Determining if a sequence of functions is a Cauchy sequence?

Question

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?

Answer 1

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.

Answer 2

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.