Editing Distribution (mathematics) (section)

====Trivial extensions and independence of ''C''<sup>''k''</sup>(''K'')'s topology from ''U''====
{{anchor|Omitting the open set from notation}}

Suppose <math>U</math> is an open subset of <math>\R^n</math> and <math>K \subseteq U</math> is a compact subset. By definition, elements of <math>C^k(K)</math> are functions with domain <math>U</math> (in symbols, <math>C^k(K) \subseteq C^k(U)</math>), so the space <math>C^k(K)</math> and its topology depend on <math>U;</math> to make this dependence on the open set <math>U</math> clear, temporarily denote <math>C^k(K)</math> by <math>C^k(K;U).</math> 
Importantly, changing the set <math>U</math> to a different open subset <math>U'</math> (with <math>K \subseteq U'</math>) will change the set <math>C^k(K)</math> from <math>C^k(K;U)</math> to <math>C^k(K;U'),</math><ref group="note">Exactly as with <math>C^k(K;U),</math> the space <math>C^k(K; U')</math> is defined to be the vector subspace of <math>C^k(U')</math> consisting of maps with [[#support of a function|support]] contained in <math>K</math> endowed with the subspace topology it inherits from <math>C^k(U')</math>.</ref> so that elements of <math>C^k(K)</math> will be functions with domain <math>U'</math> instead of <math>U.</math> 
Despite <math>C^k(K)</math> depending on the open set (<math>U \text{ or } U'</math>), the standard notation for <math>C^k(K)</math> makes no mention of it. 
This is justified because, as this subsection will now explain, the space <math>C^k(K;U)</math> is canonically identified as a subspace of <math>C^k(K;U')</math> (both algebraically and topologically). 

It is enough to explain how to canonically identify <math>C^k(K; U)</math> with <math>C^k(K; U')</math> when one of <math>U</math> and <math>U'</math> is a subset of the other. The reason is that if <math>V</math> and <math>W</math> are arbitrary open subsets of <math>\R^n</math> containing <math>K</math> then the open set <math>U := V \cap W</math> also contains <math>K,</math> so that each of <math>C^k(K; V)</math> and <math>C^k(K; W)</math> is canonically identified with <math>C^k(K; V \cap W)</math> and now by transitivity, <math>C^k(K; V)</math> is thus identified with <math>C^k(K; W).</math> 
So assume <math>U \subseteq V</math> are open subsets of <math>\R^n</math> containing <math>K.</math>

Given <math>f \in C_c^k(U),</math> its {{em|'''trivial extension''' to <math>V</math>}} is the function <math>F : V \to \Complex</math> defined by:
<math display=block>F(x) = \begin{cases}
f(x) & x \in U, \\
0 & \text{otherwise}.
\end{cases}</math>
This trivial extension belongs to <math>C^k(V)</math> (because <math>f \in C_c^k(U)</math> has compact support) and it will be denoted by <math>I(f)</math> (that is, <math>I(f) := F</math>). The assignment <math>f \mapsto I(f)</math> thus induces a map <math>I : C_c^k(U) \to C^k(V)</math> that sends a function in <math>C_c^k(U)</math> to its trivial extension on <math>V.</math> This map is a linear [[Injective function|injection]] and for every compact subset <math>K \subseteq U</math> (where <math>K</math> is also a compact subset of <math>V</math> since <math>K \subseteq U \subseteq V</math>), 
<math display=block>\begin{alignat}{4}
I\left(C^k(K; U)\right) &~=~ C^k(K; V) \qquad \text{ and thus } \\
I\left(C_c^k(U)\right) &~\subseteq~ C_c^k(V).
\end{alignat}</math>
If <math>I</math> is restricted to <math>C^k(K; U)</math> then the following induced linear map is a [[homeomorphism]] (linear homeomorphisms are called {{em|[[TVS-isomorphism]]s}}):
<math display=block>\begin{alignat}{4}
 \,& C^k(K; U) && \to \,&& C^k(K;V) \\
   & f                  && \mapsto\,&& I(f) \\
\end{alignat}</math>
and thus the next map is a [[topological embedding]]:
<math display=block>\begin{alignat}{4}
 \,& C^k(K; U) && \to \,&& C^k(V) \\
   & f                  && \mapsto\,&& I(f). \\
\end{alignat}</math>
Using the injection
<math display=block>I : C_c^k(U) \to C^k(V)</math>
the vector space <math>C_c^k(U)</math> is canonically identified with its image in <math>C_c^k(V) \subseteq C^k(V).</math> Because <math>C^k(K; U) \subseteq C_c^k(U),</math> through this identification, <math>C^k(K; U)</math> can also be considered as a subset of <math>C^k(V).</math>
Thus the topology on <math>C^k(K;U)</math> is independent of the open subset <math>U</math> of <math>\R^n</math> that contains <math>K,</math>{{sfn|Rudin|1991|pp=149-181}} which justifies the practice of writing <math>C^k(K)</math> instead of <math>C^k(K; U).</math>