Cauchy sequence

[Neoma]

We consider the space C^0 ([-1,1]) of continuous functions from [-1,1] to \mathbb{R} supplied with the following norm:

||f||_1 = \int_{-1}^{1} |f(x)| dx

a. Show that ||.||_1 defines indeed a norm.

b. Show that the sequence of functions (f_n), where


\begin{align*}
f_n(x) &= -1, \quad & -1 \leq{x} \leq{\frac{-1}{n}} \\
\ &= nx, \quad & \frac{-1}{n} \leq{x} \leq{\frac{1}{n}} \\
\ &= 1, \quad & \frac{1}{n} \leq{x} \leq{1}
\end{align*}

is a Cauchy-sequence with respect to the given norm.

c. Show that C^0 ([-1,1]) is not complete with respect to the given norm.

I figured out a. myself, by showing this norm satisfies the properties of a norm, but I can't find out how to tackle b. and c.

[Timbuqtu]

b) I assume you can compute ||f_m - f_n||_1 for arbitrary m > n > 0. Now f_n is called a Cauchy-sequence if for all \epsilon >0 there exists a N>0 such that: ||f_m - f_n||_1 < \epsilon for all m>n>N. You can do the estimates yourself.

c) Does the sequence converge to a continuous function? A set with a given norm is called closed if all Cauchy-sequences converge to an element in the same set.

[Neoma]

I knew the definition of a Cauchy sequence, but I still can't find the solution.

b) I assume you can compute ||f_m - f_n||_1 for arbitrary m > n > 0.I think I can't, given a certain n and f_n I can find f_{n+1} and I see that once n approaches infinity f_n becomes either -1 or 1, but I don't know how to work from there. In fact this is all quite new to me.

[Timbuqtu]

When we assume m > n:

\begin{align*}|f_m(x) -f_n(x)| &= 0 & x > 1/n \\
\ &= 1-nx & 1/m < x < 1/n \\
\ &= (m-n)x & 0 \leq x < 1/m \end{align*}

and |f_m(-x) -f_n(-x)| = |f_m(x) -f_n(x)| . So:

||f_m - f_n||_1 = \int_{-1}^{1} |f_m(x)-f_n(x)| dx = 2 \int_{0}^{1} |f_m(x)-f_n(x)| dx = 2 ( \int_{0}^{1/m} (m-n)x dx + \int_{1/m}^{1/n} (1-n x) dx ) =
= 2 ( \frac{m-n}{2 m^2} + 1/n - 1/m - \frac{n}{2 n^2} + \frac{n}{2 m^2}) = 1/n-1/m < 1/n

Let \epsilon > 0 . Take N > 1/\epsilon , then for all n,m > N, we have ||f_m - f_n||_1 < 1/N < \epsilon .

Now we have proven that f_n is indeed a Cauchy-sequence.