Showing incompleteness of metric space by specifying non-convergent Cauchy sequence

Question

I have been looking at the space of continuous functions over a compact interval $C([0,2])$ equiped with the the integral norm of absolut values $\| \cdot \|_1$. I read a counterexample that showed how this is not a Banach space. The author gave a Cauchy sequence of functions $f_k$: for any $k\in \mathbb{N}$ over the intervall $[1-\frac{1}{k}, 1+\frac{1}{k}]$ linearily increasing values from $0$ to $1$, else constant. It is evident that $f_k$ converges towards $f$, the characteristic function of $[1,2]$, which is not in $C([0,2])$. But is it really evident?

What are all arguments needed to conclude that $f_k$ is not convergent in $C([0,2])$? As far as I understand, just showing $\|f_k - f\|_1 \rightarrow 0$ is insuffienct, as $f \notin C([0,2])$ and we have not shown that there does not exist a limit function in $C([0,2])$.

My idea: Since $(C([0,2]), \| \cdot \|_1)$ is a normed space, and any normed space is a metric space, there exists a completion of the space, where the limits of Cauchy sequnces exist and are unique. But how would I know that $f$ is in that complete space?

Clarification: My approach is to extend the normed space at hand to $M := C([0,2]) \cup \{ f\}$ and prove that is a metric space with the distance function $d(x,y):=\|x-y\|_1$. Since I know $f_k \rightarrow f$ in $(M,d)$ and limits of Cauchy sequences are unique in metric spaces, I now have a unique limit w.r.t. the norm at hand. The question is, if I can go back to extend this notion of uniqueness of the limit to the normed space $C([0,2])$ and from there conclude incompleteness.

More generally, I'd like to abstract a procedure to show incompleteness of a normed or metric space.

Answer 1

Assume that there exists $g \in C([0,2])$ such that $\|f_n-g\|_1 \to 0$. Let $\varepsilon \in \langle 0,1\rangle$ be arbitrary.

We have $$\int_0^{1-\varepsilon} |f_n(x)-g(x)|\,dx \le \int_0^2 |f_n(x)-g(x)|\,dx = \|f_n-g\|_1 \to 0.$$ For all $n\in\Bbb{N}$ such that $\frac1n < \varepsilon$ we moreover have $f_n|_{[0,1-\varepsilon]} \equiv 0$ so $$\int_0^{1-\varepsilon} |g(x)|\,dx = 0\implies g|_{[0,1-\varepsilon]} \equiv 0$$ Since $\varepsilon$ was arbitrary we conclude $g|_{[0,1\rangle} \equiv 0$.

Similarly, let $\varepsilon \in \langle 0,1\rangle$ be arbitrary.

We have $$\int_{1+\varepsilon}^2 |f_n(x)-g(x)|\,dx \le \int_0^2 |f_n(x)-g(x)|\,dx = \|f_n-g\|_1 \to 0.$$ For all $n\in\Bbb{N}$ such that $\frac1n < \varepsilon$ we moreover have $f_n|_{[1+\varepsilon,2]} \equiv 1$ so $$\int_0^{1-\varepsilon} |1-g(x)|\,dx = 0\implies g|_{[1+\varepsilon,2]} \equiv 1$$ Since $\varepsilon$ was arbitrary we conclude $g|_{\langle 1,2]} \equiv 1$.

Now, $$\lim_{x\to1^-} g(x) = 0 \ne 1 = \lim_{x\to 1^+} g(x)$$ which contradicts continuity of $g$ at the point $x=1$.

Answer 2

Note that the above argument shows that $f_k$ is a Cauchy sequence in $C([0,2], \|\cdot\|_1)$. Now we can argue by contradiction: assume that there exists $g\in C([0,2])$ such that $\lim_{k\rightarrow\infty} f_k=g$ in $\|\cdot\|_1$. You have shown that also $\lim_{k\rightarrow\infty} f_k=\chi_{[1,2]}$ in $\|\cdot\|_1$. Thus by the uniqueness of the limit in $L^1$ we have $g=\chi_{[1,2]}$ a.e. which is a contradiction.