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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.

Friedrich
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  • Hi Zarathustra,Why the convergence to the characteristic function is not evident? It is! – Piquito May 12 '20 at 15:53
  • Well in $C([a,b])$ w.r.t. $| \cdot |_1$ it is not, since it is not an element of that space. – Friedrich May 12 '20 at 15:55
  • Precisely. Then your space is not complet. – Piquito May 12 '20 at 15:57
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    Is this really sufficient? How do I know that there is no other possible element in the space, which could be the limit. – Friedrich May 12 '20 at 15:58
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    You know that the distance between $\chi_{[1,2]}$ and the $f_k$ tends to $0$. Suppose there is a continuous function limit of the $f_k$. A few of "epsilontic" reasoning could lead you to two distinct limits wich is forbidden by the axiom of separability $d(x,y)=0 \iff x=y$. I thing you are right: you need more than the geometric intuition. So you have to apply some of the lucidity of Friedrich. – Piquito May 12 '20 at 16:08
  • Separability is only assumed for elements to the space, whence, a priori, we cannot be sure if the same holds for $\chi_{[1,2]}$ – Friedrich May 12 '20 at 16:22
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    You have the obvious isometric embedding of $\bigl(C([0,2]), \lVert,\cdot,\rVert_1\bigr)$ into $L^1([0,2])$. You know that in $L^1$ the sequence converges to the equivalence class of the characteristic function $\chi_{[1,2]}$. Thus if the sequence converged to $g$ in $C([0,2])$, we'd have $g = \chi_{[1,2]}$ almost everywhere. But every continuous function differs from $\chi_{[1,2]}$ on a set of positive measure. – Daniel Fischer May 12 '20 at 18:11
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    @Friedrich.- After the invoked "epsilontic" reasoning it must be understood you are in another context. Anyway read please last line on my comment. – Piquito May 13 '20 at 16:24

2 Answers2

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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 $x=1$.

mechanodroid
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  • Thank you for the very nice proof of the claim. I have two questions regarding the fact that I have a priori detected an element, which is a $| \cdot |_1$-limit, but it is not contained in the space $C([0,2])$. Firstly, is it possible to know a priori whether this candidate is contained in the completion of the space w.r.t. to the norm? Or secondly regardless, is it necessary to exclude the possibility that there already is an adequate limit in the space? – Friedrich May 12 '20 at 17:30
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    Yes, this element $f \notin C([0,2])$ is contained in the completion $X$ of $C([0,2])$ because it is a limit in $X$ of a sequence $(f_n)_n$ in $C([0,2])$ and $\overline{C([0,2])} = X$. If there were another limit $g \in C([0,2])$ then of course $g \in X$ as well so $f_n \to f$ and $f_n \to g$ in $X$. Uniqueness of the limit in $X$ implies $f = g$, which is a contradiction since $f \notin C([0,2])$. – mechanodroid May 12 '20 at 17:38
  • I have clarified my approach in the post above. If I am not mistaken, you are, too, referring to the completion of the metric space. I have a hard time grasping the concept that we can be sure that an arbitrary element (especially an element foreign of $C([a,b])$) finds itself in $X$. So, as long as the value of the metric w.r.t. $M$ (the underlying space and $f$) is well-defined, your line of argumentation holds up? (In this example we could even say that the norm mapping is well definied for $f$). – Friedrich May 12 '20 at 18:05
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    @Friedrich Ok, If I understood your argument correctly, it goes like this: notice that $(f_n)_n$ converges in $|\cdot|_1$ to an element $f \notin C([0,2])$ (btw. do you realize that this statement as stated doesn't even make sense? The norm $|f-f_n|_1$ is not defined if $f-f_n\notin C([0,2])$). Define the metric space $M$ as in your question and then $f_n \to f$ in $M$. Now, assume that $(f_n)_n$ converges in $C([0,2])$. – mechanodroid May 12 '20 at 18:19
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    Hence there exists $g \in C([0,2])$ such that $f_n \to g$ in $C([0,2])$. But also of course $g \in M$ and $f_n \to g$ in $M$ (the metric is exactly the same). Therefore $f_n \to f$ and $f_n \to g$ in $M$ so by the uniqueness of the limit in the metric space $M$ it follows $f = g$. But this is a contradiction since $f \notin C([0,2])$ and $g \in C([0,2])$. Therefore $(f_n)_n$ does not converge in $C([0,2])$. – mechanodroid May 12 '20 at 18:19
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    @Friedrich To make the first part more formal, define $M$ as a set as you did but set $$d(f,g) = \int_0^2 |f-g|$$ "manually" and show that it is well-defined. Then the argument is ok. However, this all can be circumvented by considering the normed completion $L^1([0,2])$ of $C([0,2])$ in place of $M$, or even better, proceeding like in my answer which works only within $C([0,2])$. – mechanodroid May 12 '20 at 18:22
  • Is this a standard result that $L^1([a,b])$ is the normed completion of $C([a,b])$? – Friedrich May 12 '20 at 18:25
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    @Friedrich Yes, you only need that $L^1([a,b])$ is complete and that $C([a,b])$ is dense in $L^1([a,b])$ (for the latter e.g. look here). – mechanodroid May 12 '20 at 18:28
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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.

Tony419
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    This is working in the completion, though, which can be a bit unsatisfying. One can instead show that this sequence "avoids" all fixed elements of $C([0,2])$ by a direct calculation not using $L^1$, but the details are annoying. – Ian May 12 '20 at 15:54
  • Yes, this answer assumes knowledge of the completed space. What if we only use the existence of the completion, following my idea? I did run into walls expanding the space to a metric space via $f$. – Friedrich May 12 '20 at 16:05
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    @Friedrich Basically you want to prove that $(\forall f \in C([0,2])(\exists \varepsilon > 0)(\forall K \in \mathbb{N})(\exists k \geq K) : | f_k - f |1 > \varepsilon$. This can in fact be done directly. Given $f \in C([0,2])$, you can find a $\delta>0$ such that if $|x-1|<\delta$ then $|f(x)-f(1)|<1/4$. Then let $k^*=\lceil 1/\delta \rceil$. Then examine $\int{1-1/k^}^{1+1/k^} |f_k(x)-f(x)| dx$. You can show directly that this behaves like $1/k^$ when $k \geq k^$. The intuition is that no continuous $f(x)$ can change so fast near $x=1$ as to follow $f_k(x)$ for large $k$. – Ian May 12 '20 at 16:51
  • @Vahe You are associating the elements of $C([0,2])$ with equivalence classes in $L^1$, each containing a continuous representant. Devoid of further argumentation, it is not clear that there is no continous representant in $\chi_{[1,2]}\in L^1$. This makes things rather complicated. – Friedrich May 12 '20 at 17:02
  • @Friedrich The equivalence class of $\chi_{[1,2]}\in L^1$ doesn't contain a continuous representant. – Tony419 May 12 '20 at 17:09
  • @Vahe your claim should not be left out in a proof, hence making the approach via $L^1$ a little unhandy – Friedrich May 12 '20 at 17:12