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Define the norm $\|f\|$ on $C[0,1]$

$$\|f\|=\int_{0}^{1}|f(x)| \text{ }dx.$$

Define $f_n(x)$ as $$f_n(x)=\begin{cases} nx & \text{if } 0\leq x\leq \frac{1}{n}\\ 1 & \text{if }\frac{1}{n}<x\leq 1 \end{cases} $$

Question 1:

I can see from the graph that $f_n$ is converging to a function which is not continuous. But how do I prove that $f_n$ is actually not converging in $C[0,1]$ with respect to this norm.

Question 2:

What is the completion of $C[0,1]$ with respect to this norm? Is it the set of integrable functions defined on $[0,1]$?

Edit:

The sequence $f_n$ is convergent to the function $1$ in this norm.

learning_math
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    That's not a norm, because for $f$ defined as $f(x)=-1$, you have $$\int_0^1 f(x)dx=-1$$ – 5xum Jul 12 '17 at 12:05
  • Thanks for pointing it out. I have edited the question. – learning_math Jul 12 '17 at 12:07
  • Q1. One way to show that a sequence is not converging in a given norm is to show that it is not a Cauchy sequence. Q2. You have a slight misstatement. It is the set of (equivalence classes of) integrable functions defined on $[0,1]$. Not "defined on $C[0,1]$". – GEdgar Jul 12 '17 at 12:10
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    Your $f_n$ converges to the constant function $1$ in this norm. – Anthony Carapetis Jul 12 '17 at 12:12
  • @GEdgar Thanks for pointing out the misstatement. About your suggestion on Q1: the sequence $f_n$ is Cauchy, so we are powerless. – learning_math Jul 12 '17 at 12:13
  • Anyway, the completion you get is known as $L^1([0,1])$, and it's almost the space of integrable functions that you suggest, but with one caveat: we have to identify two functions whenever they differ on a set of measure zero. For example the pointwise limit of your $f_n$ and the constant function $1$ are representatives of the same element of $L^1$. – Anthony Carapetis Jul 12 '17 at 12:24
  • @AnthonyCarapetis Thanks for pointing out my mistake, $f_n$ indeed converges to $1$ wrt this norm. – learning_math Jul 12 '17 at 12:25

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If your sequence was meant to show that $C[0,1]$ is not complete in the integral norm, it fails to achieve that goal, because its limit is the constant function $1$. Instead, see Showing that the space $C[0,1]$ with the $L_1$ norm is incomplete.

The completion of $C[0,1]$ with respect to the norm $\|f\| = \int_0^1 |f(x)|\,dx$ is the Lebesgue space $L^1[0,1]$. The proof of this fact consists of two independent parts, which are already discussed here:

  1. Continuous functions are dense in $L^1[0,1]$
  2. $L^1[0,1]$ is complete