ∈ V ∗ and λ1 , λ2 ∈ R, then λ1 �1 + λ2 �2 ∈ V ∗ , so V ∗ is a vector space.

F1 ≥ 0, 2. supp fi ⊆ Uα , for some α, 3. For every p ∈ U , there exists a neighborhood Up of p such that Up ∪ supp fi = φ for all i > Np , � 4. fi = 1. Proof. Let Qi , i = 1, 2, 3, . . be a collection of rectangles with the properties of the previous theorem. Then the functions fQi , i = 1, 2, 3, . . have all the properties presented in the theorem, except for property 4. We now prove the fourth property. We now that fQi > 0 on Int Qi , and {Int Qi : i = 1, 2, 3, . . } is a cover of U . So, for every p ∈ U, fQi (p) > 0 for some i.

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