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I'm having trouble with an exercise that says

Let $A\subset[a,b]$ closed and consider the set $A$ from $\mathcal{C}(I)$, given by $$\{f\in\mathcal{C}(I)\mid f(t)=0\text{ for every }t\in A\}$$ where $\mathcal{C}(I)$ denotes the space of continuous functions. Prove that $A$ closed in $\mathcal{C}(I)$ with the supreme metric, but not necessary with the integral metric.

I have proved that is closed under the supreme metric, but i don't know how to prove that is not closed under the integral metric, any help would be appreciated, thanks.

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    $\begingroup$ You are using $A$ for two different things: a subset of $[a,b]$ and a subset of $\mathcal{C}(I)$. Which is it? $\endgroup$
    – Matthew Leingang
    Sep 18, 2016 at 21:34
  • $\begingroup$ Actually, that's how we get the exercise, i was actually really confused about that too, and didn't know if i was missing something or it was a error from our teacher, i will need to ask him. $\endgroup$
    – Zigisfredo
    Sep 18, 2016 at 21:44
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    $\begingroup$ Perhaps the second $A$ is an $\mathcal{A}$. $\endgroup$
    – Matthew Leingang
    Sep 18, 2016 at 21:45

1 Answer 1

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If $A=\{a\}$ is a single point, then the associated set in $C(I)$ is dense.

You can prove this by a usual approximation argument: Given any $f\in C(I)$ and $\epsilon>0$, choose $\delta$ for which $\int_a^{a+\delta}|f(t)|dt<\epsilon$ (use continuity of $f$ at $a$, if necessary, if this is not clear).

Define $g(t)=\begin{cases}f(t)&\text{ if }t\geq a+\delta\\\frac{t-a}{\delta}f(t)&\text{ if }a\leq t\leq a+\delta\end{cases}$. Then $g$ is continuous, $g(a)=0$ and $$\Vert g-f\Vert_1\leq\int_a^{a+\delta}(1+\frac{t-a}{\delta})|f(t)|dt\leq 2\epsilon.$$

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