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I am studying "real analysis and applications :theory in practice" I have some question in this books'problem. so if you know following ploblem, please teach me ! the ploblem is as follows

excercises for section 7.2 I with N

consider the piecewise linear fuctions in $C[-1,1]$ given by $f_n(x)=0$ for $-1\leq x\leq 0$, $f_n(x)=nx$ for $0\leq x\leq \frac1n$, and $f_n(x)=1$ for $\frac1n\leq x\leq 1$ with $L^1[-1,1]$ norm

(a) show that $f_n$ is Cauchy in the $L^1$-norm (already clear for your help)

(b) show that $f_n$ converges $\chi_\left(0,1\right]$, the characteric function of (0,1], in the $L^1-norm$

(c) show that $\Vert \chi_\left(0,1\right]-h\Vert_1 \gt 0$ for every h in C[-1,1]

I solved above 3 question but problem is as follows

(d) conclude that C[-1,1] is not complete in the $L^1-norm$

this is what i was trying to solve.

(1), what we need to prove this problem suffice to show there is cauchy sequence whose limit is not in C[-1,1].

(2)then suppose (a),(b) is true, (a),(b) said if we choose piecewise function like above fnfn. it converges to a vector $\chi_\left(0,1\right]$ but $\chi_\left(0,1\right]$ is not in C[-1,1] because $\chi_\left(0,1\right]$ is not continuous function on [-1,1].

(3)thus C[-1,1] is not complete.

is there any faults without using problem (c)?

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  • $\begingroup$ Could you write down an expression for $\|f_n-f_m\|_1?$ $\endgroup$ – mfl Sep 2 '17 at 14:06
  • $\begingroup$ you mean , write definiton of cauchy in $L^1-norm$? $\endgroup$ – fivestar Sep 2 '17 at 14:17
  • $\begingroup$ I mean $\|f_n-f_m\|_1=\int_{-1}^1 |f_n(x)-f_m(x)|dx?$ $\endgroup$ – mfl Sep 2 '17 at 14:18
  • $\begingroup$ yes that's right . sorry I could not write because I was not used to writing formulas. $\endgroup$ – fivestar Sep 2 '17 at 14:20
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Let $m > n \in \mathbb{N}$. You can calculate $\|f_m-f_n\|_1$ as the area of the shaded triangle with height $1$ and side length $\frac{1}{n} - \frac{1}{m}$.

$$\|f_m-f_n\|_1 = \frac{1}{2n} - \frac{1}{2m} \xrightarrow{m, n \to \infty} 0$$

enter image description here

Your proof of the $(b)$ part is essentially correct. Notice that you don't have to look at $[-1,0]$ and $(0, 1]$ separately, since you can do the separation directly in the integral, exactly like you did in your case $(2)$. And you have a typo: $\chi_{(-1,1]}$ in the integral should be $\chi_{(0,1]}$.

Again, we can use geometry: $\|f_n - \chi_{(0, 1]}\|_1$ is the area of the shaded triangle with side length $\frac{1}{n}$ and height $1$ so:

$$\|f_n - \chi_{(0, 1]}\|_1 = \frac{1}{2n} \xrightarrow{n\to\infty}0$$ enter image description here

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  • $\begingroup$ Just curious: How do you draw this picture? could you tell me the tool you use? Thx $\endgroup$ – Yuhang Sep 2 '17 at 15:15
  • $\begingroup$ @Yuhang Chen I used TikZ with the package tkz-euclide. I just manually defined the coordinates of some points, added labels, and connected them with lines. I played around with this code a while ago to get something similar, and today I found this for the shading. Note that there are probably much better ways to plot graphs of functions (pgfplots, for example). $\endgroup$ – mechanodroid Sep 2 '17 at 16:48
  • $\begingroup$ @Yuhang Chen Thank you for more detailed explanation. I have more question. if $\lim_{n,m \to \infty} \Vert f_m-f_n\Vert_1=0$ then $f_n$ is cauchy? $\endgroup$ – fivestar Sep 3 '17 at 10:01
  • $\begingroup$ Yes. It means $\forall\varepsilon > 0$ $\exists n_0 \in \mathbb{N}$ such that $m, n \geq n_0 \implies \|f_m - f_n\|_1 < \varepsilon$, which is exactly the definition of Cauchyness for $(f_n)_{n=1}^\infty$ in $L^1$ norm. $\endgroup$ – mechanodroid Sep 3 '17 at 11:06
  • $\begingroup$ @mechanodroid I understand what you described!. thank you ! then if you don't mind, could you check above problem (b) that I solved? $\endgroup$ – fivestar Sep 3 '17 at 11:52
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If $0<m\le n$ we have that

$$\|f_n-f_m\|_1=\int_{-1}^1 |f_n(x)-f_m(x)|dx=\int_{0}^{1/n} (nx-mx)dx+\int_{1/n}^{1/m} (1-mx)dx.$$ That is

$$\|f_n-f_m\|_1=\dfrac{n-m}{2mn}=\dfrac{1}{2m}-\dfrac{1}{2n}<\dfrac{1}{2m}.$$

Now, for any $\epsilon>0$ there exists $N\in\mathbb{N}$ ($N>\dfrac{1}{2\epsilon}$) such that $n,m\ge N\implies \|f_n-f_m\|_1<\epsilon.$ This shows that $(f_n)$ is a Cauchy sequence.

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  • $\begingroup$ Writing the last fraction as $\frac 1{2m} - \frac 1{2n}$ makes it evident the sequence is Cauchy, I suppose. $\endgroup$ – Pedro Tamaroff Sep 2 '17 at 14:40
  • $\begingroup$ @PedroTamaroff Sure. I have followed your suggestion. Thank you. $\endgroup$ – mfl Sep 2 '17 at 14:49
  • $\begingroup$ thank you . I understand perfectly ! $\endgroup$ – fivestar Sep 2 '17 at 14:49

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