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Let $X=C^b(\mathbb{R})$ be the space of continuous, bounded functions in $\mathbb{R}$. $X$, equipped with the norm $\|f\|_\infty=\sup_{x\in\mathbb{R}}|f(x)|$ is a Banach space. What can you say about these subspaces, are they Banach spaces, too?

(i) $\mathcal{K}(\mathbb{R})=\{f\in X : \exists N\;\forall |x|\geq N \; f(x)=0 \}$

(ii) $C_0(\mathbb{R})=\{f\in X : \lim_{|x|\to\infty}f(x)=0 \}$

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  • $\begingroup$ The second one is clearly a Banach space, since it is closed in $X$. But what about the first one? $\endgroup$ – Marc Jan 16 '14 at 21:35
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    $\begingroup$ What can you say about the sequence with terms $f_n(x)={1\over 1+x^2}\cdot\chi_{[-n,n]}$? $\endgroup$ – David Mitra Jan 16 '14 at 21:40
  • $\begingroup$ @copper.hat Ah, of course. Thanks. $\endgroup$ – David Mitra Jan 16 '14 at 22:16
  • $\begingroup$ @DavidMitra the sequence you are talking about is not of continuous functions, so it wouldn't work... $\endgroup$ – Marc Feb 11 '14 at 16:58
  • $\begingroup$ What about this: Let $A_n=\{f\in\mathcal{K}(\mathbb{R}):f(x)=0\,\forall |x|\geq n\}$. Then $A_n\subset\mathcal{K}(\mathbb{R})$ and $\mathcal{K}(\mathbb{R})=\bigcup_{n=0}^\infty A_n$. Could we use Baire's Category Theorem? $\endgroup$ – Marc Feb 11 '14 at 17:00
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Hint: closure of $\mathcal{K}(\mathbb{R})$ is $C_0(\mathbb{R})$. $$ \phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{} $$

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