# Topological isomorphism between $C^{\infty}(\mathbb{R}) = \lim_{\leftarrow}{C^{k}([-k, k])}$

Let $X = C^{\infty}(\mathbb{R})$ be the space of smooth functions on a real line considered as a locally convex topological vector space, endowed with the family of seminorms

$$p_{m}(\varphi) = \max \{ |\varphi^{k}(t)| : k = 0, 1, \ldots, m; |t| \leq m \}$$ for $m \in \mathbb{N}$.

In this setting, the space above is a complete metrizable locally convex space.

Let $X_{n} = C^{n}([-n, n])$ be the space of all $n$ times continously differentiable functions on $[-n, n]$ endowed with norm $$q_{n}(\varphi) = \max \{ \max_{t} {|\varphi^{(i)}(t)| : i = 0, 1, \ldots, n } \}$$ for $n \in \mathbb{N}$

How to prove that $$C^{\infty}(\mathbb{R}) = \lim_{\leftarrow} {C^{n}([-n, n])}$$

?

(of course, if the topological isomorphism does exist (does it?))

• I don't think that you really have a projection here, because of the support condition. – Arnaud Mortier Jan 28 '18 at 16:52
• @ArnaudMortier Why? If we take a smooth function, then we can always consider its restriction to $[-n, n]$ via $f \mapsto f|_{[-n, n]}$? – hyperkahler Jan 28 '18 at 16:54
• But then it's not continuously differentiable on $\mathbb{R}$ supported on $[-n, n]$. It is rather continuously differentiable on $[-n, n]$. – Arnaud Mortier Jan 28 '18 at 16:56
• @ArnaudMortier Yes, you're right. Fixed. – hyperkahler Jan 28 '18 at 17:00

In some cases this is overkill, but here I think it is clarifying to be very precise: A locally convex projective (or inverse) spectrum does not only consist of locally convex spaces $X_n$ (in your case even Banach spaces $X_n=C^n([-n,n])$) but of a sequence of spaces $X_n$ together with linear continuous linking maps $\varrho_m^n: X_m\to X_n$ for all $n\le m$ such that $\varrho_n^n=id_{X_n}$ and $\varrho_m^n\circ \varrho_k^n=\varrho_k^n$ for $n\le m\le k$, in your case the restriction maps $f\mapsto f|_{[-n,n]}$. The projective limit is then $$X_\infty=\{(x_n)_{n\in\mathbb N}\in \prod_{n\in\mathbb N}X_n: \varrho_m^n(x_m)=x_n \text{ for all } n\le m\}$$ endowed with the relative topology of the product which is the same as the initial topology with respect to the maps $\varrho_\infty^n:X_\infty\to X_n$, $(x_k)_{k\in\mathbb N} \mapsto x_n$. A fundamental system of semi-norms is thus given by $p_\infty((x_k)_{k\in\mathbb N})=p(x_n)$ where $n\in\mathbb N$ and $p$ ranges over all continuous semi-norms of $X_n$ (in your case one only needs the single norms of $X_n$).
Now, the isomorphism you are looking for is $C^\infty(\mathbb R)\to X_\infty$, $f\mapsto (f|_{[-n,n]})_{n\in\mathbb N}$.