Proving the L¹ norm is a norm and completeness of C⁰([-1,1])

[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

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

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:

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

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.