Show the space X is incomplete

Question

A problem comes from the "Optimization by vector space methods". Luenberger p.34

Let $X$ be
1. the space of continuous functions on [0,1]
2. its norm is defined by $||x|| = \int^1_0|x(t)| dt$
So, 1. $X$ is a normed linear space.

To show $X$ is incomplete, it define a sequence of elements in $X$ by the equation:

with the graph:

At first, it proves that the sequence is Cauchy since $||x_n - x_m || = \frac {1}2|1/n -1/m|\rightarrow 0$

And it says, there is no continuous function to which the sequence converges.

My questions are
1. How to obtain $\frac {1}2|1/n -1/m|$
2. What does the last sentence mean?
3. How to see the completeness depends critically on the definition of norm?

Answer 1

1. $\int_0^{1/2} x_n dt$ is just the area of a triangle with base $\frac{1}{n}$ and height $1$

2. Short way: assume $x_n\rightarrow g$ in your ($L^1$) norm, and assume $x_n\rightarrow h$ pointwise (you should be able to explicitly write out this $h$), then we must have $h = g $ a.e. (why?). So is $h$ continuous? If not continuous, can we remove the discontinuity by modifying $h$ on a measure zero set?

3. A normed ( or metric) space $X$ is complete if every Cauchy sequence is convergent in $X$. Here the norm (or metric) determines whether $x_n$ is Cauchy or not, so with the $\max$ norm, the $x_n$ in the problem is not a Cauchy sequence.

Edit: Let me say more about (2), to say $x_n \rightarrow g$ in your norm, it means that $$\lim_n \int_0^1 |x_n - g| dt = 0.$$ Now you will see that $x_n$ converges pointwise to $\chi_{[\frac{1}{2}, 1]}$. By Lebesgue dominated convergence theorem, we have $$0 = \lim_n \int_0^1 |x_n - g| dt = \int_0^1 \lim_n |x_n - g| dt = \int_0^1 |\chi_{[\frac{1}{2}, 1]}-g| dt = \int_0^\frac{1}{2} |0-g| dt + \int_\frac{1}{2}^1 |1-g| dt.$$ If $g$ is continuous, then both $|0-g|$ and $|1-g|$ are nonnegative continuous functions, what can you say about them if you know their integrals are zero.

Answer 2

Hints:

(1) You can calculate the norm doing the integral.

(2) What function can be the limit?

(3) With the $\sup$ norm the space is complete.

Answer 3

$\|x_n-x_m\|$ mean:

$$\|x_n-x_m\|=\int_{0}^{1}|x_n(t)-x_m(t)|dt$$

Let $n \leq m$ (case $n \geq m$ is symmetric), then when $t \in [0,\frac{1}{2}-\frac{1}{n}]\cup [\frac{1}{2},1]$ you have $x_m(t)=x_n(t)$, so:

$$\|x_n-x_m\|=\int_{0}^{1}|x_n(t)-x_m(t)|dt=\int_{\frac{1}{2}-\frac{1}{n}}^{\frac{1}{2}}|x_n(t)-x_m(t)|dt$$

You probably can prove that $\|x_n-x_m\|=|1/n-1/m|$, but it's easier when we note:

$$\|x_n-x_m\| \leq 1$$

So:

$$\int_{\frac{1}{2}-\frac{1}{n}}^{\frac{1}{2}}|x_n(t)-x_m(t)|dt \leq \int_{\frac{1}{2}-\frac{1}{n}}^{\frac{1}{2}}1dt=\frac{1}{n}$$

It's proof that $\{x_n\}_{n=1}^{\infty}$ is a Cauchy sequence.

But we see that $\lim_{n \to \infty}x_n=x$, where $x(t)=0$ when $t \in [0,\frac{1}{2}]$ and $x(t)=1$ when $t \in [\frac{1}{2},1]$. But $x$ isb't continous function, so X is incomplete.