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Let space of schawartz $S(\mathbb{R})$, with the metric $d(f,g)$, it is a complete space metric $$d\left( f,g \right) =\sum _{ a,b\in R }{ \frac { 1 }{ { 2 }^{ a+b } } } \frac { { \left\| f-g \right\| }_{ a,b } }{ 1+{ \left\| f-g \right\| }_{ a,b } } $$ I need proof, the metric doesn't come the a norm.

I know that the following property fails; $\left\| f \right\| =0\quad \Longleftrightarrow \quad f=0$, but I can't proof.

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What is $\|\cdot\|_{a,b}$? What $R$? –  martini Oct 23 '13 at 6:35
    
$ \left\| f-g \right\| _{ a,b } =sup_{x\in \mathbb{R}}(x^{a}f^{(b)})$ for all $a,b \in \mathbb{N}x\mathbb{N}$ –  Wmmoreno Oct 23 '13 at 11:15

1 Answer 1

I guess you want to prove that there is no norm on $S(\mathbb R)$ inducing the same topology as the metric $d$. A quick solution is that $S(\mathbb R)$ is an infinite dimensional Montel space, that is, bounded sets are relatively compact. If it were normed the unit ball of the norm would be relatively compact which only holds for finite dimensional spaces.

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