1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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 $B_N(x,\varepsilon/2)+B_N(y,\varepsilon/2)\subseteq A$.

$\endgroup$

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.