E = {u: [0, 7] → Rª | u is absolutely continuous, le (0) = u(T). ù & L²([0, 71, RN)} with the inner product E= = f* [(i(1), v (1)) + (u(1), v(1))]dr for all u, v € E where (...) denotes the inner product in RN. The corresponding norm is defined by HellE = o da + \u())dr. Vus E. For every u, v E E, we define 4. V Y= fea¹)[(ù (1), v(1)) + (A(1)u(t), v(r))]dr, and we observe that, by the assumptions (A1) and (A2), it defines an inner product in E. Then, E is a separable and reflexive Banach space with the norm ||u|| =< u, u x = Vue E. ||ul|oo ≤S|| u || where ||u||∞ = max;c[0,7] |u(1)]. The question is why is norm defined in this way, please clarify? (2)
E = {u: [0, 7] → Rª | u is absolutely continuous, le (0) = u(T). ù & L²([0, 71, RN)} with the inner product E= = f* [(i(1), v (1)) + (u(1), v(1))]dr for all u, v € E where (...) denotes the inner product in RN. The corresponding norm is defined by HellE = o da + \u())dr. Vus E. For every u, v E E, we define 4. V Y= fea¹)[(ù (1), v(1)) + (A(1)u(t), v(r))]dr, and we observe that, by the assumptions (A1) and (A2), it defines an inner product in E. Then, E is a separable and reflexive Banach space with the norm ||u|| =< u, u x = Vue E. ||ul|oo ≤S|| u || where ||u||∞ = max;c[0,7] |u(1)]. The question is why is norm defined in this way, please clarify? (2)
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
Related questions
Question
![E = {u: [0, 7] → Rª | u is absolutely continuous,
le (0) = u(T), ù ≤ L²([0, 7]. R^)}
with the inner product
[*[(i(1), v (1)) + (u(1), v(1))]dr
for all u, v € E where (...) denotes the inner product in RN. The corresponding norm
is defined by
[*(\à(1³² +\u(1)³²)dt. V u € E.
< u,v>E=
||u|| E =
For every u, ve E, we define
eQ(¹) [(ù (1), v(1)) + (A(1)u(1), v(1))]dt,
and we observe that, by the assumptions (A1) and (A2), it defines an inner product in
E. Then, E is a separable and reflexive Banach space with the norm
||u|| =< u, u > ²
4. V Y=
Vue E.
||ul|∞ ≤S||u||
where ||ul|∞ = max;c[0,71 |u(1)|.
The question is why is norm
defined in this way, please clarify?
(2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1cf3e72-8d9a-4e05-9b7f-5e0dd5a4d8a9%2F0ad0d97a-ca9f-47e7-879d-d979316e295f%2Fwsao26c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:E = {u: [0, 7] → Rª | u is absolutely continuous,
le (0) = u(T), ù ≤ L²([0, 7]. R^)}
with the inner product
[*[(i(1), v (1)) + (u(1), v(1))]dr
for all u, v € E where (...) denotes the inner product in RN. The corresponding norm
is defined by
[*(\à(1³² +\u(1)³²)dt. V u € E.
< u,v>E=
||u|| E =
For every u, ve E, we define
eQ(¹) [(ù (1), v(1)) + (A(1)u(1), v(1))]dt,
and we observe that, by the assumptions (A1) and (A2), it defines an inner product in
E. Then, E is a separable and reflexive Banach space with the norm
||u|| =< u, u > ²
4. V Y=
Vue E.
||ul|∞ ≤S||u||
where ||ul|∞ = max;c[0,71 |u(1)|.
The question is why is norm
defined in this way, please clarify?
(2)
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