Let $\phi\in \mathcal{D}(\Omega):=C_c^{\infty}(\Omega)$ and $(\phi_k)_k$ a sequence of functions in $\mathcal{D}(\Omega)$

We define $\varphi_k \rightarrow \varphi$ in $\mathcal{D}(\Omega)$ as

(1) $\exists K$ compact such that $\forall k \in \mathbb{N}: \mathrm{supp}(\varphi_k), \mathrm{supp}(\varphi) \subset K$.

(2) $\forall \alpha \in \mathbb{N}^n: D^\alpha \varphi_k \rightarrow D^\alpha \varphi$ uniformly on $K$.

I was wondering what could go wrong if we get rid of point $(1)$ in the definition. Why is this part of the definition important?

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    $\begingroup$ This means there is a compact set so that the functions are all defined on the same compact set. What does (2) mean without condition (1)? $\endgroup$ – User203940 Oct 23 '20 at 14:24
  • $\begingroup$ $\sup\limits_{x\in K} \lVert D^{\alpha}\phi_k(x)-D^{\alpha}\phi(x)\rVert\to 0$ no? $\endgroup$ – roi_saumon Oct 23 '20 at 14:32
  • $\begingroup$ What is $K$ in this case? $\endgroup$ – User203940 Oct 23 '20 at 14:32
  • $\begingroup$ @User203940: not a good reason because one can simply take the sup over $\Omega$ instead of $K$. $\endgroup$ – Abdelmalek Abdesselam Oct 23 '20 at 17:56
  • $\begingroup$ Sure, I wanted them to write that though. The goal was to then explore Frechet spaces and what a norm over all of Omega would look like. $\endgroup$ – User203940 Oct 23 '20 at 17:59

If we drop the common-support-$K$ condition, and instead of requiring uniform convergence of derivatives on that common support, we simply require uniform convergence on all of $\mathbb R^n$, ... which might seem reasonable, and simpler, ... the space of test functions is no longer (sequentially) complete, which would be undesirable.

This incompleteness is similar to a simpler example, that of continuous, compactly supported functions with a single norm, the sup-norm over the whole $\mathbb R^n$. This space is not complete with respect to the corresponding metric: it is a standard exercise that the completion is the space of continuous functions going to $0$ at infinity.

Similarly, using sup-norms of all derivatives over the whole $\mathbb R^n$, the completion of test functions (with the corresponding metric attached to this countable collection of norms) can be shown to be the space of smooth functions so that they and all derivatives go to $0$ at infinity.

The "correct" topology on test functions (or even on continuous, compactly-supported functions), "correct" in the sense of being suitably complete, is more complicated than Hilbert, Banach, or Frechet, called "LF", for "(co)limit of Frechet".

  • $\begingroup$ Thank you. There is still something I don't understand. Even if $\operatorname{supp}(\phi_k),\operatorname{supp}(\phi)$ are not contained in a compact set, what would be the problem if we ask $\sup\limits_{x\in K}\lVert D^{\alpha}\phi_k(x)-D^{\alpha}\phi(x) \rVert \to 0 \forall K\text { compact}$. This would not be the same as asking $\sup\limits_{x\in \mathbb{R}^d}\lVert D^{\alpha}\phi_k(x)-D^{\alpha}\phi(x) \rVert \to 0 $ or would it? $\endgroup$ – roi_saumon Oct 24 '20 at 11:54
  • $\begingroup$ Requiring that the sups over the whole space go to $0$ is a significantly stronger condition than requiring sups over compacts go to $0$. And, without the constraint of having supports in a common compact, the space of test functions is again not complete with sups-over-whole-space: the completion is (provably) the collection of all smooth functions (with no support or growth conditions). $\endgroup$ – paul garrett Oct 24 '20 at 17:07

Paul has already given one of the main reasons for condition (1), namely completeness, but let me add a few comments collected into an answer.

One can define a locally convex topology on $\mathscr{D}(\Omega)$ using the seminorms $$ \sup_{x\in K}|\partial^{\alpha}f(x)| $$ indexed by multiindices $\alpha$ and compact subsets $K$ of $\Omega$. This topology is metrizable and one can take the completion using Cauchy sequences etc. This gives the bigger space $\mathscr{E}(\Omega)$ of all smooth functions on $\Omega$. This is a way of paraphrasing Paul's answer.

The definition of convergent sequences with (1) and (2) is the "wrong" definition. The "right" definition proceeds by first defining the topology of $\mathscr{D}(\Omega)$ and then saying that $\varphi_k\rightarrow\varphi$ iff for all open set $U$ in $\mathscr{D}(\Omega)$ which contains $\varphi$, there exists $k_0$, such that for $k\ge k_0$, $\varphi_k\in U$. The topology is a bit tricky to define, but not that difficult. It is explained here:

Doubt in understanding Space $D(\Omega)$

It turns out (as a theorem rather than a definition) that the "wrong" and "right" definitions are equivalent.

Finally, note that condition (1) is not so much tied to convergence but rather to boundedness. If a subset $A\subset \mathscr{D}(\Omega)$ is bounded, then there must exist a compact $K$ such that $\forall\varphi\in A, {\rm supp}(\varphi)\subset K$. Moreover, if a sequence converges then the set formed by the terms of the sequence and the limit is bounded and thus (1) holds.


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