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Topic: Complete set (Read 1219 times) |
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wonderful
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Complete set
« on: Sep 8th, 2008, 4:19pm » |
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Let S be the set of all continous function on [0,1], with norm ||f(x)||= integral |f(x)| from 0 to 1. Show that S is not complete.
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| « Last Edit: Sep 16th, 2008, 4:10pm by wonderful » |
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Obob
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Re: Complete set
« Reply #1 on: Sep 8th, 2008, 11:12pm » |
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The norm should be ||f||= int |f|. The completion of C[0,1] with respect to this norm is the space L^1[0,1] of all measurable functions f such that int |f| < infty, with norm given by ||f||=int |f|. For instance, the function g given by g(x) = 0 for x<1/2 and g(x) = 1 for x>=1/2 is the limit of a sequence of continuous functions with respect to this norm (just take a sequence of piecewise linear functions which start off at 0, shoot up to 1 more and more steeply near 1/2, and stay at 1 from then on). Since C[0,1] with this norm is a sub-normed space of L^1[0,1] and since there is a sequence in C[0,1] converging to an element of L^1[0,1]-C[0,1], C[0,1] cannot be complete.
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wonderful
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Re: Complete set
« Reply #2 on: Sep 16th, 2008, 4:09pm » |
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Excellent Obob!
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