Let $U \subseteq \mathbb R^n$ be open and $f \in C_c (U).$ Let $V = U \cap B (0,1)$ be a non-empty open subset of $\mathbb R^n$ contained in $U.$ Can we conclude that $f \in C_c (V)\ $?
The support of $f$ on $V$ is given by $$\overline {\left \{x \in V\ |\ f(x) \neq 0 \right \}} = \overline {\left \{x \in U\ |\ f(x) \neq 0 \right \} \cap V} \subseteq \text {supp}\ (f) \cap \overline V$$ and, hence the support is compact since the RHS of the inclusion is compact as it's the intersection of a closed set with a compact set and hence the support of $f$ on $V,$ being a closed subset of a compact set, is indeed compact but there is no guarantee that the support is contained in $V.$ So $f$ might not be in $C_c (V),$ according to my logic. But our professor said that it's in $C_c (V).$ I am not getting his point. Could anyone kindly make it clear to me?
Thanks!
No. Consider $U = B(0,2)$, and $f:U\to \Bbb R$ continuous compactly supported and satisfying $f(x)=1$ for all $ x \in \overline B(0,3/2)$. (Such a function exists by Urysohn's lemma.) Then $f\equiv 1$ on $B(0,1)=V$, so its restriction to $V$ is not compactly supported.