8
$\begingroup$

Let $\Omega$ be a nonempty open subset of $\mathbb{R}^n$, and $\mathcal{D}(\Omega)$ the set of test functions (infinitely differentiable functions $f:\Omega \rightarrow \mathbb{C}$ with compact support contained in $\Omega$).

For every compact $K \subseteq \Omega$, let $\mathcal{D}_K$ be the locally convex topological vector space of infinitely differentiable function $f:\Omega \rightarrow \mathbb{C}$ whose support lies in $K$, with the topology $\tau_K$ induced by the system of norms ($N=0,1,2,\dots$): \begin{equation} \left| \left| f \right| \right|_{N} = \max \{ \left| D^{\alpha}f(x) \right| : x \in \Omega, | \alpha | =0,1,\dots, N \}, \end{equation} where $\alpha=(\alpha_1,\dots,\alpha_n)$ is a multi-index and $|\alpha|=\alpha_1 + \dots + \alpha_n$.

The usual topology of $\mathcal{D}(\Omega)$ is defined as the strongest topology among all those topologies on $\mathcal{D}(\Omega)$ that (i) make $\mathcal{D}(\Omega)$ a locally convex topological vector space and such that (ii) the inclusion $i_K: \mathcal{D}_K \hookrightarrow \mathcal{D}(\Omega)$ is continuous for every compact $K \subseteq \Omega$. In the language of Bourbaki, $\tau$ is called the "locally convex final topology" of the family of topologies $(\tau_K)$ of the spaces $(\mathcal{D}_K)$ with respect to family of linear maps $(i_K)$.

I have two questions.

(Q1) Can we find a set $V \subseteq \mathcal{D}(\Omega)$, such that $V \cap \mathcal{D}_K \in \tau_K$ for all compact $K \subseteq \Omega$, but $V \notin \tau$?

(Q2) Can we find $V \subseteq \mathcal{D}(\Omega)$, with $0 \in V$, such that $V \cap \mathcal{D}_K \in \tau_K$ for all compact $K \subseteq \Omega$, and there is no $W \subseteq V$, with $0 \in W \in \tau$?

Clearly a positive answer to (Q2) implies that also (Q1) has a positive answer. Note that (Q1) is equivalent to ask whether $\tau$ coincides or not with the final topology $\tau'$ on $\mathcal{D}(\Omega)$ with respect to the family of inclusions $i_K: \mathcal{D}_K \hookrightarrow \mathcal{D}(\Omega)$, where $K$ is any compact subset of $\Omega$. So we have $\tau \subseteq \tau'$ and (Q1) can maybe be given a positive indirect answer, by proving that $\tau$ and $\tau'$ do not share the same properties. To give a positive answer to (Q2) seems to be more difficult.

$\endgroup$
  • $\begingroup$ (Q1) in categorical terms: the topology of the inductive limit (more appropriate than final topology since the $(\mathcal{D}_K)_{K\ \text{compact}}$ are not disjoint of each other) in the category of topological vector spaces is finer than that in the category of locally convex topological spaces (assuming the open subsets are constructed in the same way). Answer, the inductive limit in the category of topological vector spaces does not exist (NOTE (1) of the answer). $\endgroup$ – Noix07 Mar 17 at 18:02
7
$\begingroup$

Finally, I found the answers to my two questions, and they are both positive as I conjectured.

Take a sequence of compact sets $(K_m)_{m=0}^{\infty}$ in $\Omega$, each one with nonempty interior, such that:

(i) $K_m$ is contained in the interior of $K_{m+1}$ for each $m=0,1,\dots$;

(ii) $\cup_{m=0}^{\infty} K_m = \Omega$.

Let $(x_m)_{m=0}^{\infty}$ be a sequence in $\Omega$ such that $x_m$ lies in the interior of $K_m$ and $x_m \notin K_{m-1}$ (with $K_{-1}=\emptyset$). Define the set

\begin{equation} V = \{ f \in \mathcal{D}(\Omega) : \left| f(x_{|\alpha|}) D^{\alpha} f(x_0) \right| < 1, | \alpha |=0,1,2, \dots \}. \end{equation}

Let $K \subseteq \Omega$ be a compact set. Since only finitely many of the $x_m$'s belong to $K$, it is immediate to see that $V \cap \mathcal{D}_K \in \tau_K$. Assume that $V$ contains some $\tau$-open set containing 0. Then, since $\mathcal{D}(\Omega)$ with the topology $\tau$ is a locally convex topological vector space, there would exist a convex balanced set $W \subseteq V$ such that $W \in \tau$. So $W \cap \mathcal{D}_K \in \tau_K$ for each compact $K \subseteq \Omega$.

