E =le (0) =with the inner product{u: [0, 7] → RN | u is absolutely continuous,u(T), ù € L²([0, 7]. RN))[[(i(1), v (1)) + (u(1), v(1))]drfor all u, v € E where (...) denotes the inner product in RN. The corresponding normis defined byAU. VYTE=||u||E= [(lù(1³² +\u(1)1³²)dr. V u € E.For every u, ve E, we define=S² eQ(¹) [(ù (1), v(1)) + (A(1)u(1), v(t))]dt,4. V Y=and we observe that, by the assumptions (A1) and (A2), it defines an inner product inE. Then, E is a separable and reflexive Banach space with the norm||u|| =< u, u > ² Vue E.||ul|oo ≤S|| u ||where ||ul|∞ = maxic[0,71 |u(t)].The question is why is normdefined in this way, please clarify?(2)

Given as, E={u:o,T→ℝn u is absolutely continuous. u(0)=u(T).u∈L2o,T,,ℝN}˙ the inner product…

Answered: -E= T | bartleby

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter7: Distance And Approximation

Section7.1: Inner Product Spaces

Problem 12EQ

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