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(d) Still assuming the parallelogram law, show that for any a a € R, ||ax + y||²||ax - y||² = a(||x + y||² − ||x – y||²). (Hint. For a € N, this follows from part (c) by induction. Show it next for reciprocals of natural numbers, a = 1/n, then for (positive) rational numbers. Use continuity to conclude it is true for all a > 0, then deal with a ≤ 0.) Cu

(d) Still assuming the parallelogram law, show that for any a a € R, ||ax + y||²||ax - y||² = a(||x + y||² − ||x – y||²). (Hint. For a € N, this follows from part (c) by induction. Show it next for reciprocals of natural numbers, a = 1/n, then for (positive) rational numbers. Use continuity to conclude it is true for all a > 0, then deal with a ≤ 0.) Cu

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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(d) Still assuming the parallelogram law, show that for any a a € R, ||ax + y||²||ax - y||² = a(||x + y||² − ||x – y||²). (Hint. For a € N, this follows from part (c) by induction. Show it next for reciprocals of natural numbers, a = 1/n, then for (positive) rational numbers. Use continuity to conclude it is true for all a > 0, then deal with a ≤ 0.) Cu
Transcribed Image Text:(d) Still assuming the parallelogram law, show that for any a a € R, ||ax + y||²||ax - y||² = a(||x + y||² − ||x – y||²). (Hint. For a € N, this follows from part (c) by induction. Show it next for reciprocals of natural numbers, a = 1/n, then for (positive) rational numbers. Use continuity to conclude it is true for all a > 0, then deal with a ≤ 0.) Cu
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