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I have the following proof in my notes:

$(C[0,1],\| \cdot\|_1)$ is not complete

Consider the sequence of functions $f_n(t)= \begin{cases} \frac{1}{\sqrt t} &, & 1/n \le t \le 1 \\ \sqrt n &,&0\le t \le \frac{1}{n} \end{cases}$

It can easily be shown that it is Cauchy in $\lVert \cdot \rVert_1$ So we have that $\forall \varepsilon \exists N_\varepsilon$ such that $\lVert f_n -f \rVert_\infty \le \varepsilon \forall m,n >N_\varepsilon$

Suppose $f_n$ converges, Then $\exists f \in C[0,1] $such that $\lVert f_n-f\rVert_1 \to 0.$

Then $f_{n_k} \to f \text{ a.e in} [0,1]$

$ f(t)=\lim_{k \to \infty} f_{n_k}(t)= \frac{1}{\sqrt t} \text{ a.e in} [0,1]$. Absurd.

I'm just failing to understand the last line. Can someone please explain

  1. what is the absurd?

  2. And how does pointwise convergence follows from a.e convergence (last line). Am I supposed to understand this $f_{n_k} \to f \text{ a.e in} [0,1]$ as convergence in the real numbers, with the absolute value for every fixed t ( $|f_{n_k}(t) -f(t) | \text{ a.e in} [0,1]$ ) ?

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    $\begingroup$ $1/{\sqrt t}$ is not continuous on $[0,1]$. $\endgroup$ Jan 18, 2022 at 11:50
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    $\begingroup$ Since it happens almost everywhere, you can find a sequence $t_n$ which tends to $0$ such that $f(t_n)=\frac{1}{\sqrt{t_n}}$ for all $n$. This means $f$ can't be continuous at $0$, a contradiction. $\endgroup$
    – Mark
    Jan 18, 2022 at 11:52
  • $\begingroup$ @DavidC.Ullrich But $1/\sqrt t$ is a.e continuous in[0,1], right?, which is what $\lim_{k \to \infty} f_{n_k}(t)= \frac{1}{\sqrt t} \text{ a.e in} [0,1]$ says. So the limit exists $\endgroup$ Jan 18, 2022 at 11:59
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    $\begingroup$ It's not an element of $C[0,1]$! So the sequence is not convergent in $C[0,1]$. $\endgroup$ Jan 18, 2022 at 12:00
  • $\begingroup$ If two continuous functions on $(0,1]$ are equal a.e. then they are equal on all of $(0,1]$. $\endgroup$ Jan 18, 2022 at 16:39

1 Answer 1

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  1. The limit function $f$ does not belong to $\left(C([0, 1]), \|\cdot\|_{\ell^1}\right),$ creating the contradiction since $f$ was supposed to belong to this space. More precisely, $f$ is equal to $1/\sqrt{t}$ a.e, but by continuity, $$f(0) = \lim_{t \to 0} f(t) = +\infty,$$ and therefore $f$ may not be continuous at $0,$ since continuous functions on $[0,1]$ are in particular finite at $0,$ a contradiction.
  2. a.e convergence means precisely that the pointwise convergence is valid a.e.
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