complete matric space

**Kat-M** · August 12th 2009, 02:45 PM

Let $X$ be the set of all polynomial functions of degree 4 or less on $[0,1]$ . If $f=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ and $g=b_0+b_1x+b_2x^2+b_3x^3+b_4x^4$ , we define $d_2(f(x),g(x))=\sum_{n=0}^4 \mid a_n-b_n \mid$ .
Further, we define $\rho_2(f(x),g(x))=\frac{d_2(f(x),g(x))}{1+d_2(f(x) ,g(x))}$ .

Will $(X,\rho_2)$ be a complete metric space?

**Enrique2** · August 13th 2009, 12:35 AM

Since $X$ is finite dimensional and $d_2$ is a norm it follows that $(X,d_2)$ is a Banach space.

If $(x_n)_n$ is a Cauchy sequence in $(X,\rho_2)$ , then for $\varepsilon>0$ there exist $n_0$ such that $\rho_2(x_n,x_m)<\varepsilon$ for $n,m>n_0$ . A direct calculation shows that this implies

$d_2(x_n,x_m)<\frac{\varepsilon}{1-\varepsilon}.$

Taking $\varepsilon$ small enough (the quotient in the last inequality goes to 0 as $\varepsilon$ goes to 0), this shows that $(x_n)_n$ is a Cauchy sequence in $(X,d_2)$ . Hence there
$x\in X$ such that $x_n$ tends to $x$ in $(X,d_2)$ . But this implies that $(x_n)_n$ tends to $x$ in $(X,\rho_2)$ , since
$\rho_2\leq d_2$ .

Thus the answer is yes if you show that $\rho_2$ is actually a metric.

**Kat-M** · August 13th 2009, 10:25 AM

how do you show that a cauhy sequence is convergent in $(X,d_2)$ or $(X, \rho_2)$ ?

**Enrique2** · August 13th 2009, 12:51 PM

Originally Posted by Kat-M

how do you show that a cauhy sequence is convergent in $(X,d_2)$ or $(X, \rho_2)$ ?

Showing that $d_2$ is a norm, on $X$ , since X is finite dimensional, you obtain that $(X,d_2)$ is complete. You can consult any book related to topological vector spaces, in Robertson and Robertson's book "Topological vector spaces" for instance. Perhaps you are not allowed to use this fact?

The completenes of $\rho_2$ follow from the given argument above, Cauchy in $\rho_2$ implies Cauchy in $d_2$ , then convergence in $d_2$ and $\rho_2\leq d_2$ .

I believe that in the same book I have mentioned you can find the clues to show that $\rho_2$ defines a metric.

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