Showing that $C[0,1]$ is not complete

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Let $n,m\in\mathbb{N}$ and $n\geq m$.
Then $|f_n(x)-f_m(x)|\le 1$ on $[\frac{1}{2}-\frac{1}{m},\frac{1}{2}+\frac{1}{m}]$, since $0\le |f_i(x)|\le 1$ for all $i$.
This means that $$||f_n-f_m||_p=\left(\int_{\frac1{2}-\frac{1}{m}}^{\frac1{2}+\frac{1}{m}}|f_n(x)-f_m(x)|^pdx\right)^{1/p}\le \left(\int_{\frac1{2}-\frac{1}{m}}^{\frac1{2}+\frac{1}{m}}1dx\right)^{1/p}=\left(\frac{2}{m}\right)^{1/p}.$$ For $n,m\rightarrow\infty$, we see that $||f_n-f_m||_p\rightarrow 0$. Thus $(f_n)$ is Cauchy.

Suppose $f_n$ has a limit $f$ in $C[0,1]$. Then $\int_{0}^{1}|f(x)-f_n(x)|^pdx\leq||f-f_n||_p\rightarrow 0$ as $n\rightarrow\infty$. This gives us that $$f(x) = \begin{cases} 0 & \text{if $0 \le x < \frac1{2}$}\\ 1 & \text{if $ \frac {1}{2} \le x \leq 1 $} \end{cases}$$ which is not continuous.

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Updated on April 13, 2020

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  • Admin
    Admin over 3 years

    Let $C[0,1]$ have the following norm for finite $p$: $$||f||_p=\left(\int_0^1|f(x)|^pdx\right)^{1/p}$$ I want to show that $C[0,1]$ is not complete for this norm.

    How do I do this?

    What I know:
    So I want to find a Cauchy sequence that does not converge in $C[0,1]$. My idea was $$f_n(x) := \begin{cases} 0 & \text{if $0 \le x < \frac1{2}-\frac{1}{n}$}\\ \frac{n}{2}(x-\frac{1}{2}+\frac{1}{n}) & \text{if $\frac {1}{2}-\frac{1}{n} \le x \le \frac {1}{2} + \frac {1}{n}$}\\ 1 & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $} \end{cases}$$ It is clear that this converges to $$f(x) := \begin{cases} 0 & \text{if $0 \le x < \frac1{2}$}\\ 1 & \text{if $ \frac {1}{2} <x \leq 1 $} \end{cases}$$ which is not continuous. The point is now to prove that $f_n$ is Cauchy, but I have no idea how to prove this. Can you help?

    Edit:
    According to me, $||f_n-f_m||_p=\left(\int_{\frac1{2}-\frac{1}{m}}^{\frac1{2}+\frac{1}{m}}|f_n(x)-f_m(x)|^pdx\right)^{1/p}$. BUt how do I continue from here?

    • Prahlad Vaidyanathan
      Prahlad Vaidyanathan about 7 years
      Have you considered explicitly computing $\|f_n - f_m\|$?
    • Admin
      Admin about 7 years
      @PrahladVaidyanathan Please see edit, is that correct?
    • Daniel Fischer
      Daniel Fischer about 7 years
      It's sufficient to find an upper bound for $\lVert f_n - f_m\rVert$ that converges to $0$ when $m,n \to \infty$. Use a bound for $\lvert f_n(x) - f_m(x)\rvert$ on $\bigl[\frac{1}{2} - \frac{1}{m}, \frac{1}{2} + \frac{1}{m}\bigr]$ if $n \geqslant m$.
    • Admin
      Admin about 7 years
      @DanielFischer What for bound can I use?
    • Daniel Fischer
      Daniel Fischer about 7 years
      Since $0 \leqslant f_k(x) \leqslant 1$, you can bound the difference by $1$. Looking a bit closer, you can find a smaller bound, but that doesn't matter.
    • Admin
      Admin about 7 years
      @DanielFischer But why does $||f_n-f_m||$ converge to $0$ if $|f_n(x)-f_m(x)|$ is bounded?
    • Daniel Fischer
      Daniel Fischer about 7 years
      It's not that simple. But look at your expression for $\lVert f_n - f_m\rVert_p$, and estimate that by replacing the integrand $\lvert f_n(x) - f_m(x)\rvert^p$ by a constant upper bound.
    • Admin
      Admin about 7 years
      @DanielFischer Would I then get an upper bound of $\left(\frac{2}{m}\right)^{1/p}$?
    • Daniel Fischer
      Daniel Fischer about 7 years
      Yes (assuming $n \geqslant m$). And that gives you Cauchy.
    • Admin
      Admin about 7 years
      @DanielFischer I am going to write an answer for this, would you mind to check it when it's finished?
    • Admin
      Admin about 7 years
      @DanielFischer Please see my answer, it still seems a little too simple for my taste...
    • Daniel Fischer
      Daniel Fischer about 7 years
      You also should explicitly say that $f_n(x) = f_m(x)$ for $\bigl\lvert x-\frac{1}{2}\bigr\rvert \geqslant \frac{1}{m}$. But showing that the sequence is Cauchy is easy. What is harder is giving a correct proof that the sequence doesn't converge in $C([0,1])$ if you don't have any Lebesgue theory you can use. It's surprisingly hard to not omit any steps in that.
    • Admin
      Admin about 7 years
      @DanielFischer I thought that was the easy part, what can go wrong in that proof?
    • Daniel Fischer
      Daniel Fischer about 7 years
      Leaving out details. Most students leave out details. It's so obvious that one is very tempted to short-cut it.
    • Admin
      Admin about 7 years
      @DanielFischer Have I left out any details?
    • Daniel Fischer
      Daniel Fischer about 7 years
      Yes. Why does it follow that $f(x) = 0$ for $x < 1/2$ and $f(x) = 1$ for $x > 1/2$? It's not hard to give a correct argument for that, but one has to see that it's necessary.