Gaussian Processes (GP) are widely viewed as practical ways to implement Gaussian Measures (GM) numerically. In fact, in many contexts it seems that to each GP corresponds a GM and vice versa, see, e.g., this. The focus there is mainly on GP/M on infinite dimensional spaces. In particular, they consider the separable Frechet spaces $C(I),\,C^k(I),\,AC(I)$, where $I$ is a real interval. Separability is an important assumption that ensures the fact that

The Borel sigma algebra on a Frechet space $X$ coincides with the smallest sigma algebra which makes continous linear functionals on $X$ measurable (see top p.3 here)

Replacing the real interval with a bounded open set in $\mathbb R^n$ changes nothing in their approach. My question is whether the result holds also for $C^\infty$, let's say on the unit cube -- open or closed. I could not adjust the proof for $C^k$ (obviously), nor could I find a reference, despite scouring the internet.

Does anyone know the answer for the question in the title? Thank you!

  • $\begingroup$ Perhaps provide a reference for $C^\infty([0,1]^n)$, since $[0,1]^n$ isn't an open set in $\mathbb R^n$. $\endgroup$
    – GEdgar
    Commented May 9, 2023 at 17:20
  • $\begingroup$ Consider the the same question either with $(0,1)^n$ or take the family of seminorms $p_\alpha(f)=\sup_{x\in[0,1]^n}|\partial^\alpha f(x)|$, where $\alpha$ is a multi index $\endgroup$
    – Sobolev
    Commented May 9, 2023 at 17:55
  • $\begingroup$ My first thought would've been that $C^\infty[0,1]^n$ is separable by a $C^k$ version of Weierstrass' polynomial approximation. Is it not? $\endgroup$
    – Jose27
    Commented May 9, 2023 at 18:27
  • 1
    $\begingroup$ For example, what does $\partial^\alpha f(x)$ mean on a boundary point $x$ of $[0,1]^n$? When we talk of $C^\infty\big((0,1)^n\big)$, we allow unbounded functions, and we have seminorms using compact subsets $K \subset (0,1)^n$. [Shouldn't someone named Sobolev know that?] $\endgroup$
    – GEdgar
    Commented May 9, 2023 at 18:55
  • $\begingroup$ @GEdgar For $Q\subseteq\mathbb R^n$ and $x\in Q$, you can define directional derivatives $D_rf(x)=\lim\limits_{t\to 0} (f(x+tr)-f(x))/t$ whenever $0$ is in the closure of $\{r\in\mathbb R\setminus\{0\}: x+tr\in Q\}$. For a cube, all directional derivatives make sense. $\endgroup$
    – Jochen
    Commented May 10, 2023 at 17:54

2 Answers 2


Yes, $C^\infty([0,1]^n)$ is a very nice nuclear Fréchet space which, in particular, is separable.

The definition of differentiability is no problem since, at all points $x$ of the cube $Q=[0,1]^n$ and for each direction $r\in S^{n-1}$, either $x+tr\in Q$ for all small $t>0$ or $x-tr\in Q$ for all small $t>0$. (As pointed out by GEdgar, this is nonsense. But the partial derivatives exist. In general, it is important that enough directional derivatives exist to define the Fréchet derivative uniquely.)

The topology of $C^\infty(Q)$ is the topology of uniform convergence on $Q$ of all partial derivatives. That this topology is completely metrizable is a standard argument based on the fundamental theorem of calculus (for $f_n\to f$ such that $\partial_j f_n\to g_j$ uniformly, the limit $f$ is partially differentiable with $\partial_f=g$). Also nuclearity is, essentially, a consequence of the FTC, alternatively, one can use Arzelá-Ascoli to prove that $C^\infty(Q)$ is Fréchet-Schwartz which is enough for separability (but slightly weaker than nuclearity). You can also show directly that (restrictions to $Q$ of) polynomials are dense in $Q$, e.g., using Weierstrass's theorem as suggested in the comment of Jose27. As a reference I would recommend the book Introduction to Functional Analysis of Meise and Vogt.

A simple instance of Whitney's extension theorem implies that the restriction operator $C^\infty(\mathbb R^n)\to C^\infty(Q)$, $f\mapsto f|_Q$ is surjective and even has a continuous linear right inverse.

For general compact sets $Q$ there is a notion of smooth functions on $Q$ in terms of Whitney jets. The definition is somwhat elaborated but it turns out that the restriction operators from $C^\infty(\mathbb R^n)$ to $C^\infty(Q)$ (which is usually denoted as $\mathscr E(Q)$) is surjective but the existence of a continuous linear right inverse (an extension operator) is a notorously difficult question about the geoemtry of $Q$.

  • $\begingroup$ So, if $Q=[0,1]^2$ and $x$ is the origin $(0,0)$ and $r$ points into the second quadrant, then $x+tr$ and $x-tr$ are both outside $Q$ for all $t>0$. $\endgroup$
    – GEdgar
    Commented May 10, 2023 at 18:00
  • $\begingroup$ Ooops. Nevertheless, the partial derivatives both exist. $\endgroup$
    – Jochen
    Commented May 10, 2023 at 19:19

Thanks everyone for the nice answers and insights!!

Since the original question spawned a second question, I am writing an answer trying to address this secondary question:

I am sorry that @GEdgar struggled so much with the definition of $C^\infty([0,1]^n)$. For me $C^m(\bar\Omega)$ for open and bounded $\Omega\subset\mathbb R^n$ is always given by restrictions of $C^m(\mathbb R^n)$ functions; here we can include $m=\infty$. Its Banach ($m<\infty$) or Frechet ($m=\infty$) space structure is given by the semi norms $p_\alpha(f)=\sup_{x\in \Omega}|\partial^\alpha F(x)| $ for $|\alpha|<m+1$, where $F\big |_\Omega=f$. There is no problem of well definedness because all extensions $F$ and their derivatives coincide in $\Omega$. It can thus be identified with a linear subspace of $C^m(\Omega)$. The completeness of $C^m(\bar\Omega)$ follows from the Whitney extension theorem.

It seems to me that one can also define $$ C^\infty(\bar\Omega)=\{f\in C^\infty(\Omega)\colon \partial^\alpha f\text{ is uniformly continuous}\} $$ and then the semi norms give the right topology.

Defining $C^\infty(K)$ for an arbitrary compact set $K$ requires more work because the well definedness of the seminorms of $f$ in terms of the extensions $F$ is no longer so clear. Then one needs to use an intrinsic definition as in the last paragraph of @Jochen's comprehensive answer.


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