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I'm trying to construct a norm on the space $\mathcal{D}(\Omega) := \{ f \in C^\infty(\Omega) | f $ is compactly supported on $ \Omega \}$ where $\Omega$ is an open subset of $\mathbb{R}$. I want this norm to include, somehow, the $L^\infty$-norms of all the derivatives of the smooth function to which it is applied. Specifically, I want to be able to encapsulate the statement

"$f_n$$^{(m)} \rightarrow f^{(m)} ($as $n \rightarrow \infty)$ uniformly for all non-negative values of $m$."

as the statement

"$\|f_n - f\| \rightarrow 0$ $(n \rightarrow \infty)$.",

where $\|\cdot\|$ denotes my desired norm.

So far I've considered trying to write it as something along the lines of

\begin{align}\|f\| := \sum\limits_{m \in \mathbb{N}} \frac{1}{m!} \|f^{(m)}\|_\infty, \end{align} where $\|\cdot\|_\infty$ denotes the usual $L^\infty$-norm. I am very unsure of the validity of this sort of "definition", as I can't see how to prove that this (or a related/similar) series converges. Any help on this would be greatly appreciated! Thanks in advance.

Stromael
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    Typically, $C^\infty(\Omega)$ contains unbounded functions. So you can only use $L^\infty$ norms on a (rather small) subspace of that. – Daniel Fischer Jul 22 '13 at 12:14
  • You cannot go very far without requesting some boundedness assumption on $f$ and its derivatives. – Giuseppe Negro Jul 22 '13 at 12:14
  • Alternatively, you can define a "local" version of your topology, by weakening the notion of convergence so that it is uniform only on compact subsets of $\Omega$. – Giuseppe Negro Jul 22 '13 at 12:15
  • However, $C^\infty(\Omega)$ is a fine Fréchet space with the topology of compact convergence in all derivatives. – Daniel Fischer Jul 22 '13 at 12:16
  • Side note: The comments by me and those by Daniel Fischer are saying exactly the same thing. – Giuseppe Negro Jul 22 '13 at 12:17
  • I have edited the question slightly. It now matches the structure as given in Numerical Models for Differential Problems, A. Quarteroni, although it is a specific case of the definition he gave (which was for multi-index partial differentiation in $\mathbb{R}^n$). Does the fact the functions have compact support aid in the construction of such a norm? Forgive me if this should be obvious, but I'm still quite new to the field. – Stromael Jul 22 '13 at 14:18

2 Answers2

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I'm not an expert in functional analysis, but I think this answers your question.

The topology of $\mathcal{D}(\Omega)$ is not induced by a norm. There is a discussion of this in chapter 1 of Rudin's Functional Analysis book. Briefly, a topological vector space whose topology is induced by a norm is locally bounded (proof: the open unit ball is bounded). So we want to show that $\mathcal{D}(\Omega)$ is not locally bounded. There is a result stating that a locally bounded topological vector space satisfying the Heine-Borel property (every closed and bounded set is compact) is finite-dimensional. So it only remains to show the Heine-Borel property for either $\mathcal{D}(\Omega)$ or $C^\infty(\Omega)$.

If you like, I can provide more details from Rudin's book.

youler
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  • I have not assumed any topology on $\mathcal{D}(\Omega)$. It is simply the vector space of all continuous, real-valued functions on $\Omega$ which have compact support. Also, I'm not sure in what way your ideas help in constructing a norm with the desired property that: a function sequence has that every sequence, of derivatives of its terms of a fixed order ($m$), is uniformly convergent in $\mathcal{D}(\Omega)$ if and only if it is convergent under the topology of this norm. – Stromael Jul 22 '13 at 17:34
  • In a metric space (which a normed vector space certainly is), specifying the limits of sequences determines a topology, so you are implicitly assuming a topology on $\mathcal{D}(\Omega).$ Moreover, even if you were only assuming a vector space structure, then the problem is trivial since every infinite dimensional vector space admits some norm. To be clear, what I'm saying is that a norm of the type you're looking for doesn't exist. – youler Jul 22 '13 at 17:52
  • What if the space is restricted somewhat, e.g., consider only sequences of functions from the space $L^\infty (\Omega) \cap \mathcal{D} (\Omega)$? – Stromael Jul 22 '13 at 19:29
  • Don't take my word for it, but I think the same argument should apply. In particular I think $L^\infty(\Omega)\cap\mathcal{D}(\Omega)$ should still satisfy the Heine-Borel property, since both $L^\infty(\Omega)$ and $\mathcal{D}(\Omega)$ do (I think). – youler Jul 22 '13 at 20:20
  • A minor note: when one writes the symbol $\mathcal{D}(\Omega)$ it usually denotes the space of continuous functions with compact support equipped with a topology whose induced notion of convergence is $$u_n \to u\ \iff\ \forall D\subset\subset \Omega,\ \forall \alpha\in \mathbb{N},\ \sup_{x\in D}\lvert \partial^\alpha u_n(x)-\partial^\alpha u(x)\rvert\to 0.$$ – Giuseppe Negro Jul 22 '13 at 22:22
  • Where $D\subset \subset \Omega$ means that $D$ is a bounded open set whose closure is in $\Omega$. I wrote all this to point out that this topology does not come from any metric (one says that it is not metrizable), let alone from a norm. – Giuseppe Negro Jul 22 '13 at 22:25
  • I proved today that $\mathcal{D}(\Omega) \subset L^\infty(\Omega) \cap L^2 (\Omega)$. Still struggling to make it work. I need to show that for any $f \in \mathcal{D}(\Omega)$, there exists $M>0$ such that $|f^{(n)}|_\infty<M$ for all $n \in \mathbb{N} \cup {0}$. – Stromael Jul 23 '13 at 22:16
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If $\Omega$ is open then there is no norm on $C^\infty(\Omega)$, it is a classical Montel space.

In addition, for compact $K\subset\Omega$ we may put $\sup_{x\in K}|f^{(k)}(x)|$ is finite, but $\sup_K\sup_{x\in K}|f^{(k)}(x)|$ may be infinite.

AD - Stop Putin -
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  • Thank you for the comment. I have edited the question, now. Does this help in construction of the norm? – Stromael Jul 22 '13 at 14:19
  • $\mathcal{D}(\Omega)$ is also a Montel space. You can read about in Hörmander "An introduction to the theory of linear partial differential operators Vol I" or Rudin "Functional Analysis". – AD - Stop Putin - Jul 22 '13 at 17:31