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Consider the metric space $$B = \{ f \in C[0,1] : \int_a^b \left| f(x) \right| dx \leq 1\},$$ where $d(f,g) = \int_0^1 \left| f(x) - g(x) \right|dx$.

I'm trying to show that this metric space is not complete. I have proved that the metric space is not totally bounded. I did this by showing that there existed at least one sequence with a subsequence that was not Cauchy.

I'm aware that a metric space is compact $\Leftrightarrow$ Complete + Totally bounded.

Similarly, I know that compactness $\implies$ completeness, but $\neg$(compact) does not imply $\neg$(complete).

Can't seem to get the logic out.

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  • $\begingroup$ What you want is a Cauchy sequence in $(B, d)$ that does not converge in $(B, d)$. $\endgroup$ – Andrew D. Hwang May 22 '16 at 0:49
  • $\begingroup$ Jordan, you are not trying to prove that this space is not compact, rather that it is not complete. Here are some avenues to explore: (1) Try finding a Cauchy sequence in this space which does not converge (2) If you can find the completion of this space, you will then show that the completion and the original space are not the same (or isomorphic). $\endgroup$ – Rado May 22 '16 at 0:52
  • $\begingroup$ Okay, so let $(f_n) \subset B$ be a sequence such that $$\forall \epsilon > 0 \ \exists N \in \mathbb{N} \ \text{such that} \ m,n > N \implies \int_0^1 \left| f_n(x) - f_m(x) \right| dx < \epsilon.$$ Now, suppose $f_n \to f$ as $n \to \infty$. Thus, consider that $$\int_0^1 \left| f_m(x) - f_n(x) \right| dx = \int_0^1 \left| f_m(x) +f(x) - f(x) - f_n(x) \right|dx \leq \int_0^1 \left| f_m(x) - f(x) \right| dx + \int_0^1 \left| f_n(x) - f(x) \right|dx = 2\epsilon.$$ By the triangle inequality... How does this arrive at a contradiction? $\endgroup$ – user319128 May 22 '16 at 1:45
  • $\begingroup$ @Jordan you showed nothing there, just that $\epsilon<2\epsilon$... $\endgroup$ – YoTengoUnLCD May 22 '16 at 1:49
  • $\begingroup$ How would you suggest I approach the problem? $\endgroup$ – user319128 May 22 '16 at 1:50
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Let $n \in \mathbb{N}$, and define $ f_n(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 0 & \text{if } 0 \leq x \leq 1/2\\ 2nx - n & \text{if } 1/2 < x \leq (n + 1)/(2n)\\ 1 & \text{if } (n + 1)/(2n) < x \leq 1\\ \end{array}\right.$

Then $\{f_n\}$ is a Cauchy sequence whose limit is $ f(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 0 & \text{if } 0 \leq x \leq 1/2\\ 1 & \text{if } 1/2 < x \leq 1\\ \end{array}\right.$

because $d(f,f_n) = \int_0^1 |f(x) - f_n(x)| dx = 1/(4n)$.

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  • $\begingroup$ $f_n$ being convergent already implies that it is Cauchy. $\endgroup$ – Math1000 May 22 '16 at 3:33
  • $\begingroup$ @Steven There is gap: For each $n \in \mathbb{N}$, the integral $\int_0^1 f_n(x) dx$ is not bounded by $1$. Even for $n=2$ this fails; the integral is $51/4$. $\endgroup$ – Very Confused Aug 24 '20 at 22:51

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