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Proving that $C_{c}(R)$ is not dense in $L^{\infty}(R)$.

My attempt: Let $f=\chi_{(0,1)}$ be in $L^{\infty}(R)$. Then $||f||_{\infty}=1$. Let there be a function $g \in C_{c}(R)$, such that $||f-g||_{\infty}<1/2$.

$$ \{x \in R : sup(|f(x)-g(x))| \} < \epsilon$$

Thus $ |g(1)| < 1/2$, $ |g(0)| < 1/2$, and for $ 0<x<1, |g(x)-1| < 1/2$.

$$-1/2 < g(1) <1/2$$ $$ -1/2 < g(0) <1/2$$ $$ -1/2 < g(x)-1<1/2= 1/2 < g(x) < 3/2$$

Let $x_{n}=1/n, x_{n} \to 0$. By Continuity $g(x_{n}) \to g(0)$. But $1/2< g(x_{n})<3/2$, so as $n \to \infty$, $$1/2 \leq g(0) \leq 3/2$$ But this contradicts that $g(0)<1/2$. Hence proved.

Please let me know if its correct.

hashirama
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    The norm on $L^\infty$ is not the $\sup$ norm (which might not be defined for such functions) but the $\operatorname{essup}$ norm (which introduces an "almost everywhere" aspect). For continuous functions they're equal but for functions like $\chi_{(0,1)}$ you need to be more careful. In particular $|g(0)|$ and $|g(1)|$ do not directly have to be lesser than $1/2$ here, though with the continuity of $g$ you can probably say something. However you are on the right track: instead of $1/2$, pick $1/3$ and look at what happens when you approach $0$ from the left and what happens from the right. – Bruno B Apr 22 '23 at 14:15
  • Remember that for example $\chi_{(0,1)}$, $\chi_{[0,1)}$, $\chi_{(0,1]}$ and $\chi_{[0,1]}$ are the "same function" in regards to $L^p$ and $L^\infty$ spaces, since it's technically spaces of equivalence classes of functions. – Bruno B Apr 22 '23 at 14:21
  • @OliverDíaz Separability rules out countable dense subset, but I am confused whether uncountable dense subset can be there. I am not sure whether $C_{c}(R)$ is uncountable.. – hashirama Apr 22 '23 at 15:08
  • @BrunoB I am not able to figure out how the almost everywhere part comes into the picture. I get that $m{x| \chi_{[0,1]}-g(x) > 1/3}=0$. But if I were to take $x_{n} \to 0$, I get that $\chi_{[0,1]}-g(x)=-g(0)$ as the Left Hand Limit and $1-g(0)$ as the Right Hand Limit. So clearly, the sum of these two functions isn't continuous. I am stuck here. – hashirama Apr 22 '23 at 15:25
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    I think the simplest answers is : if a sequence of continuous functions is convergent uniformly ( i.e. in $|\cdot|_\infty $ norm) the limit function is continuous. – Ryszard Szwarc Apr 22 '23 at 15:25
  • If $C_c$ were dense, one would be able to find a countable dense set in $L_\infty$ since $C_\inty$ is separable in the uniform topology. – Mittens Apr 22 '23 at 15:30
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    If $C_c(\mathbb{R})$ denotes the functions with compact supports, the answer is even simpler: the function $f\equiv 1$ cannot be approximated by functions in $C_c(\mathbb{R}).$ – Ryszard Szwarc Apr 23 '23 at 05:19
  • @RyszardSzwarc thanks! I have a much clearer understanding now. – hashirama Apr 23 '23 at 13:47
  • @OliverDíaz I see, thanks for the help! – hashirama Apr 23 '23 at 14:26

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