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.**

**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.

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.