Proposition 18. Norms are conver. Proof. Let || - || be a norm on a vector space V. Then for all x, y € V and t € [0, 1], ||tx + (1 – t)y|| ≤ ||tx|| + ||(1 – t)y|| = |t|||x|| + |1 − t|||y|| = t||x|| + (1 – t)||y||| where we have used respectively the triangle inequality, the homogeneity of norms, and the fact that t and 1- t are nonnegative. Hence ||-|| is convex. 0
Proposition 18. Norms are conver. Proof. Let || - || be a norm on a vector space V. Then for all x, y € V and t € [0, 1], ||tx + (1 – t)y|| ≤ ||tx|| + ||(1 – t)y|| = |t|||x|| + |1 − t|||y|| = t||x|| + (1 – t)||y||| where we have used respectively the triangle inequality, the homogeneity of norms, and the fact that t and 1- t are nonnegative. Hence ||-|| is convex. 0
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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![Proposition 18. Norms are conver.
Proof. Let || || be a norm on a vector space V. Then for all x, y = V and t = [0, 1],
||tx + (1 - t)y|| ≤ ||tx|| + ||(1 – t)y|| = |t|||x|| + |1 − t|||y|| = t||x|| + (1 – t) ||y||
where we have used respectively the triangle inequality, the homogeneity of norms, and the fact that
t and 1-t are nonnegative. Hence ||- || is convex.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F56b4ab23-f6d5-4206-a94e-63cad8267fd4%2F7a7839a6-5b99-4ea0-a3c9-f9b2b4aa50c5%2Fps45rht_processed.png&w=3840&q=75)
Transcribed Image Text:Proposition 18. Norms are conver.
Proof. Let || || be a norm on a vector space V. Then for all x, y = V and t = [0, 1],
||tx + (1 - t)y|| ≤ ||tx|| + ||(1 – t)y|| = |t|||x|| + |1 − t|||y|| = t||x|| + (1 – t) ||y||
where we have used respectively the triangle inequality, the homogeneity of norms, and the fact that
t and 1-t are nonnegative. Hence ||- || is convex.
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