# Proving that a space of compactly supported functions forms a topological vector space

Let $\mathcal{D}_K$ be the space of all on $K\subseteq \mathbb{R}$ compactly supported, infinitely differentiable functions.

I have shown for $N \in \mathbb{N}$, $\epsilon > 0$ and $U_{N,\epsilon} := \{f \in \mathcal{D}_K: \max_{0\leq i \leq N} ||f^{(i)}||_\infty < \epsilon\}$, that the sets $f + U_{N,\epsilon}$ with $f\in \mathcal{D}_K$ form a basis of a topology on our space.

I would now like to show that with this topology, our space beocmes a topological vector space. However, I am having troubles showing the continuity of the addition and multiplication.

Since I have a basis of my topology, it would seem helpful to use the fact that I only have to show that preimages of basis sets are open. I tried to use this fact by showing that $$\{(g_1,g_2) \in (\mathcal{D}_K \times \mathcal{D}_K) : g_1+g_2 \in h + U_{N,\epsilon}\} \in \mathcal{T} \times \mathcal{T}$$ but this did not really get me anywhere. Any hints would be greatly appreciated!

## 1 Answer

This has nothing to do with the particular space $\mathcal D_K$. You have a vector space $X$ and a sequence $(p_N)_{N\in\mathbb N}$ of norms with corresponding balls $B_N(x,r)=\{y\in X: p_N(x-y)<r\}$. They define a topology where $A\subseteq X$ is open if, for every $a\in A$ there are $N\in\mathbb N$ and $\varepsilon>0$ such that $B_N(a,\varepsilon)\subseteq A$. Continuity of multiplication with scalars and addition then just follows from $p_N(tx)=|t|p_N(x)$ and the triangle inequality, respectively: For the latter, consider $x,y\in X$ and an open set $A$ containing $x+y$. For some $N$ and $\varepsilon>0$ you then have $B_N(x+y,\varepsilon)\subseteq A$ and this yields $B_N(x,\varepsilon/2)+B_N(y,\varepsilon/2)\subseteq A$.