Proving Normality of ||.||_1 & Cauchy-Sequence & Completeness in C^0([-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

<br /> \begin{align*}<br /> f_n(x) &amp;= -1, \quad &amp; -1 \leq{x} \leq{\frac{-1}{n}} \\<br /> \ &amp;= nx, \quad &amp; \frac{-1}{n} \leq{x} \leq{\frac{1}{n}} \\<br /> \ &amp;= 1, \quad &amp; \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 &gt;0 there exists a N>0 such that: ||f_m - f_n||_1 &lt; \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)| &amp;= 0 &amp; x &gt; 1/n \\<br /> \ &amp;= 1-nx &amp; 1/m &lt; x &lt; 1/n \\<br /> \ &amp;= (m-n)x &amp; 0 \leq x &lt; 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 &lt; 1/n

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

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