0
$\begingroup$

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?))

$\endgroup$
  • $\begingroup$ I don't think that you really have a projection here, because of the support condition. $\endgroup$ – Arnaud Mortier Jan 28 '18 at 16:52
  • $\begingroup$ @ArnaudMortier Why? If we take a smooth function, then we can always consider its restriction to $[-n, n]$ via $f \mapsto f|_{[-n, n]}$? $\endgroup$ – hyperkahler Jan 28 '18 at 16:54
  • $\begingroup$ But then it's not continuously differentiable on $\mathbb{R}$ supported on $[-n, n]$. It is rather continuously differentiable on $[-n, n]$. $\endgroup$ – Arnaud Mortier Jan 28 '18 at 16:56
  • $\begingroup$ @ArnaudMortier Yes, you're right. Fixed. $\endgroup$ – hyperkahler Jan 28 '18 at 17:00
1
$\begingroup$

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 $C^\infty(\mathbb R)\to X_\infty$, $f\mapsto (f|_{[-n,n]})_{n\in\mathbb N}$.

$\endgroup$
1
$\begingroup$

Unless there is a subtlety that I missed, I would say that the isomorphism that you are looking for is tautological. You can have a look there if that helps: Methods of constructing topological vector spaces. Your settings are similar to those of Proposition 2.1.6.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.