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 Kright}

$$

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) capleft{phi: v_{K}^{i}(phi-psi)<rright}

$$

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 $mathscr{D}(U, Y)=bigcupleft{mathscr{D}_{K}(U, Y): Kright.$ is a compact subset of $left.Uright}$

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 $K=K_1 cup K_2$?