The space $C[a,b]$ with $\|f\|:=\int_a^b |f(x)|d(x)$ is a normed space but not a Banach space

Question

Let $-\infty\lt a\lt b \lt \infty$. Let $C[a,b]$ is space of continuous functions and a function $\|.\|:C[a,b]\to R$ is given by $\|f\|:=\int_a^b |f(x)|d(x)$

a). Show that $\|\cdot\|$ is a norm in $C[a,b]$

b). Show that $C[a,b]$ with this norm is not a Banach space.

Well the 1st question is kind of easy. Non-negativity, Homogenity, and Triangle inequality is easily proven. At the 2nd question I have a problem.

I know the $$C[0,1]$$ space with Norm $$\|f\| = \int_0^1|f(x)|d(x)$$ is not complete that means not Banach-Space. But how do I prove the generality?

Does anyone has a suggestion?

Thanks

You should add your thoughts or ideas - then one knows where to start helping. Moreover I guess the upper bound of the Integral should be $b$. — T'x, Apr 11 '16 at 14:14
Well frankly i have no idea. i started with the Question a and couldn't prove non negativiy, symmetry and trianlge inequality — suugii19, Apr 11 '16 at 14:20
it seems the Question b can be easily derived from a. By the way thanks — suugii19, Apr 11 '16 at 14:21
OK, start with non-negativity. What do you have to do to show $|f| \ge 0$ if $f \in C[a,b]$ ?? — GEdgar, Apr 11 '16 at 14:24
I think you can manage to show $||x|| = 0 \Leftrightarrow x = 0$ and $||\lambda x || = |\lambda| ||x||$ — T'x, Apr 11 '16 at 14:25
well you say Integration of the Absolute Value is always positive — suugii19, Apr 11 '16 at 14:36
See also this question and the questions which are linked there. — Martin Sleziak, Apr 12 '16 at 05:14
Well my Brothers-in-arms, i made some of you angry because of my stupidity. Sorry for that :) . After few hours consideration and Research on Google. I made this please tell me if i made a mistake — suugii19, Apr 12 '16 at 14:35
b). Homogeneity comes from the $$||\lambda f|| = \int_a^b|\lambda f(x)| = \int_a^b|\lambda||f(x)|d(x) = |\lambda|\int_a^b|f(x)|d(x) = |\lambda|||f(x)||$$ — suugii19, Apr 12 '16 at 14:49
Triangle inequality comes like this $$||f+g|| = \int_a^b|f(x)+g(x)|d(x) \le \int_a^b(|f(x)|+|g(x)|)dx = \int_a^b|f(x)|d(x) + \int_a^b|g(x)|d(x) = ||f|| + ||g||.$$Here i used linearity — suugii19, Apr 12 '16 at 14:56

pmichel31415 · Accepted Answer · 2016-04-12T16:58:16.740

2

The first question is straight forward if you apply the definition of the norm, and I encourage you to do it on your own in order to assimilate the concept of norm.

For the second question, you need to produce a sequence of functions $(f_n\in C[a,b]^{\mathbb N})$ such that $f_n$ converges (with respect with the norm defined in the exercise) towards a function that is not in $C[a,b]$.

Hint : (Assuming, without loss of generality, that a=-1, b=1) You may want to try with $f_n:x\longrightarrow 1 \mathrm {\ if\ }x<0,(1 - nx)$ if $0\leqslant x<\frac 1 n$, $0$ otherwise.

This sequence is Cauchy with respect to the $L_1$ norm in $C[a,b]$ but you can prove that its limit is not cotinuous (it is $\mathbb {1}_{[-1,0]}$).

edited Apr 12 '16 at 16:58

answered Apr 11 '16 at 14:31

pmichel31415

1,207

2

Your sequence $f_n$ converges in norm to the constant function $0$ because $|f_n|=1/2 n.$ – DanielWainfleet Apr 12 '16 at 06:26
1

It is better to look at piecewise linear functions that approximate for example the indicator of $[1/2,1]$. Take for $f_n$ the function connecting the points $(0,0)-(1/2-1/n,0)-(1/2,1)-(1,1)$. They are Cauchy in integral norm but they do not converge. To see this take a continuous function $f$ and assume $f_n$ converges to $f$ in integral norm. Then show that $f$ is the indicator (which is a contradiction). To do this use continuity of the integral. – Apr 12 '16 at 16:40
Yes sorry for the mistake, I updated my answer. Thank you for your feedback. – pmichel31415 Apr 12 '16 at 16:59

The space $C[a,b]$ with $\|f\|:=\int_a^b |f(x)|d(x)$ is a normed space but not a Banach space

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