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Consider $C([0, 1])$, the linear space of continuous complex-valued functions on the interval $[0, 1]$, with the norm

$$\displaystyle\lVert f\rVert_1 = \int_0^1 \lvert f(x)\rvert \,dx.$$

I have to show that $C([0, 1])$ is not complete with respect to this norm. I have found the following example from a book.

Let $f_n \in C[0,1]$ be given by

$$f_n(x) := \begin{cases} 0 & \text{if $0 \le x \le \frac1{2}$}\\ n(x-\frac{1}{2}) & \text{if $\frac {1}{2} < x \le \frac {1}{2} + \frac {1}{n}$}\\ 1 & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $} \end{cases}$$

How to prove that $f_n$ is a Cauchy sequence with respect to $\lVert \cdot\rVert_1$?

If I use basic definition then I have to prove that $\lVert f_n - f_m\rVert_1 < \epsilon$ $\forall n, m > N$. But I am finding it difficult to prove this.

Please help me to understand how to prove that $f_n$ is Cauchy sequence in $C([0, 1])$.

Thanks

mathscrazy
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  • "I am confused with whether I have to choose $f_n = 0$, $1$ or $n(x-\tfrac12)$." I don't understand this. Each $f_n$ is a continuous function on the unit interval given by the formula provided. Nowhere is there a choice involved. – kahen May 26 '13 at 12:29
  • The proof goes as almost all proofs in basic real analysis: "Let $\epsilon > 0$ be given. We are to find a natural number $N$ (depending on $\epsilon$) such that when $m,n \geq N$, then [$\ldots$]" - The only challenge lies in picking $N$ large enough. I suggest drawing a picture so you can see how to pick a good $N$. – kahen May 26 '13 at 12:36
  • Please take a look at my edit to see the corrections I made to your TeX. Particularly note \lvert\lvert -> \lVert. – kahen May 26 '13 at 12:45
  • See also: http://math.stackexchange.com/questions/814305/let-v-c0-1-cdot-with-f-int-01fx-dx-consider-the and http://math.stackexchange.com/questions/1479443/c0-1-with-l1-norm-is-not-banach-space – Martin Sleziak Apr 03 '16 at 01:37
  • But how to prove that $f_n$ is not convergent with respect to this norm ? – learning_math Aug 15 '16 at 21:38
  • @mathscrazy I know its been years since, but I am wondering what book did you find this example at? – Tiger Blood Oct 04 '16 at 19:46
  • @Ugo I just forgot from which book I got this problem but will search it for you. Thanks – mathscrazy Oct 05 '16 at 10:18
  • The other answers here are incomplete. For a complete answer, see https://math.stackexchange.com/a/375234/52912 – PatrickR Mar 08 '20 at 07:06

2 Answers2

9

Let $m\leq n$ both natural numbers, then $$\|f_n-f_m\|_1 = \int_0^1 |f_n(x)-f_m(x)|\,\mathrm{d}x $$ $$ = \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n}}(n-m)\left(x-\frac{1}{2}\right)\,\mathrm{d}x + \int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}\left(1-m\left(x-\frac{1}{2}\right)\right)\,\mathrm{d}x.$$

Now try to bound these integrals for $n,m\geq N$.

EternalBlood
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Abel
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    Thanks for the answer. Why we need to break the inteval from ${\frac{1}{2}+\frac{1}{n}}$ to ${\frac{1}{2}+\frac{1}{m}}$? Do I have ot integrate this integral and then only bound of this integral can be found. – mathscrazy May 26 '13 at 12:43
  • Because between $\frac{1}{2}+\frac{1}{n}$ and $\frac{1}{2}+\frac{1}{m}$, $f_n(x) = 1$, whereas $f_m(x) = m(x-\frac{1}{2})$. Both integrals are easy to compute, so you could just do that. I believe you'll end up with $\frac{1}{2m}-\frac{1}{2n}$, which is quite easy to bound. – Abel May 26 '13 at 12:46
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    Thanks a lot. I understood fully now. Later on It could be bounded by $\frac{1}{2m}< \frac{1}{N} < \epsilon $. – mathscrazy May 26 '13 at 12:53
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Its correct but some part is still left. You have to show that it is not complete. For this we need to show that limit point is not continuous. Suppose function f(x) is the limit. Observe that it is not right continuous at x=1/2. hence it is not complete

Satyam Bhartiya
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