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)?

  • $\begingroup$ 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

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

| cite | improve this answer | |
  • $\begingroup$ 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

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


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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.