Then for each m, there exists a positive integer $N(m)$ and $\epsilon(m) > 0$ such that the set

\begin{equation} U_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{N(m)} < \epsilon(m) \right \} \end{equation}

is contained in $W \cap \mathcal{D}_{K_m}$. Let $m=N(0)+1$. Then the interior of $K_m$ contains the point $x_{N(0)+1}$, so that there exists $g \in U_m$ such that $|g(x_{N(0)+1})| > 0$. Now note that for any $M > 0$, we can find $f \in U_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = N(0)+1$. This in turn implies that for any $c \in (0,1)$, we can find $f \in U_0$ such that $cf+(1-c)g$ does not belong to $V$. So $W$ is not convex, against the hypothesis. This shows that (Q2), and so (Q1), has a positive answer.

NOTE (1). Actually, this example also shows that $\mathcal{D}(\Omega)$ with the topology $\tau'$ is not even a topological vector space. Indeed, if it were, then we should be able to find $S \in \tau'$ such that $S + S \subseteq V$. Again, we could find then for each $m$ a positive integer $P(m)$ and $\delta(m) > 0$ such that the set \begin{equation} T_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{P(m)} < \delta(m) \right \} \end{equation} is contained in $S \cap \mathcal{D}_{K_m}$. Choose $m=P(0)+1$, so that the interior of $K_m$ contains $x_{P(0)+1}$. Then there exists $g \in T_m$ such that $|g(x_{P(0)+1})| > 0$. As before, note for any $M > 0$, we can find $f \in T_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = P(0)+1$. This in turn implies that there exists $f \in T_0$ such that $f+g \notin V$. QED

NOTE (2). We can prove in the same way as before that for every $f \in V$, there is no $U \in \tau$ such that $f \in U$ and $U \subseteq V$. Assume there exists. Then, being $\mathcal{D}(\Omega)$ with the topology $\tau$ a locally convex space, we can find a convex balanced set $W \in \tau$ such that $f + W \subseteq U$. Again, we could find then for each $m$ a positive integer $N(m)$ and $\epsilon(m) > 0$ such that the set \begin{equation} U_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{N(m)} < \epsilon(m) \right \} \end{equation} is contained in $W \cap \mathcal{D}_{K_m}$. Choose then $m=N(0)+1$, so that the interior of $K_m$ contains the point $x_{N(0)+1}$. Then there exists $g \in U_m$ such that $|g(x_{N(0)+1})| > 0$ and $|g(x_{N(0)+1})| < | \varphi(x_{N(0)+1})|$ if $| \varphi(x_{N(0)+1})| > 0$. Now note that for any $M > 0$, we can find $f \in U_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = N(0)+1$. This in turn implies that for any $c \in (0,1)$, we can find $f \in U_0$ such that $\varphi + cf+(1-c)g$ does not belong to $V$, which gives a contradiction, since $cf+(1-c)g \in W$.

$\endgroup$
  • $\begingroup$ The idea of the proof is taken from the answer of fweth in the post Topology on the space of test function. Since the discussion in that post is quite confused and there are several errors in the exposition of fweth counterexample, I have preferred to write it down carefully here, generalizing it to the case of any nonempty open $\Omega \subseteq \mathbb{R}^n$. $\endgroup$ – Maurizio Barbato Aug 12 '16 at 10:33
  • $\begingroup$ The statement that for any $M > 0$, we can find $f \in U_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $| \alpha|=N(0)+1$ seemed quite obvious to me at first glance. Anyhow, proving it is not immediate. For a proof, see the post Smooth functions. $\endgroup$ – Maurizio Barbato Sep 6 '17 at 8:18
  • $\begingroup$ An attempt to disprove the claims breaks on the following point. $\exists\, U_1 \subset V \cap \mathcal{D}_{K_1}$. Mimicking the construction of a convex neighborhood of the inductive topology, we look for at an open $U_2\subset \mathcal{D}_{K_2} $ (as in the answer, convex) such that $U_1 \subset U_2 \subset V\cap \mathcal{D}_{K_2}$. With the given norms/semi-norms it seems to be impossible. One may think of replacing it by an equivalent family or just take a $U_2 \supset U_1$, then the intersection $U_2 \cap V$ and finally find a convex subset out of it, which still contains $U_1$. $\endgroup$ – Noix07 Mar 19 at 14:54
  • $\begingroup$ (The intention is to take $\bigcup_i U_i$). At first I thought it was possible but now I represent myself $V$ as fig tree (that would be symmetric with an imaginary trunk underground) and the roots are just the transition to other trunks (in the horizontal plane). $U_1$ is for example the trunk (although as a subset of a subspace it should be flat...) and as soon as $U_2\supset U_1$ contains an element positioned on a "root", $U_2$ can no more be both convex and inside $V$. As for a proof? cf. answer above! $\endgroup$ – Noix07 Mar 19 at 15:23

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.