A variation of Urysohn's lemma.

Question

Let $\Omega \subseteq \mathbb R^{N}$ be open and $K_1, K_2 \subseteq \Omega$ be compact subsets of $\Omega$ with $K_1 \cap K_2 = \varnothing.$ Then there exists a continuous function $f \in C_c (\Omega)$ with compact support contained in $\Omega$ such that $|f (x)| \leq 1$ for all $x \in \Omega$ and $f = 1$ on $K_1$ and $f = - 1$ on $K_2.$

This is almost an application of Urysohn's lemma except for the compactly supported part. Having obtained such a continuous function on $\Omega$ can we modify it further so that the resultant function is also bounded by $1$ and takes all the prescribed values on the compact sets and is compactly supported?

Answer 1

Edit: The version of Urysohn's lemma that I'm using is Theorem 2.12 in Rudin's Real and Complex Analysis. It says:

Suppose $\Omega$ is a locally compact Hausdorff space, $U$ is open in $\Omega$, $K \subset U$ and $K$ is compact. Then there exists an $f \in C_c(X)$ such that

• $0 \leq f(x) \leq 1$ for all $x \in \Omega$.
• $f(x) = 1$ for all $x \in K$.
• $\text{supp}(f)$ is a compact subset of $U$.

I would apply Urysohn's lemma twice.

$K_1$ is a compact subset of $U_1 := \Omega \setminus K_2$. By Urysohn's lemma, there exists a continuous $g_1: \Omega \to \mathbb R$ such that

• $0 \leq g_1(x) \leq 1$ for all $x \in \Omega$;
• $g_1(x) = 1$ for all $x \in K_1$;
• $\text{supp}(g_1)$ is a compact subset of $U_1$.

Similarly, $K_2$ is a compact subset of $U_2 := \Omega \setminus K_1$, so by Urysohn's lemma, there exists a continuous $g_2 : \Omega \to \mathbb R$ such that

• $0 \leq g_2(x) \leq 1$ for all $x \in \Omega$;
• $g_2(x) = 1$ for all $x \in K_2$;
• $\text{supp}(g_2)$ is a compact subset of $U_2$.

Finally, we define the function $f : \Omega \to \mathbb R$ by $$ f(x) = g_1 (x) - g_2(x).$$

Observe that

• $f$ is continuous, since $ g_1$ and $g_2$ are continuous.
• $|f(x)| \leq 1$ for all $x \in \Omega$, since $0 \leq g_1(x) \leq 1$ and $0 \leq g_2(x) \leq 1$ for all $x \in \Omega$.
• $f_1(x) = 1$ for all $x \in K_1$, since $g_1(x) = 1$ and $g_2(x) = 0$ for all $x \in K_1$.
• $f_1(x) = -1$ for all $x \in K_2$, since $g_1(x) = 0$ and $g_2(x) = 1$ for all $x \in K_2$.
• $\text{supp}(f)$ is compact, since $\text{supp}(f)$ is a closed subset of the compact set $\text{supp}(g_1) \cup \text{supp}(g_2)$.