Test functions with "wrong" topology not locally convex?

Question

I didn't find it in any book, although it seems that this should be standard: Endow the space $C^\infty_c(\mathbb{R})$ of compactly supported functions with the inductive topology coming from the embeddings $$ \mathcal{D}_K \longrightarrow C^\infty_c(\mathbb{R}).$$ (Here $\mathcal{D}_K$ is the set of all smooth functions (on $\mathbb{R}$) with support contained in $K$, endowed with its usual Fréchet topology.)

This means that a set $U$ is open in $C^\infty_c(\mathbb{R})$ iff $U \cap \mathcal{D}_K$ is open in $\mathcal{D}_K$ for all compact subsets $K$ of $\mathbb{R}$.

Now show that this topology is not locally convex, i.e. find an open neighborhood of zero that is not the union of absolutely convex, absorbent sets.

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Edit

I believe that (contrary to the claim of Peter Michor below) the final topology w.r.t. the injections $\mathcal{D}_K \longrightarrow \mathcal{D} := C^\infty_c(\mathbb{R})$ is a vector space topology after all. (Edit2: The mistake in this argument was pointed out by Peter Michor in his edit below.)

Let $$ \alpha : \mathcal{D} \times \mathcal{D} \longrightarrow \mathcal{D}, ~~~~~~~~~ \mu: \mathbb{R} \times \mathcal{D} \longrightarrow \mathcal{D}$$ denote addition and scalar multiplication. Let $U \subseteq \mathcal{D}$ be open, i.e. $U \cap \mathcal{D}_K$ is open for all compact $K$. Then $$\alpha^{-1}(U) \cap \mathcal{D}_K \times \mathcal{D}_K = \alpha^{-1}((U \cap \mathcal{D}_K) \cup (U \setminus \mathcal{D}_K)) \cap \mathcal{D}_K \times \mathcal{D}_K = (\alpha^{-1}(U \cap \mathcal{D}_K) \cup \alpha^{-1}(U \setminus \mathcal{D}_K) ) \cap \mathcal{D}_K \times \mathcal{D}_K = (\alpha^{-1}(U \cap \mathcal{D}_K) \cap \mathcal{D}_K \times \mathcal{D}_K) \cup \underbrace{(\alpha^{-1}(U \setminus \mathcal{D}_K) \cap \mathcal{D}_K \times \mathcal{D}_K)}_{=0} = (\alpha|_{\mathcal{D}_K})^{-1}(U \cap \mathcal{D}_K) $$ which is open in $\mathcal{D}_K$ as addition is continuous on $\mathcal{D}_K$. Similarly $$ \mu^{-1}(U) \cap \mathbb{R} \times \mathcal{D}_K = \mu^{-1}((U \cap \mathcal{D}_K) \cup (U \setminus \mathcal{D}_K)) \cap \mathbb{R} \times \mathcal{D}_K = (\mu^{-1}(U \cap \mathcal{D}_K) \cup \mu^{-1}(U \setminus \mathcal{D}_K) ) \cap \mathbb{R} \times \mathcal{D}_K = (\mu^{-1}(U \cap \mathcal{D}_K) \cap \mathbb{R} \times \mathcal{D}_K) \cup \underbrace{(\mu^{-1}(U \setminus \mathcal{D}_K) \cap \mathbb{R} \times \mathcal{D}_K)}_{=0} = (\mu|_{\mathcal{D}_K})^{-1}(U \cap \mathcal{D}_K) $$ which is open in $\mathcal{D}_K$ because scalar multipliation is continuous on $\mathcal{D}_K$.

In both cases, the underbraced term is zero because $\mathcal{D}_K$ is closed under addition and scalar amultiplication, respectively.

Answer 1

The inductive topology you describe in the category of topological spaces is not locally convex -- it equals the final topology with respect to all smooth curves in $C^\infty_c(\mathbb R)$; there are also many other descriptions. See section 4 in

• Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997.

But the inductive topology in the category of locally convex vector spaces is of course locally convex; it the finest locally convex topology which is coarser than the first one.

Edit (enlarged later answering a comment of OP).

The remark after proposition 4.26 on page 46 of the source cited above says, that the the direct limit topology in the category of topological spaces, is NOT a vector space topology. This topology equals the $c^\infty$-topology, the final topology with respect to all smooth curves, because: It is the final topology with respect to the inclusion of Frechet spaces. Frechet spaces carry the $c^\infty$-topology. Each smooth curve in $\mathcal D$ locally lifts to a step in the direct limit, since this is a strict direct limit. The arguments given in that source start with lemma 4.20.

