Let $C[0,\infty]$ be the space of all continuous functions on $[0, \infty ]$ with metric $$ \phi(\omega_1, \omega_2) = \sum^{\infty}_{n=1} \frac{1}{2^n}\max_{0{\leq} t {\leq} n}(|\omega_1(t)-\omega_2(t)|\wedge 1)$$ where $n \in \mathbb{N}$. Show that the space $C[0,\infty]$ under $\phi$ is complete and separable.

I proceeded as follows: suppose I have a Cauchy sequence $\{\omega_n\}$ in $C[0,\infty]$. Its limit exists in $\phi$ as $\phi$ is bounded. Let the limit be $\omega$, now I have to show that $\omega $ belongs to $C[0,\infty]$.

  • $\begingroup$ I think the definition of $\phi$ should be something of the form $$ \phi(\omega_1, \omega_2) = \Sigma^{\infty}_{n=1} (1/2^n)\cdot\max_{0\leq t \leq n}(|\omega_1(t)-\omega_2(t)|\wedge 1)$$ $\endgroup$ – Orest Bucicovschi Sep 11 '14 at 9:20
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    $\begingroup$ Yes, the $\phi$ as it exists is pretty strange. The part after the $*$ does not mention $n$. Do you really want functions even continuous at $\infty$? If so, there is no need to try $ \wedge 1$ in there. More conventional would be the space $C[0,\infty)$ where we do not require functions continuous (or even defined) at $\infty$ itself. $\endgroup$ – GEdgar Sep 11 '14 at 14:31

Notice that for each $n$, $\sup_{0\leqslant x\leqslant N}|\omega_n(x)-\omega(x)|\to 0$ as $n$ goes to infinity and a uniform limit of continuous functions is continuous. This proves that $\omega$ is continuous on $[0,N]$ for each $N$, hence $\omega$ is continuous on $[0,\infty)$.

To conclude the proof of completeness, it remains to check that $\phi(\omega_n,\omega)\to 0$.

For separability, notice that $C[0,N]$ is separable for each $N$.


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