##### real analysis question

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Apr 28th, 2015

||.||_1 is an integral of a function's absolute value.

Here |(f_n - f)(x)| = {0, when x from [0; 1/2] U [1/2+1/n; 1] and = n(x-1/2) for x from (1/2; 1/2+1/n).
It is a linear function at (1/2; 1/2+1/n), = 0 at 1/2 and =1 at 1/2+1/n, so the integral is the area of a right triangle,

(1/2)*1*1/n = 1/(2n). Yes, this is tends to 0 when n tends to infinity.

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We can compute the integral directly:
n*int(x-1/2) = n*(x^2/2 - x/2)  with x from 1/2 to 1/2+1/n.
This is equal to (1/2)n*[(1/2+1/n)^2 - (1/2)^2 - (1/2+1/n) + 1/2] =
= (n/2)*[1/n + 1/n^2 + 1/4 - 1/4 -1/2 - 1/n + 1/2] = (n/2)*(1/n^2) = 1/(2n).

Apr 28th, 2015

I have a question how does this show that c is not complete in 1 norm?

Apr 28th, 2015

I thought that was a comment.

Yes, this shows. We have functions fn which are continuous on [0; 1] (is it clear for you?) and also in L1. They are tend _by the norm of L1_ to the function f (we prove that), but f isn't continuous.

Because fn -> f by ||.||1, that sequence is fundamental with respect to ||.||1. But the sequence doesn't converge to something in C[0; 1] (although it does converge _pointwise_ to f). This is the contradiction with the definition of norm's completeness. So the C[0; 1] is not a closed subspace of L1[0; 1].

Is this the answer you need?

Apr 29th, 2015

Yes i think its a little more clear, when you say norm completeness what do you mean? Is it like completeness in C[0,1] in the vertical axis? Is it because there exists all of its limit points?

Apr 30th, 2015

Norm completeness is a standard notion (see wiki, for example http://en.wikipedia.org/wiki/Banach_space).
It means that if a sequence (x_n) of elements of a normed space X is Cauchy (fundamental) (||x_n - x_m|| --> 0, m,n-->inf) then there is exists a limit of it (x_n --> x from X, n-->inf).

Note that for any metric (and therefore norm) having a limit implies fundamentality.

Apr 30th, 2015

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Apr 28th, 2015
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Apr 28th, 2015
Oct 20th, 2017
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