This answers your question. In fact, addition is not jointly continuous, but scalar multiplication is.

The proof amounts to the following fact: You find closed linear subspaces in $E,F$ in $\mathcal D$, one of which is Frechet (like one $\mathcal D_K$), and the other one is isomorphic to $\mathbb R^{(\mathbb N)}$ (the direct sum of countably many copies of the real line), and a bilinear bounded mapping $E\times F\to \mathbb R$ which is not jointly continuous, like the the evaluation $\mathbb R^{\mathbb N}\times \mathbb R^{(\mathbb N)}\to \mathbb R$.

Now, I hope the following clarifies your thinking: The final topology (NOT locally convex topology) with respect to all embeddings $\mathcal D_K\times \mathcal D_K$ into $\mathcal D\times \mathcal D$ is strictly finer than the product topology of the final topologies on each copies of $\mathcal D$. This follows from a careful reading of the references above. The proof in your edit seems to show that addition is continuous for the final topology of these inclusions on $\mathcal D\times \mathcal D$ which is finer that the product topology.

By the way: Terry Tao's reference exactly answered your question.

Answer 2

Here is a more direct argument than that of the other answer: Let us consider the function $$ \Phi : C_c^\infty (\mathbb{R}) \to \mathbb{R}, f \mapsto \sum_{n=0}^\infty |f^{(n)} (0)| \cdot |f(n)| . $$ It is easy to see that $\Phi$ is continuous when restricted to each of the sets $\mathcal{D}_K$ for $K \subset \mathbb{R}$ compact. Indeed, if $K \subset [-N,N]$ and $f \in \mathcal{D}_K$, then $\Phi(f) = \sum_{n=0}^N |f^{(n)}(0)| \cdot |f(n)|$, where each of the summands depends continuously on $f$ (with respect to the usual topology on $\mathcal{D}_K$). Hence, $\Phi : C_c^\infty(\mathbb{R}) \to \mathbb{R}$ is continuous with respect to the inductive limit topology.

Now, let us assume towards a contradiction that the inductive limit topology on $C_c^\infty(\mathbb{R})$ is locally convex. Since $\Phi(0) = 0$, there is then an open convex neighorhood $U \subset C_c^\infty(\mathbb{R})$ of the zero function such that $|\Phi(f)| < 1$ for all $f \in U$. Since $U \cap \mathcal{D}_{[-1,1]}$ is open, there is some $M \in \mathbb{N}$ such that $$ V := \Big\{ f \in C_c^\infty(\mathbb{R}) \quad\colon\quad \mathrm{supp}(f) \subset [-1,1] \text{ and } \max_{0 \leq \ell \leq M} \| \partial^\ell f \|_{L^\infty} \leq \frac{1}{M} \Big\} \subset U . $$ Next, since $U \cap \mathcal{D}_{[-(M+5), M+5]}$ is open, there is some $N \in \mathbb{N}$ such that $$ W := \Big\{ f \in C_c^\infty(\mathbb{R}) \quad\colon\quad \mathrm{supp}(f) \subset [-(M+5),M+5] \text{ and } \max_{0 \leq \ell \leq N} \| \partial^\ell f \|_{L^\infty} \leq \frac{1}{N} \Big\} \subset U . $$

Now, fix some $f_0 \in W \subset U$ satisfying $f_0 (M+1) \neq 0$. Note that the condition $g \in V$ only restricts the size of the first $M$ derivatives $g, g', \dots, g^{(M)}$ of $g$. Therefore, it is not hard to see for each $L \in \mathbb{N}$ that there is $g_L \in V \subset U$ satisfying $|g_L^{(M+1)}(0)| \geq L$. By convexity of $U$, we have $\frac{1}{2} (f_0 + g_L) \in U$, and hence $|\Phi(\frac{1}{2} (f_0 + g_L))| < 1$. However, for $L > 2 \, |f_0^{(M+1)}(0)|$, we see because of $g_L (M+1) = 0$ (since $\mathrm{supp}(g_L) \subset [-1,1]$) that $$ \Phi \big( \tfrac{1}{2} (f_0 + g_L) \big) \geq \frac{1}{4} \cdot \Big( |g_L^{(M+1)} (0)| - |f_0^{(M+1)}(0)| \Big) \cdot |f_0 (M+1)| \geq \frac{L}{8} \cdot |f_0(M+1)| \xrightarrow[L\to\infty]{} \infty, $$ which yields the desired contradiction.