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The Assignment:

Let $V:= (C([0,1]),\|\cdot\|$) with $\|f\|:= \int_0^1|f(x)| \ dx$. Consider the continuous function $f$ for all $n \in \mathbb{N}$:

$$f_n: [0,1] \rightarrow \mathbb{R}, \ x\rightarrow \begin{cases} 1 & \text{, $\ $0 ≤ x ≤ $\frac{1}{2}$}\\[2ex] 1- (n+1)(x-\frac{1}{2}) & \text{, $\frac{1}{2} < x ≤ \frac{1}{2} + \frac{1}{n+1}$} \\[2ex] 0 & \text{, $\frac{1}{2} + \frac{1}{n+1} < x ≤ 1$} \end{cases} $$

Show with $f_n$ that $V$ is not a Banach space.

I want to show that $f_n$ is Cauchy but also that the limit function $f$ is not continuous and hence $ \notin C([0,1])$. The second part is obvious, since $f$ is not continous at $\frac{1}{2}$.

What I tried is that: Let $n≥m$: \begin{align} \|f_n - f_m\| &= \int_0^1 | f_n(x) - f_m(x) | \ dx \\ &= \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n+1}} | f_n(x) - f_m(x) | \ dx + \int_{\frac{1}{2}+\frac{1}{n+1}}^{\frac{1}{2}+\frac{1}{m+1}} | f_n(x) - f_m(x) | \ dx\\ &= \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n+1}} | (m-n) |\left(x-\frac{1}{2}\right) \ dx + \int_{\frac{1}{2}+\frac{1}{n+1}}^{\frac{1}{2}+\frac{1}{m+1}} 0 - (1 - (m+1)\left(x-\frac{1}{2})\right) \ dx \\ &= ( n-m)\left[\frac{1}{2}x^2 - \frac{1}{2}\right]_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n+1}} + (...) = (...) \end{align}

Are my calculations and my idea correct so far? Since the calculating by hand gets messy with the int-intervals (at least in my opinion), I used maple to get:

$$(...) = -\frac{1}{2} \frac{(m-n)(3m+2-n)}{(m+1)(n+1)^2}$$

Which doesn't seem right, which is why I was wondering if I made a typo or I did a mistake before.

I'd appreciate any help.

Davide Giraudo
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Nhat
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  • Well. I think it will be easier to consider $||f_n - f_m|| \leq ||f_n - f|| + ||f_m - f||$ (note that $||f_n - f||$ still makes sense. –  May 29 '14 at 22:35
  • ... Good idea. - However, I'm still having problems determining the $N$ such that for larger $n≥m$ the inequality holds. The approach yields: $|f_n - f_m| ≤ \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n+1}} 1 - (n+1)(x-\frac{1}{2}) dx + \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{m+1}} 1 - (m+1)(x-\frac{1}{2}) dx = \frac{1}{2} \frac{m+n+2}{(n+1)(m+1)} < \frac{m+n+2}{(n+1)(m+1)}$. How do I go further from here? – Nhat May 29 '14 at 22:47
  • You dont have to be so precise here. You can approximate $\int_{\frac 12}^{\frac 12 + \frac{1}{n+1}}1- (n+1) (x-\frac 12) dx \leq \frac{1}{n+1}$ and similarly for the second term, so $||f_n - f_m|| \leq \frac{1}{n+1} + \frac{1}{m+1}$. –  May 29 '14 at 22:53
  • I'm not seeing it, why can you approximate the integral like that? – Nhat May 29 '14 at 22:57
  • I just forget about the term $(n+1)(x-\frac 12)$ as it is positive on $[\frac 12, \frac 12 + \frac{1}{n+1}]$ –  May 29 '14 at 22:58
  • Thanks a lot, John! Too bad, I cannot close this thread with your answer. – Nhat May 29 '14 at 23:07
  • Possible duplicate of $C[0,1]$ with $L^1$ norm is not Banach space. –  Oct 14 '15 at 09:37
  • See also: http://math.stackexchange.com/questions/402841/c0-1-is-not-complete-space-with-respect-to-norm-lvert-f-rvert-1-int – Martin Sleziak Apr 03 '16 at 01:37

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