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I want to check that $(C[0, 1], \|\cdot\|_1)$ is not a Banach space, where $$\|f\|_1 = \int_0^1 |f(x)|\,{\rm d}x.$$

I took $(f_n)_{n \geq 1}$ a sequence in $C[0, 1]$ given by: $$f_n(x) = \begin{cases} nx, &\text{ if } x < \frac{1}{n} \\ 1, &\text{ if } x \geq \frac{1}{n} \end{cases} $$

If $m > n$, I computed: $$\|f_m - f_n\|_1 = 1 - \frac{n+1}{m} + \frac{1}{n} + \frac{n}{2}\left( \frac{1}{m^2}-\frac{1}{n^2} \right)$$

It is intuitive that the sequence is Cauchy, but I don't know how to exactly formalize this using the definition of a Cauchy sequence. Sending $m,n$ to $+\infty $ at the same time intuitively gives zero, but it doesn't seem rigorous to me.

How can I formalize this $m,n \to \infty$?

To proving that the sequence does not converge also seems intuitive, we would get a sort of vertical line - not a function. But I'm also not sure of how to approach this - the only thing that comes to mind is some kind of contradiction - suppose that $f_n \to f$, and find $\epsilon > 0 $ such that $\|f_n - f\| > \epsilon $ for infinite values of $n$.

Is the above idea in the right track?

Thanks.

Cookie
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Ivo Terek
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    How come you get that? The norm is equal to the difference of the areas of two triangles with height $1$ and bases $1/n$ and $1/m$ respectively. That would give $\left|\frac{1}{2m}-\frac{1}{2n}\right|$. The expression that you computed $\to1+\frac{1}{2n}$ when $m\to\infty$. That wouldn't be a Cauchy sequence. – Alamos May 02 '15 at 16:49
  • Well.. I COULD have messed up there. I supposed $m>n$ and wrote $$(f_m - f_n)(x) = \begin{cases} (m-n)x , \text{ if } x \leq1/m \ 1 - nx, \text{ if } 1/m < x < 1/n \ 0, \text{ if } x \geq 1/n \end{cases}$$ – Ivo Terek May 02 '15 at 16:52
  • Then I computed $$\int_0^{1/m} (m-n)x,{\rm d}x + \int_{1/m}^{1/n} 1-nx,{\rm d}x.$$ – Ivo Terek May 02 '15 at 16:53
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    Don't you get $(m-n)\frac{1}{2m^2}+\left(\frac{1}{n}-\frac{1}{m}\right)-n\left(\frac{1}{2n^2}-\frac{1}{2m^2}\right)=\frac{1}{2n}-\frac{1}{2m}$? – Alamos May 02 '15 at 16:56
  • I'll redo the calculation, give me two seconds. – Ivo Terek May 02 '15 at 16:57
  • Yeah, I had messed up the first integral. I got the same result as you now, thanks for pointing! – Ivo Terek May 02 '15 at 17:01
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    Rather than actually going to the trouble of computing that integral and then looking for bounds on the answer, it is simpler just to note that $0 \le |f_n(x) - f_m(x)| \le 1$ for $0 \le x \le 1/n$, and $|f_n(x) - f_m(x)| = 0$ otherwise, and therefore $|f_n - f_m| \le 1/n$. – Nate Eldredge May 02 '15 at 17:03
  • @Alamos what was wrong with what you had written? It really seems that we would have the thing converging pointwise to that discontinuous function. – Ivo Terek May 02 '15 at 17:20
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    No, we don't. See my edits, please. – Martin Argerami May 02 '15 at 17:20
  • Oh, I got it. Not used to think about convergence in the integral norm. I'm convinced. Sad that that sequence wouldn't work. – Ivo Terek May 02 '15 at 17:24
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    @IvoTerek Yes, it does converge pointwise to that discontinuous function, but it also converges pointwise a.e. to the function $1$. So, we don't really get a contradiction. – Alamos May 02 '15 at 17:54
  • Oh, yeah. Makes sense.. the integral does not sense the change in a set of zero measure. I should have seen that haha. Thanks! – Ivo Terek May 02 '15 at 17:57
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    Possible duplicate of $C[0,1]$ with $L^1$ norm is not Banach space. –  Oct 14 '15 at 09:38

2 Answers2

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I think you are making your estimate too convoluted, probably because you try to calculate $\|f_n-f_m\|_1$ exactly. What I would do: assume $m>n$; then $f_m=f_n=1$ for $x>1/n$, and so $$ \|f_m-f_n\|_1=\int_0^{1/n}|f_m-f_n|\leq\frac2n $$ since $|f_n|\leq1$, $|f_m|\leq1$. This shows that the sequence is Cauchy, because given $\varepsilon>0$, if $m>n>2/\varepsilon$, then $\|f_m-f_n\|_1<\varepsilon$.

Edit: The problem with your example is that the limit is in C$[0,1]$: it is the constant function $1$. Indeed, $$ \|1-f_n\|_1=\int_0^{1/n}(1-nx)\,dx=\frac1n-\left.\frac{nx^2}2\right|_0^{1/n}=\frac1{2n}\to0. $$

Edit2: To achieve what you want, an $\|\cdot\|_1$-Cauchy sequence that does not converge in $C[0,1]$, you can try something like $$ f_n(t)=\begin{cases}0,&\ 0\leq t\leq 1/2-1/2n,\\ nt\,&\ 1/2-1/2n\leq t\leq 1/2+1/2n,\\ 1,&\ 1/2+1/2n\leq t\leq 1 \end{cases} $$

Martin Argerami
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Hint: consider $m>n, m = n + h$, and prove that the expression converges to $0$ at a speed independent of $h$. In other words: prove that

$$ \sup_{h>0} \left\| f_{n+h} - f_n \right\| \to 0 $$when $n\to\infty$.

mookid
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