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For a topological space $X$, the space of smooth functions with compact support (denoted by $C^{\infty}_0(X)$) is dense in the space of continuous functions vanishing at infinity (denoted $C_{\infty}(X)$). Now if $X$ is compact (as is $\mathbb T$) then the spaces $C_0(X)$ and $C_{\infty}(X)$ and $C(X)$ coincide. Then does it hold that $C^{\infty}(\mathbb T)$ is dense in $C(\mathbb T)$? Here, $\mathbb T$ is the 1-dimensional Torus.

I'm trying to show that $C^{\infty}(\mathbb T)$ is an operator core for an operator whose domain is a dense subset of $C(\mathbb T)$.

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  • $\begingroup$ What is $\mathbb{T}$? And what is smooth function on general topological space? Or is $X$ manifold? $\endgroup$ – tom Sep 13 '15 at 17:16
  • $\begingroup$ Welcome to our site! $\endgroup$ – kjetil b halvorsen Sep 13 '15 at 17:17
  • $\begingroup$ Sorry I should have said - $\mathbb T$ is the 1-dimensional Torus. And by smooth function I mean functions that are infinitely differentiable. $\endgroup$ – Ifan Dafydd Sep 13 '15 at 17:19
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Yes. Indeed, $C^\infty(\mathbb T)=C^\infty_0(\mathbb T)$ since $\mathbb T$ is compact, so trivially any smooth function on $\mathbb T$ vanishes outside of a compact subset. Hence $C^\infty(\mathbb T)$ is dense in $C_\infty(\mathbb T)$, which as you pointed out is equal to $C(\mathbb T)$.

As an aside, I prefer to use the notation $C^\infty_c$ for smooth with compact support and $C_0$ for vanishing at infinity to avoid using both $C^\infty$ and $C_\infty$ to mean very different things. Also as @tom pointed out, you do need further assumptions on $X$ in order for differentiation to be defined, but this is not a problem for $\mathbb T$.

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