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24.2 Lemma Let X be an inner product space and ƒ € X'. (a) Let {u₁, 2,...} be an orthonormal set in X. Then Σlf(un)1² ≤||f||². Proof: (a) For m = 1,2,..., let ym = Σf(un)un. n=1 Since {u₁,, um} is an orthonormal set, m ||ym||² = (ym, Ym) = f(un)² = 3m, say. -[if n=1 m Since f(ym) = f(un)f(un) = ßm and [ƒ(3m)| ≤ ||f|| ||ym||, we s n=1 that m≤|f||√3m, that is, ßm ≤ ||f||². Letting m→→∞ (if the set denumerable), we obtain

24.2 Lemma Let X be an inner product space and ƒ € X'. (a) Let {u₁, 2,...} be an orthonormal set in X. Then Σlf(un)1² ≤||f||². Proof: (a) For m = 1,2,..., let ym = Σf(un)un. n=1 Since {u₁,, um} is an orthonormal set, m ||ym||² = (ym, Ym) = f(un)² = 3m, say. -[if n=1 m Since f(ym) = f(un)f(un) = ßm and [ƒ(3m)| ≤ ||f|| ||ym||, we s n=1 that m≤|f||√3m, that is, ßm ≤ ||f||². Letting m→→∞ (if the set denumerable), we obtain

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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24.2 Lemma Let X be an inner product space and ƒ € X'. (a) Let {u₁, ₂,...} be an orthonormal set in X. Then Σlf(un)1² ≤||f||². Proof: (a) For m=1,2,..., let Ym = Σf(un)un. n=1 Since {u₁,, um} is an orthonormal set, m m ||3m||² = (ym, Ym) = |ƒ(un)|² = ßm, say. n=1 Since f(ym) = [ƒ(un)ƒ (un) = ßm and |ƒ(ym)| ≤ ||ƒ|| ||ym||, we see n=1 that 3m ≤||f||√m, that is, ßm ≤ ||f||². Letting m→ ∞o (if the set is denumerable), we obtain Σlf(un)1² ≤||f||².
Transcribed Image Text:24.2 Lemma Let X be an inner product space and ƒ € X'. (a) Let {u₁, ₂,...} be an orthonormal set in X. Then Σlf(un)1² ≤||f||². Proof: (a) For m=1,2,..., let Ym = Σf(un)un. n=1 Since {u₁,, um} is an orthonormal set, m m ||3m||² = (ym, Ym) = |ƒ(un)|² = ßm, say. n=1 Since f(ym) = [ƒ(un)ƒ (un) = ßm and |ƒ(ym)| ≤ ||ƒ|| ||ym||, we see n=1 that 3m ≤||f||√m, that is, ßm ≤ ||f||². Letting m→ ∞o (if the set is denumerable), we obtain Σlf(un)1² ≤||f||².
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