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In my course material I have the following notation:

$$f\in C^{\infty}_0(\Omega, \mathbb{R}),$$

where $\Omega \subset\mathbb{R}^n$ is a bounded open set. I was wondering what does this notation mean? What is the set $f$ belongs into?

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2 Answers 2

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I'm not sure if this is 100% standard, but I've always understood this notation to mean: $f$ is a function from $\Omega$ to $\mathbb{R}$ which has compact support (the subscript $0$), and is smooth/infinitely differentiable (the superscript $\infty$). I just checked, for example, that Stein and Shakarchi use this notation in their Functional Analysis.

I have also seen $C_c^{\infty}(\Omega, \mathbb{R})$ to mean the same thing. For example, Rudin uses this notation in Real and Complex Analysis.

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  • $\begingroup$ Thank you =) Appreciate it $\endgroup$
    – jjepsuomi
    Commented Jan 19, 2015 at 20:44
  • $\begingroup$ I have seen the notation to mean exactly this (in several cases), so I'm just acknowledging your answer. $\endgroup$
    – Eff
    Commented Jan 19, 2015 at 20:47
  • $\begingroup$ It is a hundred percent standard notation, e.g. in Hörmander linear PDE $\endgroup$ Commented Jan 19, 2015 at 21:20
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The zero subscript usually means vanishing at infinity, in a very general context. Vanishing at infinity means that for every $\varepsilon$, there is a compact set $K$ such that the function is smaller than $\varepsilon$ outside $K$.

In other words, $C_0(X)$ is the closure of $C_c(X)$ (compactly supported continuous functions) under uniform convergence topology, for a general (usually locally compact Hausdorff) space $X$, and for a smooth manifold $M$ (like an open subset of an Euclidean space), the family $C^\infty_0(M)$ is defined in the analogous manner, as the uniform closure of $C_c^\infty(M)$.

The ${\bf R}$ in $C_0^\infty(M,{\bf R})$ merely reasserts the fact that the functions are into reals (which is implicit in the notation of the previous paragraph, and under the all conventions that I'm aware of).

So in this case, this is most likely the set of smooth functions vanishing on the boundary of $\Omega$ (this is because $\Omega$ is open in ${\bf R}^n$, which means that compact subsets are bounded away from the boundary).

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