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Consider the following set F of functions from $[0,1]$ to $\mathbb{R}$:

$$ F = \left\{f: [0,1] \rightarrow \mathbb{R} \mid f \ \text{is continuous}, \ \ \exists \delta \in (0,1] , \forall x \in [0, \delta] : f(x) = 0 \right\}$$ We consider the supremum metric $d_{\infty}$ on $F$, that is $d_{\infty} (f,g) = \sup_{x \in [0,1]} |f(x) - g(x)|$. Is $(F, d_{\infty})$ complete?

My attempt: I think it is complete. I let $(f_n)$ be a Cauchy sequence in $F$. I need to show that it converges to a limit which is also in $F$. Since for all $n, m$ and all $x$ we have $$|f_n(x) - f_m(x)| \leq d_{\infty} (f_n, f_m)$$ we know that $f_n$ is a Cauchy sequence in $\mathbb{R}$. Since $\mathbb{R}$ is complete, $(f_n)$ converges to some $f$. We now need to show that $d_{\infty} (f_n, f) \to 0$ and that $f \in F$.

For the first part, I was trying to show that $f_n(x)$ converges uniformly to $f(x)$ in $\mathbb{R}$. From this it would follow that $d_{\infty} (f_n, f) \to 0$. But I'm not sure how to do this. I know that $f_n$ is uniformly continuous (since it is continuous on a compact set). How do I use this, and the fact that $f_n \in F$ to get $|f_n(x) - f(x)|$ smaller than some $\epsilon$? I cannot use $$|f_n(x) - f(x)| \leq d_{\infty} (f_n, f)$$ because I don't know yet that $f \in F$.

Help with this problem is appreciated.

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Hint: Your $F$ is a subspace of $(C([0,1]), d_\infty)$ and it is well known that this space is complete. Hence, you have to figure out whether $F$ is closed in this space.

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  • $\begingroup$ Ah, thank you. I found it I think. I took a sequence $f_n$ which converges to a $f$ by the $d_{\infty}$ metric. To show $F$ is closed I need $f \in F$. But since $f_n \to f$ on the $d_{\infty}$ metric, we have $f_n \to f$ uniformly in $\mathbb{R}$. Then since each $f_n$ is continuous, then the limit function is also continuous. Proving that $f$ satisfies the other property was now also easy. $\endgroup$ – Kamil Jun 22 '17 at 12:18
  • $\begingroup$ @Kamil: No, $f \in F$ does not follow! You only can check $f(0) = 0$ and, in fact, every continuous $f$ with $f(0) = 0$ is the limit of a sequence in $F$. $\endgroup$ – gerw Jun 22 '17 at 14:29
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On a "high level" description your space is the same as $C_c(\,(0,1]\,)$, that is compactly supported continuous functions on $(0,1]$ which is not complete.

For example the sequence $f_n(x)=\begin{cases}x-\frac1n & x≥\frac1n\\ 0& x<\frac1n\end{cases}$ lies in your space. It converges uniformly to the function $f(x)=x$, but this is not in your space.

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