By J. L. Morris
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Extra info for Conference on the Numerical Solution of Differential Equations
CHAPTER 2. HILBERT SPACES 34 Proof: Let h ∈ H, deﬁne cj (h) = h, ej , and for n ∈ N let sn (h) = n j=1 cj ej ∈ H. 1: 0 ≤ ||h − sn (h)||2 n = h− n cj ej , h − j=1 n = ||h||2 − cj ej j=1 |cj |2 . j=1 This implies ∞. n 2 j=1 |cj | ≤ ||h||2 for every n and therefore ∞ 2 j=1 |cj | < We therefore obtain a linear map T : H → l2 (N) mapping h to the sequence (cj (h))j . Since nj=1 |cj (h)|2 ≤ ||h||2 we infer that ||T h|| ≤ ||h|| for every h ∈ H. For h in the span of (ej )j we furthermore have ||T h|| = ||h||.
Ek+1 are orthonormal. If H is ﬁnite-dimensional, this procedure will produce a basis (ej ) in ﬁnitely many steps and then stop. If H is inﬁnite-dimensional, it will not stop and will thus produce a sequence (ej )j∈N . By construction we have span(ej )j = span(aj )j , which is dense in H. Therefore (ej )j∈N is an orthonormal basis. 2 Suppose H is an inﬁnite-dimensional separable preHilbert space; and let (ej ) be an orthonormal basis of H. Then every element h of H can be represented in the form ∞ cj ej , h = j=1 where the sum is convergent in H, and the coeﬃcients cj satisfy ∞ |cj |2 < ∞.
3. ORTHONORMAL BASES AND COMPLETION T✲ V H ❅ T 35 ❅ S ❅ ❘ ❄ H It is customary to consider a pre-Hilbert space as a subspace of its completion by identifying it with the image of the completion map. Again this theorem also holds for nonseparable spaces, but we prove it only for separable ones. Proof: Let V be a separable pre-Hilbert space. If V is ﬁnitedimensional, then V is itself a Hilbert space, and we can take T equal to the identity. 2. We have to show that T (V ) is dense in H = 2 (N). Let f ∈ 2 (N), and for n ∈ N let fn ∈ 2 (N) be given by fn (j) = f (j) if j ≤ n, 0 if j > n.