I'm reading "Pseudo-differential Operators and the Nash-Moser Theorem" and at the top of on page 8 they write: "Finally, we note that if $u \in C^0(\Omega)$, then the support of $u$ defined above coincides with the closure of $\{x \in \Omega \mid u(x) \neq 0 \}$." $\Omega$ here is an open subset of $\mathbb{R}^n$.

The support of a distribution $u$ was defined on the previous page as "the complement in $\Omega$ of the points in the neighbourhood of which $u$ is zero".

Question 1: What's the difference between $C(\Omega)$ and $C^0(\Omega)$?

Question 2: From $\operatorname{supp}{u} \subset \Omega$ I deduce that we think of $u$ here as a distribution on a subset of reals. How does this make sense? If my test functions are constants then the integral of $cu$ might not be finite.

Thanks for your help.

  • $\begingroup$ Test functions are usually defined as smooth with compact support, so the integral exists. $\endgroup$
    – WimC
    Commented Feb 23, 2012 at 10:46
  • $\begingroup$ Yes I know. But if the test functions are a subset of $\mathbb{R}$ then that's false. $\endgroup$ Commented Feb 23, 2012 at 11:17
  • $\begingroup$ Can you give an example? (Also, I don't have my copy of that book handy, but doesn't it have a table of notations near the end?) $\endgroup$ Commented Feb 23, 2012 at 11:36
  • $\begingroup$ @WillieWong Yes it does have a table but $C^0$ is not on it. An example of what? $\endgroup$ Commented Feb 23, 2012 at 11:53
  • $\begingroup$ An example of a distribution represented by a function $u$ such that its support is contained in some open set $\Omega$ and such that its integral against a test function is not finite. $\endgroup$ Commented Feb 23, 2012 at 12:13

1 Answer 1


For question 1: I don't know where you see the notation $C(\Omega)$, so I cannot tell you the difference. But on the first page of Chapter 0 of the French original of Alinhac and Gerard, it is stated (I translate)

Given $\Omega$ an open set in $\mathbb{R}^n$. If $k$ is a non-negative integer, we designate by $C^k(\Omega)$ the space of functions $k$-times continuously differentiable on $\Omega$, taking values in $\mathbb{C}$.

So in particular $C^0(\Omega)$ refers to the space of continuous complex valued functions on $\Omega$.


$C^k_0(\Omega)$ denotes the subspace of $C^k(\Omega)$ whose elements vanish outside a compact subset of $\Omega$, while $C^k(\overline{\Omega})$ is formed by the restriction to $\Omega$ of elements of $C^k(\mathbb{R}^n)$.


A distribution on an open set $\Omega$ is a linear functional $u$ over $C^\infty_0(\Omega)$ satisfying the following continuity property: for every compact subset $K\subset \Omega$, there exists an integer $m$ and a constant $C$ such that for all $\psi\in C^\infty_0(\Omega)$ that vanishes on the complement of $K$, $$|\langle u,\psi\rangle| \leq \sup_{x\in K} \sup_{|\alpha|\leq m} |\partial^\alpha \psi(x)| $$

Hence, for question 2:

If you identify a function $u\in C^0(\Omega)$ as a linear functional on $C^\infty_0(\Omega)$ given by $\langle u,\psi\rangle = \int_{\Omega} u\psi \mathrm{d}x$, the space of test functions you are looking at have compact support in $\Omega$. In particular, $\psi \equiv c$ is not an element of $C^\infty_0(\Omega)$ and hence not a test function.

  • $\begingroup$ I may have not entirely understood your question 2. If the above doesn't answer your question, can you give a concrete example of what you think can go wrong? $\endgroup$ Commented Feb 23, 2012 at 12:33
  • $\begingroup$ Thank you! I'll look at it tomorrow. I can't look at it right now. $\endgroup$ Commented Feb 24, 2012 at 0:42

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