# Closeness of a family of functions

I am studying chapter 4 of Geometric Measure Theory book by H. Federer and I have some questions from the following part:

Assuming that $$X, Y$$ are Banach spaces with $$\operatorname{dim} X<\infty$$ and that $$U$$ is an open subset of $$X$$ we let $$\mathscr{E}(U, Y)$$ be the vectorspace of all functions of class $$\infty$$ mapping $$U$$ into $$Y$$. For each nonnegative integer $$i$$ and each compact subset $$K$$ of $$U$$ we define the seminorm $$v_{K}^{i}(\phi)=\sup \left\{\left\|D^{\prime} \phi(x)\right\|: 0 \leq j \leq i \text { and } x \in K\right\}$$ whenever $$\phi \in \mathscr{E}(U, Y)$$.

1. Why is $$v^i _K$$ a seminorm? For this, I have to prove that if $$v_{K}^{i}(\phi)=0$$, then the mentioned $$\sup$$ is also $$0$$, but how can I show this?

The family of all seminorms $$v_{K}^{i}$$ induces a locally convex, translation invariant Hausdorff topology on $$\mathscr{E}(U, Y) ;$$ basic neighborhood of any $$\psi \in \mathscr{E}(U, Y)$$ are the sets $$\mathscr{E}(U, Y) \cap\left\{\phi: v_{K}^{i}(\phi-\psi) corresponding to all $$i, K$$ and all $$r>0 .$$ We let $$\mathscr{E}^{\prime}(U, Y)$$ be the vectorspace of all continuous real valued linear functions on $$\mathscr{E}(U, Y),$$ and we endow $$\mathscr{E}^{\prime}(U, Y)$$ with the weak topology generated by the sets $$\mathscr{E}^{\prime}(U, Y) \cap\{T: a corresponding to all $$\phi \in \mathscr{E}(U, Y)$$ and all $$a, b \in \mathbf{R} .$$ Defining $$\operatorname{spt} \phi=U \cap \operatorname{Clos}\{x: \phi(x) \neq 0\} \text { for } \phi \in \mathscr{E}(U, Y)$$ spt $$T=U \sim \bigcup\{W: W$$ is open, $$T(\phi)=0$$ whenever $$\phi \in \mathscr{E}(U, Y) \text { and } \operatorname{spt} \phi \subset W\}$$ for $$T \in \mathscr{E}^{\prime}(U, Y),$$ we observe that spt $$T$$ is compact because $$T \leq M \cdot v_{K}^{i}$$ for some $$i, K$$ and some $$M<\infty$$ Thus we find that $$\mathscr{E}^{\prime}(U, Y)$$ is the union of its closed subsets $$\mathscr{E}_{K}^{\prime}(U, Y)=\mathscr{E}^{\prime}(U, Y) \cap\{T: \operatorname{spt} T \subset K\}$$ corresponding to all compact subsets $$K$$ of $$U$$. It may also be shown that all members of any convergent sequence in $$\mathscr{E}^{\prime}(U, Y)$$ belong to some single set $$\mathscr{E}_{K}^{\prime}(U, Y)$$ For each compact subset $$K$$ of $$U$$ we define $$\mathscr{D}_{K}(U, Y)=\mathscr{E}(U, Y) \cap\{\phi: \operatorname{spt} \phi \subset K\}$$ and observe that $$\mathscr{D}_{K}(U, Y)$$ is closed in $$\mathscr{E}(U, Y) .$$

1. How can I observe this? Actually, I don't understand the topology of these spaces. I read about the topology induced by seminorms, but still don't understand how to prove that this is close!

Then we consider the vectorspace $$\mathscr{D}(U, Y)=\bigcup\left\{\mathscr{D}_{K}(U, Y): K\right.$$ is a compact subset of $$\left.U\right\}$$ with the largest topology such that the inclusion maps from all the sets $$\mathscr{D}_{K}(U, Y)$$ are continuous; accordingly a subset of $$\mathscr{D}(U, Y)$$ is open if and only if its intersection with each $$\mathscr{D}_{K}(U, Y)$$ belongs to the relative topology of $$\mathscr{D}_{K}(U, Y)$$ in $$\mathscr{E}(U, Y)$$.

1. And finally, why is $$\mathscr{D}(U, Y)$$ a vectorspace? For this, I'm gonna take $$f \in \mathscr{D}_{K_1}(U, Y)$$ and $$g \in \mathscr{D}_{K_2}(U, Y)$$. Now, to prove that $$af+g \in \mathscr{D}(U, Y)$$, can I say that it lies in $$K=K_1 \cup K_2$$?
• 3 questions, please split these and make them more precise. I think MO is not for learning textbooks. Question 1 is completeley not understandable, at least for me. It seems to be definition. – Dieter Kadelka 2 days ago