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 $$ v_{K}^{i}(\phi)=\sup \left\{\left\|D^{j} \phi(x)\right\|: 0 \leq j \leq i \text { and } x \in K\right\} $$ whenever $\phi \in \mathscr{E}(U, Y)$.
- 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 $$ \mathscr{E}(U, Y) \cap\left\{\phi: v_{K}^{i}(\phi-\psi)<r\right\} $$ 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 $$ \mathscr{E}^{\prime}(U, Y) \cap\{T: a<T(\phi)<b\} $$ 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 $\mathscr{E}(U, Y) .$
- 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 $\mathscr{E}(U, Y)$.
- 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 $K=K_1 \cup K_2$?
To show that $v_K^i$ is a seminorm you only need to show that i) $v_K^i(\phi+\psi) \leq v_K^i(\phi)+v_K^i(\psi)$ and ii) $v_K^i(c\phi) = |c| v_K^i(\phi)$ for all $\phi,\psi \in \mathscr{E}(U, Y)$.
Show that if $\phi_k \in \mathscr{D}_K(U,Y)$ and $\phi_k \to \phi$ in the topology of $\mathscr{E}(U,Y)$ then $\phi \in \mathscr{D}_K(U,Y).$ The only thing you need to show for this is that $\operatorname{spt}\phi \subseteq K.$
Yes, you only need to show that $\operatorname{spt}(af+g)$ is compact. Everything else follows from $\mathscr{E}(U,Y)$ being a vector space.