Consider the spaces LP = LP (Rd) for 0 < p <∞ with Lebesgue measure. (a) Show that if ||ƒ +9||LP ≤ ||f||LP+ ||g||LP for all f and g then it is necessary to have p≥ 1. E.g. if 0 < p < 1 then give an example where this fails. (b) Consider LP LP(R) for 0 < p < 1 with Lebesgue measure. Show that there are no bounded linear functionals on this space. In other words if l is a linear functional l : L² → R (or C) and l satisfies |l(ƒ)| ≤ M ||f||LP(R) for all f € LP(R) and for some M > 0 then l = 0. = (To prove (a), first prove that if 0 < p < 1 and x, y> 0 then x² + y² > (x+y)³. For (b), let F be defined by F(x) = l(Xx) where X is the characteristic function of [0, x] and consider F(x) – F(y).)
Consider the spaces LP = LP (Rd) for 0 < p <∞ with Lebesgue measure. (a) Show that if ||ƒ +9||LP ≤ ||f||LP+ ||g||LP for all f and g then it is necessary to have p≥ 1. E.g. if 0 < p < 1 then give an example where this fails. (b) Consider LP LP(R) for 0 < p < 1 with Lebesgue measure. Show that there are no bounded linear functionals on this space. In other words if l is a linear functional l : L² → R (or C) and l satisfies |l(ƒ)| ≤ M ||f||LP(R) for all f € LP(R) and for some M > 0 then l = 0. = (To prove (a), first prove that if 0 < p < 1 and x, y> 0 then x² + y² > (x+y)³. For (b), let F be defined by F(x) = l(Xx) where X is the characteristic function of [0, x] and consider F(x) – F(y).)
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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![Consider the spaces LP = LP (Rd) for 0 < p < ∞ with Lebesgue measure.
(a) Show that if || ƒ + 9||L² ≤ ||f||LP+ ||9||LP for all f and g then it is necessary
to have p≥ 1. E.g. if 0 < p < 1 then give an example where this fails.
р
(b) Consider Lº LP(R) for 0 < p < 1 with Lebesgue measure. Show that
there are no bounded linear functionals on this space. In other words if l
is a linear functional l : L² → R (or C) and l satisfies
|e(f)| ≤ M||f||LP(R)
for all ƒ € L³(R) and for some M > 0 then l = 0.
=
(To prove (a), first prove that if 0 < p < 1 and x, y> 0 then x² + y² > (x + y)².
For (b), let F be defined by F(x) = (xx) where Xx is the characteristic function
of [0, x] and consider F(x) – F(y).)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdbd07f6b-900f-495a-9a3d-8b3e4b5ea96c%2F75951c66-9e28-4762-9357-fdb1a0069fe7%2Fzbpbak_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the spaces LP = LP (Rd) for 0 < p < ∞ with Lebesgue measure.
(a) Show that if || ƒ + 9||L² ≤ ||f||LP+ ||9||LP for all f and g then it is necessary
to have p≥ 1. E.g. if 0 < p < 1 then give an example where this fails.
р
(b) Consider Lº LP(R) for 0 < p < 1 with Lebesgue measure. Show that
there are no bounded linear functionals on this space. In other words if l
is a linear functional l : L² → R (or C) and l satisfies
|e(f)| ≤ M||f||LP(R)
for all ƒ € L³(R) and for some M > 0 then l = 0.
=
(To prove (a), first prove that if 0 < p < 1 and x, y> 0 then x² + y² > (x + y)².
For (b), let F be defined by F(x) = (xx) where Xx is the characteristic function
of [0, x] and consider F(x) – F(y).)
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