Show that the space $C([a,b])$ equipped with the $L^1$-norm $||\cdot||_1$ defined by $$ ||f||_1 = \int_a^b|f(x)|dx ,$$ is incomplete.

I was given a counter example to disprove the statement:

Let $f_n$ be the sequence of functions:

$$f_n(x) = \begin{cases} 0 & x\left[a,\frac{b-a}{2}\right)\\ nx-n\frac{(b-a)}{2} & x\in\left[\frac{b-a}{2},\frac{b-a}{2}+\frac{1}{n}\right)\\ 1 & x\in \left[\frac{b-a}{2}+\frac{1}{n},b\right] \end{cases}.$$

This is a cauchy sequence that converges to a discontinuous function.

My question is:

How do I see that such a sequence of functions is cauchy? My thought was that the $||\cdot||_1$ will determine the differences in area under the curve for each function, so that $||f_n-f_m||\leq \frac{(b-a)}{2}$. Is this correct?

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    $\begingroup$ If $n < m$, then the two functions differ only on the interval $\left[\frac{a+b}{2}, \frac{a+b}{2}+\frac1n\right]$ (the $\frac{b-a}{2}$ is a typo, that number need not lie in $[a,b]$). Both function's values are between $0$ and $1$ inclusive, so $\lVert f_n f_m\rVert_1 \leqslant \frac1n$. $\endgroup$ – Daniel Fischer Oct 1 '13 at 23:12
  • $\begingroup$ I think I'd attack it in the following way (which is basically identical to @DanielFischer' suggestion although somewhat lazier): the sequence clearly converges in $L^1[0,1]$, hence it's Cauchy. $\endgroup$ – Jonathan Y. Oct 1 '13 at 23:44

No, that is not correct. You need to be able to make $\lVert f_n - f_m \rVert$ arbitrarily small for sufficiently large $n,m$. $(b-a)/2$ is a fixed number. However, you do have the right idea: try find a bound for $\lVert f_n - f_m \rVert$ for $m \geq n$ by bounding the measure of set on which the difference is nonzero, multiplied by the maximum difference between the functions on that set.


I don't think you need to be so precise in finding a counterexample, you just have to show the existence of one. First, recall that the step functions are dense in $\mathcal{L}^1([a,b])$. Then, if you slope the sides of the characteristic funtions somewhat (this can be made precise, see any book on real analysis), you get a set of continuous functions dense in $\mathcal{L}^1([a,b])$. But there are most definitely non continuous functions in this set.

You could get to the same thing using Stone-Weirstrass also, but that is even less concrete.

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    $\begingroup$ I'm not sure the OP is familiar with the facts mentioned here (seeing that continuous functions may converge pointwise to a discontinuous one is somewhat more basic, even if most do see Darboux sums in their first analysis course). $\endgroup$ – Jonathan Y. Oct 1 '13 at 23:41

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