By ed van der Geer at al Birkhaeuser

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**Extra resources for Arithmetic Algebraic Geometry**

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1. Let G be a complex vector space. G is said to be an inner product space if there exists a function ·, · : G × G → C that satisfies the following conditions: 1. f, f ≥ 0 for all f ∈ G. 2. f, f = 0 if and only if f = 0. 3. f, g = g, f for all f, g ∈ G. 4. af + bg, h = a f, h + b g, h for all f, g, h ∈ G and all a, b ∈ C. The function ·, · : G × G → C is referred to as an inner product. From properties 3 and 4 above, we have the following property too: h, af + bg = a h, f + b h, g for all f, g, h ∈ G and all a, b ∈ C.

AN0 be the corresponding scalars. For any N0 N ≥ N0 , the linear combination n=1 an en is in the subspace spanned by e1 , . . , eN . Define M = span{e1 , . . , eN }. 18, PM g = N n=1 g, en en is the best approximation for g within M , therefore N0 N n=1 g, en en − g ≤ n=1 an en − g < ǫ. 13. Let {ei }i∈I be an orthonormal system in a Hilbert space H, and let {ai }i∈I be a set of complex numbers. The series i∈I ai ei converges in H if and only if i∈I |ai |2 < ∞. Proof. If ai = 0 for more than countably many values of i, then we know that neither one of the sums converges.

Then e1 , . . , en is an orthonormal sequence, and en ∈ span{v1 , . . , vn } by construction. Thus span{e1 , . . , en } ⊆ span{v1 , . . , vn }. But since e1 . . , en are n linearly independent vectors, we must have span{e1 , . . , en } = span{v1 , . . , vn }. That completes the proof. 2. Every separable inner product space has a countable complete orthonormal system. 14, every separable Hilbert space has a countable orthonormal basis. 1. Explain why the following statement is false, and find a meaningful way to fix it: A subset S of a vector space V is convex if and only if 12 x + 12 y ∈ S for all x, y ∈ S.

### Arithmetic Algebraic Geometry by ed van der Geer at al Birkhaeuser

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