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To better train my intuition, what are some illustrative examples of function sequences in C[0,1] that are Cauchy but do not converge under the integral norm?

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A now-deleted answer suggested "example 7.13 of the following book:";. If anyone can find that example, I urge them to post it as an answer, because it might be an interesting example, and I cannot find it. – MickG Dec 23 '15 at 15:45
up vote 9 down vote accepted

You can get examples by considering elements of $L^1[0,1]$ that are not equal almost everywhere to any continuous function, and considering sequences of continuous functions converging in the $L^1$ norm to these discontinuous functions. Because convergent sequences are Cauchy and $L^1$ limits are unique up to equality almost everywhere, such sequences will be Cauchy and nonconvergent in $C[0,1]$.

E.g., let $f$ be $1$ on $[0,\frac{1}{2}]$ and $0$ elsewhere. Let $f_n$ be the continuous function that is $1$ on $[0,\frac{1}{2}]$, $0$ on $[\frac{1}{2}+\frac{1}{n},1]$, and linear on $[\frac{1}{2},\frac{1}{2}+\frac{1}{n}]$. Then because $f_n\to f$ in $L^1$, $(f_n)$ is Cauchy. (The Cauchy-ness is also easy to verify directly.) If there were a limit function $g\in C[0,1]$, you would have $g=f$ a.e.. But this is impossible, because the left-hand and right-hand limits at $\frac{1}{2}$ would not agree.

More generally, a Cauchy sequence in a metric space $X$ with completion $\overline{X}$ that does not converge in $X$ is basically the same as a sequence in $X$ that converges to an element of $\overline{X}\setminus X$. In a case like this where $X=C[0,1]$ with $L^1$ norm and $\overline{X}=L^1[0,1]$ have explicit descriptions, you can find examples by starting with an element of $\overline{X}\setminus X$, and find a sequence in $X$ converging to that element. The same idea applies to demonstrating nonconvergent Cauchy sequences in $\mathbb{Q}$, where you can take any irrational number and consider the sequence of truncated decimal expansions.

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By measure theory you can always pass to a subsequence that converges pointwise a.e. so it's usually quite easy to determine the only possible limit. Then you should try to build the examples in such a way that the pointwise (a.e.) limit has no chance of being continuous.

Here's an example of what I have in mind: $f_{n}(x) = \begin{cases} 0 & 0 \leq x \leq \frac{1}{2} \\\ nx - \frac{n}{2} & \frac{1}{2} \leq x \leq \frac{1}{2} + \frac{1}{n} \\\ 1 & \frac{1}{2} + \frac{1}{n} \leq x \leq 1 \end{cases}$ defined for $n \geq 2$. It's easy to see that it converges in the $L^{1}$-norm to the characteristic function of $[\frac{1}{2},1]$.

Added: Since an $L^{1}$-Cauchy sequence will always converge ($L^{1}$ is complete by the theorem of Riesz-Fischer (or Fréchet-Riesz)), the only thing you have to do is to pick a function in $L^{1}$ that doesn't have a continuous representative. The easiest example is a step function (that's probably the reason why all three answerers came up with essentially the same example). This suffices, because every function in $L^{1}$ is the $L^1$-limit of a sequence of continuous (even smooth) functions, as can be seen by convolving with mollifiers, for example. In other words, $C[0,1]$ (or even $C^{\infty}[0,1]$) are dense in $L^{1}$.

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How about the sequence of functions given by $f_n(x)=x^n$? This sequence is Cauchy in $(C([0,1]),\|\cdot\|_{L^1(0,1)})$ but does not converge in $C([0,1])$.

EDIT: This example doesn't work. Thanks for pointing that out. An example that does work is given by the sequence of continuous functions $f_n$ with $f_n(x)=\frac{2}{n}(x-\frac{1}{2}+\frac{1}{n})$ for $|x-\frac{1}{2}|\leq \frac{1}{n}$, $f_n(x)=0$ for $x<\frac{1}{2}-\frac{1}{n}$ and $f_n(x)=1$ for $x>\frac{1}{2}+\frac{1}{n}$. That's basically a continuous approximation of the characteristic function of the interval $[\frac{1}{2},1]$, which is not continuous.

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That's not what's asked. It converges in the $L^1$-norm (= integral norm) to zero. – t.b. Feb 13 '11 at 20:01
You're right, sorry. I'll supply a valid example... – lvb Feb 13 '11 at 20:05
Great. Now I supplied your example... – lvb Feb 13 '11 at 20:21
+1 anyway. @Jonas: Well, three people coming up with essentially the same example doesn't make us seem particularly inventive... – t.b. Feb 13 '11 at 20:24
To obtain a somewhat more fancy example one can also take the sequence of partial sums of a Fourier series of a discontinuous but square-integrable function $f$, provided $f$ does not coincide with some continuous function almost everywhere. – lvb Feb 13 '11 at 20:33

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