9-5

Question 9.5. Let G and G1 be bounded open subsets of Rn such that \bar{G_1} \subset G .Construct a function h\in\complement_0^\infty such that h =1 in G1 and h =0 outside G.
Solutions:
We can choose an open G2 such that \bar{G_1} \subset {G_2} ,and \bar{G_2} \subset {G}.
Proof. 
We use the fact that Rn is a normal space: every two disjoint closed sets of Rn have disjoint open neighborhoods. 
Note that \bar{G_1} and Gc (complement of G) are disjoint closed sets. Thus there are disjoint open sets G2, G3 such that \bar{G_1} \subset {G_2}and \bar{G^c} \subset {G_3}. Note that G2 ∩ G3 = ∅ implies G_2 \subset G_3^c and G^c \subset G_3 implies G_3^c \subset G . Further- moreG_3^c is closed. Then
\bar{G_1}\subset{G_2}\subset\bar{G_2}\subset{G_3^c}\subset{G}
Define
\epsilon_1 =inf {|x-y|:x\in G_1,y\in G_2^c}= dist(G_1,\delta G_2)
\epsilon_2 =inf {|x-y|:x\in G_2,y\in G^c}= dist(G_2,\delta G)
And let\epsilon= min\{\epsilon_1,\epsilon_2\}
By Question 4(c), we can choose K\in C_0^\infty with sup(K) = \bar{B_\epsilon(0)}. Since K is continuous with compact support, K is integrable. By multiplying with a suitable constant,we can further assume that \int\limits K=1.
Let h(x)=(\chi G_2*K)(x).Theorem 9.3 tells us that h ∈ C ∞h\in C^\infty, since \chi G_2 \in L^1 as G2 is a bounded set. 
Let x\in G_1.Then
h(x)=\int\limits_{R^n} \chi G_2(x-t)K(t)dt
=\int\limits_{B_\epsilon(0)} \chi G_2(x-t)K(t)dt
=\int\limits_{B_\epsilon(0)} K(t)dt
=1
Finally, if x\in G_c , then
h(x)=\int\limits_{R^n} \chi G_2(x-t)K(t)dt
=\int\limits_{B_\epsilon(0)} \chi G_2(x-t)K(t)dt
=0
sincex-t \not\in G_2 for t<\epsilon.(proved)